# NCERT Solutions Class 12 Maths Chapter 6

## NCERT Solutions Class 12 Mathematics Chapter 6 – Application of Derivatives

Mathematics is a subject requiring a strong conceptual understanding of its applications. Hence, students are advised to study the subject from good resources to gain in-depth knowledge about  Mathematics chapters with examples, solutions, theorems as well as exercises and to score well in exams.

The Class 12 Mathematics Chapter 6 ‘Application of derivatives’ is a part of the Calculus section of Mathematics. It covers the various applications of derivatives and their role while doing calculations. The vital topics covered in Chapter 6 Class 12 Mathematics include:

• Introduction
• Rate of Change of Quantities
• Increasing and Decreasing Functions
• Tangents and Normals
• Approximations
• Maxima and Minima
• Maximum and Minimum Values of a Function in a Closed Interval

The chapter’s important formulas are listed in the NCERT Solutions Class 12 Mathematics Chapter 6. Also, one can find easy ways to carry out calculations after referring to them thoroughly and consistently.

Extramarks is a reliable and trustworthy source for all the NCERT-related study material. One can find NCERT textbooks, NCERT solutions, NCERT exemplars, NCERT revision notes, NCERT formulas, NCERT-based mock tests and the NCERT solutions Class 12 Mathematics Chapter 6 on the Extramarks official website.

### Key Topics Covered in NCERT Solutions Class 12 Mathematics Chapter 6

NCERT Solutions Class 12 Mathematics Chapter 6 is about derivatives and their applications.

This chapter is a part of Calculus and requires the study of differential calculus. One should have a good command of it to excel in this chapter.

If you have a hold on calculus, this chapter will help students to learn to carry out calculations easily. As a result, they will be able to approach problems more logically. The entire chapter is covered thoroughly in the NCERT Solutions Class 12 Mathematics Chapter 6 and is available on the Extramarks website.

After completing Chapter 6 Mathematics Class 12, students will learn to analyse the problems with a better approach and be able to solve them easily.

Introduction

A derivative is the rate of change or the amount on which a particular function changes at one given point.

In the above-given figure, the function is represented in black colour, and a tangent is represented in red colour.

Rate of Change of Quantities

The derivative ds/dt is used to show the rate of change of distance s to the time t.

Assume a particular quantity y varies with another quantity x which satisfies y = f(x), here dy/dx  or  f’(x) shows the rate of change of y w.r.t. x and [dy/dx]x = x0 or f’(x0) shows the rate of change of y w.r.t. x at x = x0.

Now, assume the two variables x and y are varying w.r.t. another variable t i.e. x = f(t) and y = g(t)

Thus, by chain rule

dy/dx = (dy/dt)/(dx/dt) here dx/dt ≠ 0

So, the rate of change of y to x could be calculated using the rate of change of y at a given time and that of x to t.

Increasing and Decreasing Functions

The function’s derivative might determine if it is increasing or decreasing at any intervals in its domain.

Assume I will be an open interval contained in the domain of the real-valued function f. Then f is said to be

(i) increasing on I when x1 < x2 in I => f(x1) ≤ f(x2) for all x1, x2 Є I.

(ii) strictly increasing on I if x1 < x2 in I => f(x1) < f(x2) for all x1, x2 Є I.

(iii) decreasing on I if x1 < x2 in I => f(x1) ≥ f(x2) for all x1, x2 Є I.

(iv) strictly decreasing on I if x1 < x2 in I => f(x1) > f(x2) for all x1, x2 Є I.

However, some functions are neither increasing nor decreasing.

The graphical representations of all these functions are given below:

function f will be said to be increasing at x0 when there exists an interval I = (x0 – h, x0 + h), h > 0 such that for  x1, x2 ∈  I,

x1 < x2 in I => f (x1) ≤ f (x2)

With the help of a theorem listed in the NCERT Solutions Class 12 Mathematics Chapter 6, you can test for increasing and decreasing functions for a given interval.

Tangents and Normals

For the given equation of a straight line that is passing through a given point (x0, y0) having a finite slope, m is represented as y – y0 = m(x – x0)

Assume the given curve y = f(x), and the tangent of the curve at that point (x0, y0) will be

[dy/dx](x0, y0) or f’(x0)

Now, the equation of the tangent of the curve y = f(x) at (x0, y0) will be

y – y0 = f’(x0)(x – x0)

The slope of the normal to curve y = f(x) at  (x0, y0) is given by -1/ f’(x0) here f’(x0) ≠ 0

For the equation of normal to the curve, y = f(x) at (x0, y0) will be

Y – y0 = (-1/f’(x0))(x – x0)

(y – y0)* f’(x0) +(x – x0) = 0

Particular cases

(i) If the slope of the tangent line is zero, then tan θ = 0, i.e. θ = 0, which means a tangent line is parallel to the x-axis. For this case, the equation of the given tangent at the point (x0, y0) will be y = y0.

(ii) If θ -> π/2 then tan θ → ∞, that means the tangent line will be perpendicular to the x-axis, that is parallel to the y-axis. For this case, the equation of the tangent at (x0, y0) will be x = x0.

Approximations

An approximation is anything which is similar but not the same as something else.

Assume f : D => R, D Ì R, will be a given function and assume y = f (x). Assume ∆x denotes a small increment in terms of x.

Now, consider the increment in y corresponding to the increment in x will be ∆y = f (x + ∆x) – f (x).

(i) The differential of x is represented as dx = ∆x.

(ii) The differential of y is represented as dy = f’(x) dx or dy = (dy/dx) * ∆x

Maxima and Minima

In the section, you will learn the different methods of calculating a function’s maximum and minimum values in a given domain. You will also learn about the absolute maximum and minimum of a function used to find the solution to many applied problems.

You will clearly understand this topic through the definitions and theorems included in the NCERT Solutions Class 12 Mathematics Chapter 6 available on the Extramarks official website.

Maximum and Minimum Values of a Function in a Closed Interval

You will learn about the two theorems to find out the absolute maximum and minimum values of a function upon a closed interval I.

Theorem: Let f be a continuous function on an interval I = [a, b]. Then f has the absolute maximum value, as well as f attains it at least once in I. Also, f has the absolute minimum value and attains it at least once in I.

Theorem: Let f be a differentiable function on a closed interval I and let c be any interior point of I. Then

(i) f′(c) = 0 when f attains its absolute maximum value at c.

(ii) f′(c) = 0 when f attains its absolute minimum value at c.

### NCERT Solutions Class 12 Mathematics Chapter 6 Exercise &  Solutions

Find all the NCERT solutions to the exercises covered in the chapter in the NCERT Solutions Class 12 Mathematics Chapter 6. It is prepared by the subject experts while adhering to the NCERT book and following the CBSE guidelines and curriculum.

All the vital topics and key concepts are covered in a point-wise manner. NCERT Solutions Class 12 Mathematics Chapter 6 is written end-to-end, highlighting everything in detail by the faculty experts. . Therefore, it is trusted by students and teachers across the private and government schools.

You can avail of the NCERT Solutions Class 12 Mathematics Chapter 6 from the Extramarks website.

Extramarks provides one-stop solutions to all your problems. To enjoy the maximum benefit of these resources, students just need to register themselves at Extramarks’ official website and stay ahead of the pack.

Click on the below links to view exercise-specific questions and solutions for NCERT Solutions Class 12 Mathematics Chapter 6:

• Class 12 Mathematics Chapter 6: Questions and Answers

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### NCERT Solutions Class 12 Mathematics Chapter 6 Exercise & Answer Solutions

NCERT Solutions Class 12 Mathematics Chapter 6 Exercise and Answer Solutions are available for students to refer for free on the Extramarks website. The material offers step-by-step solutions that help students understand how to solve problems relating to the chapter. The solution also helps students solve complex questions in a simplified manner. Students may refer to NCERT Solutions Class 12 Mathematics Chapter 6 to strengthen their basics.

Students may refer to the links below to download exercise-specific questions and solutions for NCERT Solutions Class 12 Mathematics Chapter 6 Application of Derivatives:

### NCERT Exemplar Class 12 Mathematics

NCERT Exemplar Class 12 Mathematics book is an excellent source for students preparing for JEE Mains, NEET, MHT-CET etc. It has questions according to the topics and concepts covered in the NCERT textbook. This aids students in solving all types of questions confidently.

It has different questions with varying levels of difficulty, which helps students to improve their performance in various tests and exams and definitely builds their confidence level in the process. The questions are selected from different sources for students to prepare and improve their performance. It is also a complete source of information for CBSE students preparing for their 12th standard examinations. The students think logically about a problem after referring to the NCERT Solutions Class 12 Mathematics Chapter 6 and NCERT Exemplar.

The NCERT solutions Class 12 Mathematics Chapter 6 is prepared after analysing CBSE past years’ question papers. It contains extra questions from the NCERT Class 12 Mathematics textbook. Students can refer to NCERT Exemplar for Class 12 Mathematics for more practice and face their examinations with courage. They can assure themselves that nothing remains untouched in the chapter and will score very well in all their examinations. To get good grades in exams students must refer to multiple study resources, practice a lot of questions and stick to a study schedule and follow it religiously to come out with flying colours.

### Key Features for NCERT Solutions Class 12 Mathematics Chapter 6

Regular practice is quite necessary for the students to excel. Hence, NCERT Solutions Class 12 Mathematics Chapter 6 helps students develop the habit of regular practice and to clarify their doubts then and there, take regular tests to assess their performance and get proper feedback to step up their learning. . The key features are as follows:

• You can find all the topics covered as well as in text to end text exercises from the chapter covered in the NCERT Solutions Class 12 Mathematics Chapter 6.
• It helps students analyse the easy and difficult topics in the chapter.
• After completing the NCERT Solutions Class 12 Mathematics Chapter 6, students will become experts in solving derivatives problems.

Q.1 Find the rate of change of the area of a circle with respect to its radius r when

(a) r = 3 cm (b) r = 4 cm

Ans.

$\begin{array}{l}\mathrm{The}\mathrm{area}\mathrm{of}\mathrm{a}\mathrm{circle}\left(\mathrm{A}\right)\mathrm{with}\mathrm{radius}\left(\mathrm{r}\right)\mathrm{is}\mathrm{given}\mathrm{by},\\ \mathrm{A}={\mathrm{\pi r}}^{\mathrm{2}}\\ \mathrm{Now},\mathrm{the}\mathrm{rate}\mathrm{of}\mathrm{change}\mathrm{of}\mathrm{the}\mathrm{area}\mathrm{with}\mathrm{respect}\mathrm{to}\mathrm{its}\mathrm{radius}\\ \mathrm{is}\mathrm{given}\mathrm{by},\\ \mathrm{ }\frac{\mathrm{dA}}{\mathrm{dr}}=\frac{\mathrm{d}}{\mathrm{dr}}{\mathrm{\pi r}}^{\mathrm{2}}=2\mathrm{\pi r}\\ \left(\mathrm{a}\right)\mathrm{ }\mathrm{When}\mathrm{r}=3\mathrm{cm}\\ \mathrm{ }\frac{\mathrm{dA}}{\mathrm{dr}}=2\mathrm{\pi }\left(3\right)=6\mathrm{\pi }\\ \mathrm{Hence},\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{circle}\mathrm{is}\mathrm{changing}\mathrm{at}\mathrm{the}\mathrm{rate}\mathrm{of}6\mathrm{\pi }{\mathrm{cm}}^{\mathrm{2}}/\mathrm{cm}\\ \mathrm{when}\mathrm{its}\mathrm{radius}\mathrm{is}3\mathrm{cm}\mathrm{.}\\ \left(\mathrm{b}\right)\mathrm{ }\mathrm{When}\mathrm{r}=4\mathrm{cm}\\ \mathrm{ }\frac{\mathrm{dA}}{\mathrm{dr}}=2\mathrm{\pi }\left(4\right)=8\mathrm{\pi }\\ \mathrm{Hence},\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{circle}\mathrm{is}\mathrm{changing}\mathrm{at}\mathrm{the}\mathrm{rate}\mathrm{of}\\ \mathrm{8}\mathrm{\pi }{\mathrm{cm}}^{\mathrm{2}}/\mathrm{cm}\mathrm{when}\mathrm{its}\mathrm{radius}\mathrm{is}8\mathrm{cm}\mathrm{.}\end{array}$

Q.2 The volume of a cube is increasing at the rate of 8 cm3/s. How fast is the surface area increasing when the length of an edge is 12 cm?

Ans.

$\begin{array}{l}\mathrm{Let}\mathrm{x}\mathrm{be}\mathrm{the}\mathrm{length}\mathrm{of}\mathrm{a}\mathrm{side},\mathrm{V}\mathrm{be}\mathrm{the}\mathrm{volume},\mathrm{and}\mathrm{S}\mathrm{be}\mathrm{the}\\ \mathrm{surface}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{cube}.\\ \mathrm{Then},\mathrm{V}={\mathrm{x}}^{\mathrm{3}}\mathrm{and}\mathrm{S}= 6{\mathrm{x}}^{\mathrm{2}}\mathrm{where}\mathrm{x}\mathrm{is}\mathrm{a}\mathrm{function}\mathrm{of}\mathrm{time}\mathrm{t}\mathrm{.}\\ \mathrm{It}\mathrm{is}\mathrm{given}\mathrm{that}, \mathrm{ }\frac{\mathrm{dV}}{\mathrm{dt}}=8\mathrm{ }{\mathrm{cm}}^{3}/\mathrm{s}\\ \mathrm{Then},\mathrm{by}\mathrm{using}\mathrm{the}\mathrm{chain}\mathrm{rule},\mathrm{we}\mathrm{have}:\\ \mathrm{ }8=\frac{\mathrm{dV}}{\mathrm{dt}} =\frac{\mathrm{d}}{\mathrm{dt}}\left({\mathrm{x}}^{3}\right)\\ \mathrm{ }=3{\mathrm{x}}^{2}\frac{\mathrm{dx}}{\mathrm{dt}}\\ ⇒ \frac{\mathrm{dx}}{\mathrm{dt}}=\frac{8}{3{\mathrm{x}}^{2}} ...\left(\mathrm{i}\right)\\ \mathrm{Now}, \frac{\mathrm{dS}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\left(6{\mathrm{x}}^{2}\right)\\ \mathrm{ }=12\mathrm{x}\frac{\mathrm{dx}}{\mathrm{dt}}\\ \mathrm{ }=12\mathrm{x}×\frac{8}{3{\mathrm{x}}^{2}}\left[\mathrm{From}\mathrm{equation}\left(\mathrm{i}\right)\right]\\ \mathrm{ }=\frac{32}{\mathrm{x}}\\ ⇒\frac{\mathrm{dS}}{\mathrm{dx}}=\frac{32}{12}\left[\mathrm{Putting}\mathrm{x}=\mathrm{12}\right]\end{array}$ $\begin{array}{l}⇒\text{}\frac{dS}{dx}=\frac{8}{3}c{m}^{2}/s.\\ \text{Hence, if the length of the edge of the cube is 12 cm, then the}\\ \text{surface area is increasing}\text{\hspace{0.17em}}\text{at the rate of}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{8}{3}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\text{cm}}^{\text{2}}\text{/s}.\end{array}$

Q.3 The radius of a circle is increasing uniformly at the rate of 3 cm/s. Find the rate at which the area of the circle is increasing when the radius is 10 cm.

Ans.

$\begin{array}{l}\mathrm{The}\mathrm{area}\mathrm{of}\mathrm{a}\mathrm{circle}\left(\mathrm{A}\right)\mathrm{with}\mathrm{radius}\left(\mathrm{r}\right)\mathrm{is}\mathrm{given}\mathrm{by},\\ \mathrm{ }\mathrm{A}={\mathrm{\pi r}}^{2}\\ \mathrm{Now},\mathrm{the}\mathrm{rate}\mathrm{of}\mathrm{change}\mathrm{of}\mathrm{area}\left(\mathrm{A}\right)\mathrm{with}\mathrm{respect}\mathrm{to}\mathrm{time}\left(\mathrm{t}\right)\mathrm{is}\\ \mathrm{given}\mathrm{by},\frac{\mathrm{dA}}{\mathrm{dt}}=\frac{\mathrm{d}}{\mathrm{dt}}{\mathrm{\pi r}}^{2}\\ =2\mathrm{\pi r}\frac{\mathrm{dr}}{\mathrm{dt}}\left[\mathrm{By}\mathrm{chain}\mathrm{rule}\right]\\ =2\mathrm{\pi r}×3\left[\because \frac{\mathrm{dr}}{\mathrm{dt}}=3\mathrm{ }\mathrm{cm}/\mathrm{s}\right]\\ \therefore \frac{\mathrm{dA}}{\mathrm{dt}}=2\mathrm{\pi }\left(10\right)×3\left[\mathrm{Putting}\mathrm{r}=10 \mathrm{cm}\right]\\ =60\mathrm{\pi } {\mathrm{cm}}^{\mathrm{2}}/\mathrm{s}\\ \mathrm{Hence},\mathrm{the}\mathrm{rate}\mathrm{at}\mathrm{which}\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{circle}\mathrm{is}\mathrm{increasing}\mathrm{when}\\ \mathrm{the}\mathrm{radius}\mathrm{is}10\mathrm{cm}\mathrm{is}60\mathrm{\pi } {\mathrm{cm}}^{\mathrm{2}}/\mathrm{s}\mathrm{.}\end{array}$

Q.4 An edge of a variable cube is increasing at the rate of 3 cm/s. How fast is the volume of the cube increasing when the edge is 10 cm long?

Ans.

$\begin{array}{l}\mathrm{Let}\mathrm{the}\mathrm{length}\mathrm{of}\mathrm{a}\mathrm{side}\mathrm{be}\mathrm{x}\mathrm{and}\mathrm{V}\mathrm{be}\mathrm{the}\mathrm{volume}\mathrm{of}\mathrm{the}\mathrm{cube}.\\ \mathrm{Then}, \mathrm{ }\mathrm{V}={\mathrm{x}}^{\mathrm{3}}\\ \mathrm{Differentiating}\mathrm{both}\mathrm{the}\mathrm{sides}\mathrm{with}\mathrm{respect}\mathrm{to}\mathrm{t},\mathrm{we}\mathrm{get}\\ \frac{\mathrm{dV}}{\mathrm{dt}}=3{\mathrm{x}}^{2}\frac{\mathrm{dx}}{\mathrm{dt}}\left[\mathrm{By}\mathrm{chain}\mathrm{rule}\right]\\ =3{\mathrm{x}}^{2}\left(3\right)\left[\because \frac{\mathrm{dx}}{\mathrm{dt}}=3\mathrm{ }\mathrm{cm}/\mathrm{s},\mathrm{ }\mathrm{Given}\right]\\ =9{\mathrm{x}}^{2}\\ \mathrm{Since}, \mathrm{edge}\mathrm{of}\mathrm{cube}\left(\mathrm{x}\right)=10 \mathrm{cm}, \mathrm{so}\\ \frac{\mathrm{dV}}{\mathrm{dt}}=9{\left(10\right)}^{2}\\ =900\mathrm{ }{\mathrm{cm}}^{3}/\mathrm{s}\\ \mathrm{Hence},\mathrm{the}\mathrm{volume}\mathrm{of}\mathrm{the}\mathrm{cube}\mathrm{is}\mathrm{increasing}\mathrm{at}\mathrm{the}\mathrm{rate}\mathrm{of}\\ 900{\mathrm{cm}}^{\mathrm{3}}/\mathrm{s}\mathrm{when}\mathrm{the}\mathrm{edge}\mathrm{is}10\mathrm{cm}\mathrm{long}\mathrm{.}\end{array}$

Q.5 A stone is dropped into a quiet lake and waves move in circles at the speed of 5 cm/s. At the instant when the radius of the circular wave is 8 cm, how fast is the enclosed area increasing?

Ans.

$\begin{array}{l}\mathrm{The}\mathrm{area}\mathrm{of}\mathrm{a}\mathrm{circle}\left(\mathrm{A}\right)\mathrm{with}\mathrm{radius}\left(\mathrm{r}\right)\mathrm{is}\mathrm{given}\mathrm{by}\\ \mathrm{A}={\mathrm{\pi r}}^{2}\\ \mathrm{Therefore},\mathrm{the}\mathrm{rate}\mathrm{of}\mathrm{change}\mathrm{of}\mathrm{area}\left(\mathrm{A}\right)\mathrm{with}\mathrm{respect}\mathrm{to}\mathrm{time}\left(\mathrm{t}\right)\\ \mathrm{is}\mathrm{given}\mathrm{by},\frac{\mathrm{dA}}{\mathrm{dt}}=\frac{\mathrm{d}}{\mathrm{dt}}{\mathrm{\pi r}}^{2}\\ =2\mathrm{\pi r}\frac{\mathrm{dr}}{\mathrm{dt}}\left[\mathrm{By}\mathrm{chain}\mathrm{rule}\right]\\ =2\mathrm{\pi }\left(8\right)\left(5\right)\left[\begin{array}{l}\because \frac{\mathrm{dr}}{\mathrm{dt}}=5\mathrm{cm}/\mathrm{s},\\ \mathrm{r}=8\mathrm{ }\mathrm{cm},\mathrm{Given}\end{array}\right]\\ =80\mathrm{\pi }\\ \mathrm{Hence},\mathrm{when}\mathrm{the}\mathrm{radius}\mathrm{of}\mathrm{the}\mathrm{circular}\mathrm{wave}\mathrm{is}8\mathrm{cm},\mathrm{the}\\ \mathrm{enclosed}\mathrm{area}\mathrm{is}\mathrm{increasing}\mathrm{at}\mathrm{the}\mathrm{rate}\mathrm{of}80\mathrm{\pi }{\mathrm{cm}}^{\mathrm{2}}/\mathrm{s}\mathrm{.}\end{array}$

Q.6 The radius of a circle is increasing at the rate of 0.7 cm/s. What is the rate of increase of its circumference?

Ans.

$\begin{array}{l}\mathrm{Let}\mathrm{the}\mathrm{circumference}\mathrm{of}\mathrm{a}\mathrm{circle}\mathrm{be}\mathrm{C}\mathrm{with}\mathrm{radius}\left(\mathrm{r}\right)\mathrm{is}\\ \mathrm{given}\mathrm{byC}=\mathrm{2}\mathrm{\pi r}\\ \mathrm{Then},\mathrm{the}\mathrm{rate}\mathrm{of}\mathrm{change}\mathrm{of}\mathrm{circumference}\left(\mathrm{C}\right)\mathrm{with}\mathrm{respect}\\ \mathrm{to}\mathrm{time}\left(\mathrm{t}\right)\mathrm{is}\mathrm{given}\mathrm{by},\\ \frac{\mathrm{dC}}{\mathrm{dt}}=\frac{\mathrm{d}}{\mathrm{dt}}\mathrm{2}\mathrm{\pi r}\\ =\mathrm{2}\mathrm{\pi }\frac{\mathrm{dr}}{\mathrm{dt}}\left[\mathrm{By}\mathrm{chain}\mathrm{rule}\right]\\ ⇒ \frac{\mathrm{dC}}{\mathrm{dt}}=\mathrm{2}\mathrm{\pi }\left(0.7\right)\left[\because \frac{\mathrm{dr}}{\mathrm{dt}}=0.7\mathrm{ }\mathrm{cm}/\mathrm{s}\right]\\ ⇒ \frac{\mathrm{dC}}{\mathrm{dt}}=1.4\mathrm{\pi }\\ \mathrm{Hence},\mathrm{the}\mathrm{rate}\mathrm{of}\mathrm{increase}\mathrm{of}\mathrm{the}\mathrm{circumference} \mathrm{is}\mathrm{ }1.4\mathrm{\pi }\mathrm{ }\mathrm{cm}/\mathrm{s}\mathrm{.}\end{array}$

Q.7 The length x of a rectangle is decreasing at the rate of 5 cm/minute and the width y is increasing at the rate of 4 cm/minute. When x = 8 cm and y = 6 cm, find the rates of change of (a) the perimeter, and (b) the area of the rectangle.

Ans.

$\begin{array}{l}\mathrm{Since}\mathrm{the}\mathrm{length}\left(\mathrm{x}\right)\mathrm{is}\mathrm{decreasing}\mathrm{at}\mathrm{the}\mathrm{rate}\mathrm{of}5\mathrm{cm}/\mathrm{minute}\\ \mathrm{and}\mathrm{the}\mathrm{width}\left(\mathrm{y}\right)\mathrm{is}\mathrm{increasing}\mathrm{at}\mathrm{the}\mathrm{rate}\mathrm{of}4\mathrm{cm}/\mathrm{minute},\\ \mathrm{we}\mathrm{have}:\frac{\mathrm{dx}}{\mathrm{dt}}=-5\mathrm{cm}/\mathrm{minute} \mathrm{and} \frac{\mathrm{dy}}{\mathrm{dt}}=4\mathrm{cm}/\mathrm{minute}\\ \left(\mathrm{a}\right)\mathrm{Perimeter}\mathrm{of}\mathrm{rectangle}\left(\mathrm{P}\right)=2\left(\mathrm{x}+\mathrm{y}\right)\\ \mathrm{Differentiating}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathrm{t},\mathrm{we}\mathrm{get}\\ \frac{\mathrm{dP}}{\mathrm{dt}}=2\frac{\mathrm{d}}{\mathrm{dt}}\left(\mathrm{x}+\mathrm{y}\right)\\ =2\left(\frac{\mathrm{dx}}{\mathrm{dt}}+\frac{\mathrm{dy}}{\mathrm{dt}}\right)\\ =2\left(-5+4\right)\\ =-2\mathrm{ }\mathrm{cm}/\mathrm{minute}\\ \mathrm{Thus},\mathrm{the}\mathrm{perimeter}\mathrm{is}\mathrm{decreasing}\mathrm{at}\mathrm{the}\mathrm{rate}\mathrm{of}2\mathrm{cm}/\mathrm{min}\mathrm{.}\\ \left(\mathrm{b}\right)\mathrm{Area}\mathrm{of}\mathrm{rectangle}\left(\mathrm{A}\right)=\mathrm{xy}\\ \mathrm{Differentiating}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathrm{t},\mathrm{we}\mathrm{get}\\ \mathrm{ }\frac{\mathrm{dA}}{\mathrm{dt}}=\frac{\mathrm{d}}{\mathrm{dt}}\mathrm{xy}\\ =\mathrm{x}\frac{\mathrm{dy}}{\mathrm{dt}}+\mathrm{y}\frac{\mathrm{dx}}{\mathrm{dt}} \left[\mathrm{By}\mathrm{product}\mathrm{rule}\right]\\ =4\mathrm{x}-5\mathrm{y}\left[\begin{array}{l}\because \frac{\mathrm{dx}}{\mathrm{dt}}=-5\mathrm{cm}/\mathrm{min} \mathrm{and} \\ \frac{\mathrm{dy}}{\mathrm{dt}}=4\mathrm{cm}/\mathrm{min}\end{array}\right]\\ =4\left(8\right)-5\left(6\right)\\ =32-30\\ =2{\mathrm{cm}}^{2}/\mathrm{min}\\ \mathrm{Hence},\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{rectangle}\mathrm{is}\mathrm{increasing}\mathrm{at}\mathrm{the}\mathrm{rate}\\ \mathrm{of}2{\mathrm{cm}}^{\mathrm{2}}/\mathrm{min}\mathrm{.}\end{array}$

Q.8 A balloon, which always remains spherical on inflation, is being inflated by pumping in 900 cubic centimetres of gas per second. Find the rate at which the radius of the balloon increases when the radius is 15 cm.

Ans.

$\begin{array}{l}\mathrm{The}\mathrm{volume}\mathrm{of}\mathrm{a}\mathrm{sphere}\left(\mathrm{V}\right)\mathrm{with}\mathrm{radius}\left(\mathrm{r}\right)\mathrm{is}\mathrm{given}\mathrm{by},\\ \mathrm{V}=\frac{4}{3}{\mathrm{\pi r}}^{3}\\ \mathrm{Differentiating}\mathrm{both}\mathrm{sides},\mathrm{with}\mathrm{respect}\mathrm{to}\mathrm{t},\mathrm{we}\mathrm{get}\\ \frac{\mathrm{dv}}{\mathrm{dt}}=\frac{\mathrm{d}}{\mathrm{dt}}\frac{4}{3}{\mathrm{\pi r}}^{3}\\ =\frac{4}{3}\mathrm{\pi }×3{\mathrm{r}}^{2}\frac{\mathrm{dr}}{\mathrm{dt}}\\ 900=4{\mathrm{\pi r}}^{2}\frac{\mathrm{dr}}{\mathrm{dt}}\left[\because \frac{\mathrm{dV}}{\mathrm{dt}}=900{\mathrm{cm}}^{\mathrm{3}}/\mathrm{sec}\right]\\ ⇒ \mathrm{ }\frac{\mathrm{dr}}{\mathrm{dt}}=\frac{900}{4{\mathrm{\pi r}}^{2}}\\ =\frac{225}{\mathrm{\pi }{\left(15\right)}^{2}}\left[\because \mathrm{r}=15 \mathrm{cm}\right]\\ =\frac{225}{\mathrm{\pi }\left(225\right)}=\frac{1}{\mathrm{\pi }}\\ \mathrm{Hence},\mathrm{the}\mathrm{rate}\mathrm{at}\mathrm{which}\mathrm{the}\mathrm{radius}\mathrm{of}\mathrm{the}\mathrm{balloon}\mathrm{increases}\\ \mathrm{when}\mathrm{the}\mathrm{radius}\mathrm{is}15\mathrm{cm}\mathrm{ }\mathrm{is} \frac{1}{\mathrm{\pi }}\mathrm{ }\mathrm{cm}/\mathrm{sec}.\end{array}$

Q.9 A balloon, which always remains spherical has a variable radius. Find the rate at which its volume is increasing with the radius when the later is 10 cm.

Ans.

$\begin{array}{l}\mathrm{Volume}\mathrm{of}\mathrm{sphere}\mathrm{having}\mathrm{radius}\mathrm{r}=\frac{4}{3}{\mathrm{\pi r}}^{3}\\ ⇒\mathrm{V}=\frac{4}{3}{\mathrm{\pi r}}^{3}\\ \mathrm{Differentiating}\mathrm{with}\mathrm{respect}\mathrm{to}\mathrm{r},\mathrm{we}\mathrm{get}\\ ⇒\frac{\mathrm{dV}}{\mathrm{dr}}=\frac{\mathrm{d}}{\mathrm{dr}}\left(\frac{4}{3}{\mathrm{\pi r}}^{3}\right)\\ =\frac{4}{3}\mathrm{\pi }\frac{\mathrm{d}}{\mathrm{dr}}{\mathrm{r}}^{3}\\ =\frac{4}{3}\mathrm{\pi }×3{\mathrm{r}}^{2}\\ =4{\mathrm{\pi r}}^{2}\\ =4\mathrm{\pi }{\left(10\right)}^{2}\\ =400\mathrm{\pi }\\ \mathrm{Hence},\mathrm{the}\mathrm{volu e}\mathrm{of}\mathrm{the}\mathrm{balloon}\mathrm{is}\mathrm{increasing}\mathrm{at}\mathrm{the}\mathrm{rate}\\ \mathrm{of}400{\mathrm{\pi }}^{}\mathrm{.}\end{array}$

Q.10 A ladder 5 m long is leaning against a wall. The bottom of the ladder is pulled along the ground, away from the wall, at the rate of 2 cm/s. How fast is its height on the wall decreasing when the foot of the ladder is 4 m away from the wall?

Ans.

$\begin{array}{l}\text{Let y m be the height of the wall at which the ladder touches}\text{.}\\ \text{Also, let the foot of the ladder be x m away from the wall}\text{.}\\ \text{Then in ΔABC, by Pythagoras theorem, we have:}\\ \text{}{\text{x}}^{\text{2}}{\text{+ y}}^{\text{2}}\text{=25}\text{}\text{}\text{}\text{[Length of the ladder = 5 m]}\\ ⇒\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{y=}\sqrt{\text{25}-{\text{x}}^{\text{2}}}\\ \text{Differentiating both sides with respect to t, we get}\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{\text{dy}}{\text{dt}}\text{=}\frac{\text{d}}{\text{dt}}\sqrt{\text{25}-{\text{x}}^{\text{2}}}\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{=}\frac{-\text{2x}}{\text{2}\sqrt{\text{25}-{\text{x}}^{\text{2}}}}\frac{\text{dx}}{\text{dt}}\text{}\text{}\text{}\left[\text{By}\text{\hspace{0.17em}}\text{chain rule}\right]\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{=}-\frac{\text{x}}{\sqrt{{\text{25-x}}^{\text{2}}}}\left(\text{2}\right)\text{}\text{}\text{}\left[\frac{\text{dx}}{\text{dt}}\text{=2 cm/s}\right]\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{=}-\frac{\text{2}\left(\text{4}\right)}{\sqrt{\text{25}-{\left(\text{4}\right)}^{\text{2}}}}\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{=}-\frac{\text{8}}{\text{3}}\\ \text{Hence, the height of the ladder on the wall is decreasing at the}\\ \text{rate of}-\frac{\text{8}}{\text{3}}\text{\hspace{0.17em}}\text{cm/s}\text{.}\end{array}$

Q.11 A particle moves along the curve 6y = x3 + 2. Find the points on the curve at which the y coordinate is changing 8 times as fast as the x-coordinate.

Ans.

$\begin{array}{l}\mathrm{The}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{curve}\mathrm{is}\mathrm{given}\mathrm{as}:\\ 6\mathrm{y}={\mathrm{x}}^{3}+2\\ \mathrm{The}\mathrm{rate}\mathrm{of}\mathrm{change}\mathrm{of}\mathrm{the}\mathrm{position}\mathrm{of}\mathrm{the}\mathrm{particle}\mathrm{with}\\ \mathrm{respect}\mathrm{to}\mathrm{time}\left(\mathrm{t}\right)\mathrm{is}\mathrm{given}\mathrm{by},\\ \mathrm{6}\frac{\mathrm{dy}}{\mathrm{dt}}=\frac{\mathrm{d}}{\mathrm{dt}}\left({\mathrm{x}}^{3}+2\right)\\ \mathrm{6}\frac{\mathrm{dy}}{\mathrm{dt}}=3{\mathrm{x}}^{2}\frac{\mathrm{dx}}{\mathrm{dt}}\\ \mathrm{When}\mathrm{the}\mathrm{y}–\mathrm{coordinate}\mathrm{of}\mathrm{the}\mathrm{particle}\mathrm{changes}8\mathrm{times}\mathrm{as}\mathrm{fast}\\ \mathrm{as}\mathrm{the}\mathrm{x}–\mathrm{coordinate}\mathrm{i}.\mathrm{e}.,\mathrm{ }\frac{\mathrm{dy}}{\mathrm{dt}}=8\frac{\mathrm{dx}}{\mathrm{dt}},\mathrm{we}\mathrm{have}\\ \mathrm{6}\left(8\frac{\mathrm{dx}}{\mathrm{dt}}\right)=3{\mathrm{x}}^{2}\frac{\mathrm{dx}}{\mathrm{dt}}\\ ⇒ \mathrm{ }\left(16-{\mathrm{x}}^{2}\right)\frac{\mathrm{dx}}{\mathrm{dt}}=0\\ ⇒ 16-{\mathrm{x}}^{2}=0\\ ⇒ \mathrm{x}=±\mathrm{ }4\\ \mathrm{When} \mathrm{ }\mathrm{x}=4,\\ \mathrm{y}=\frac{{\left(4\right)}^{3}+2}{6}=11\\ \mathrm{When} \mathrm{ }\mathrm{x}=-4,\\ \mathrm{y}=\frac{{\left(-4\right)}^{3}+2}{6}\\ =-\frac{62}{6}=-\frac{31}{3}\\ \mathrm{Hence},\mathrm{the}\mathrm{points}\mathrm{required}\mathrm{on}\mathrm{the}\mathrm{curve}\mathrm{are}\mathrm{ }\left(4,11\right)\mathrm{ }\mathrm{and}\left(-4,-\frac{31}{3}\right).\end{array}$

Q.12 The radius of an air bubble is increasing at the rate of ½ cm/s. At what rate is the volume of the bubble increasing when the radius is 1 cm?

Ans.

$\begin{array}{l}\mathrm{The}\mathrm{air}\mathrm{bubble}\mathrm{is}\mathrm{in}\mathrm{the}\mathrm{shape}\mathrm{of}\mathrm{a}\mathrm{sphere}\mathrm{.}\\ \mathrm{So},\mathrm{the}\mathrm{volume}\mathrm{of}\mathrm{an}\mathrm{air}\mathrm{bubble}\left(\mathrm{V}\right)\mathrm{with}\mathrm{radius}\left(\mathrm{r}\right)\mathrm{is}\mathrm{given}\mathrm{by},\\ \mathrm{V}=\frac{4}{3}{\mathrm{\pi r}}^{3}\\ \mathrm{The}\mathrm{rate}\mathrm{of}\mathrm{change}\mathrm{of}\mathrm{volume}\left(\mathrm{V}\right)\mathrm{with}\mathrm{respect}\mathrm{to}\mathrm{time}\left(\mathrm{t}\right)\mathrm{is}\\ \mathrm{obtained}\mathrm{by}\mathrm{differentiating}\mathrm{both}\mathrm{sides}\mathrm{with}\mathrm{respect}\mathrm{to}\mathrm{t},\mathrm{so}\\ \mathrm{ }\frac{\mathrm{dV}}{\mathrm{dt}}=\frac{\mathrm{d}}{\mathrm{dt}}\left(\frac{4}{3}{\mathrm{\pi r}}^{3}\right)\\ \mathrm{ }\frac{\mathrm{dV}}{\mathrm{dt}}=\frac{4}{3}×3{\mathrm{\pi r}}^{2}\frac{\mathrm{dr}}{\mathrm{dt}}\\ \\ \mathrm{ }=4\mathrm{\pi }{\left(1\right)}^{2}\left(\frac{1}{2}\right)\left[\begin{array}{l}\because \frac{\mathrm{dr}}{\mathrm{dt}}=\frac{1}{2}\mathrm{ }\mathrm{cm}/\mathrm{s}\\ \mathrm{and}\mathrm{r}=1\mathrm{ }\mathrm{cm}\end{array}\right]\\ \mathrm{ }=2\mathrm{\pi } {\mathrm{cm}}^{\mathrm{3}}/\mathrm{s}\\ \mathrm{Hence},\mathrm{the}\mathrm{rate}\mathrm{at}\mathrm{which}\mathrm{the}\mathrm{volume}\mathrm{of}\mathrm{the}\mathrm{bubble}\mathrm{increases}\\ \mathrm{is}2\mathrm{\pi }{\mathrm{cm}}^{\mathrm{3}}/\mathrm{s}. \end{array}$

Q.13 A balloon, which always remains spherical, has a variable diameter 32(2x+1). Find the rate of change of its volume with respect to x.

Ans.

$\begin{array}{l}\mathrm{The}\mathrm{volume}\left(\mathrm{V}\right)\mathrm{of}\mathrm{a}\mathrm{sphere}\mathrm{with}\mathrm{radius}\left(\mathrm{r}\right)\mathrm{is}\mathrm{given}\mathrm{by},\\ \mathrm{V}=\frac{\mathrm{4}}{\mathrm{3}}{\mathrm{\pi r}}^{\mathrm{3}}\\ \mathrm{Diameter}\mathrm{of}\mathrm{balloon}=\frac{\mathrm{3}}{\mathrm{2}}\mathrm{ }\left(2\mathrm{x}+1\right)\\ \mathrm{So},\mathrm{the}\mathrm{radiusof}\mathrm{balloon}=\frac{\mathrm{3}}{\mathrm{4}}\mathrm{ }\left(2\mathrm{x}+1\right)\\ \therefore \mathrm{V}=\frac{\mathrm{4}}{\mathrm{3}}\mathrm{\pi }{\left\{\frac{\mathrm{3}}{\mathrm{4}}\mathrm{ }\left(2\mathrm{x}+1\right)\right\}}^{\mathrm{3}}\\ \mathrm{=}\frac{\mathrm{4}}{\mathrm{3}}\mathrm{\pi }×\frac{\mathrm{27}}{\mathrm{64}}\mathrm{ }\left(2\mathrm{x}{+1\right)}^{\mathrm{3}}\\ \mathrm{ }\mathrm{V}=\frac{\mathrm{9}}{\mathrm{16}}\mathrm{\pi }\mathrm{ }\left(2\mathrm{x}{+1\right)}^{\mathrm{3}}\\ \mathrm{Hence},\mathrm{the}\mathrm{rate}\mathrm{of}\mathrm{change}\mathrm{of}\mathrm{volume}\mathrm{with}\mathrm{respect}\mathrm{to}\mathrm{x}\mathrm{is}\\ \mathrm{given}\mathrm{by}\\ \frac{\mathrm{dV}}{\mathrm{dx}}\mathrm{=}\frac{\mathrm{9}}{\mathrm{16}}\mathrm{\pi }\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{ }\left(2\mathrm{x}{+1\right)}^{\mathrm{3}}\\ \mathrm{=}\frac{\mathrm{9}}{\mathrm{16}}\mathrm{\pi }×3\mathrm{ }\left(2\mathrm{x}{+1\right)}^{\mathrm{2}}\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{ }\left(2\mathrm{x}+1\right)\\ \left[\mathrm{By}\mathrm{chain}\mathrm{rule}\right]\\ \frac{\mathrm{dV}}{\mathrm{dx}}\mathrm{=}\frac{\mathrm{27}}{\mathrm{16}}\mathrm{\pi }\mathrm{ }\left(2\mathrm{x}{+1\right)}^{\mathrm{2}}×2\\ ⇒\frac{\mathrm{dV}}{\mathrm{dx}}\mathrm{=}\frac{\mathrm{27}}{\mathrm{8}}\mathrm{\pi }\mathrm{ }\left(2\mathrm{x}{+1\right)}^{\mathrm{2}}\\ \mathrm{Thus},\mathrm{the}\mathrm{rate}\mathrm{of}\mathrm{change}\mathrm{of}\mathrm{volume}\mathrm{with}\mathrm{respect}\mathrm{to}\mathrm{x}\mathrm{is}\\ \frac{\mathrm{27}}{\mathrm{8}}\mathrm{\pi }\mathrm{ }\left(2\mathrm{x}{+1\right)}^{\mathrm{2}}\mathrm{.}\end{array}$

Q.14 Sand is pouring from a pipe at the rate of 12 cm3/s. The falling sand forms a cone on the ground in such a way that the height of the cone is always onesixth of the radius of the base. How fast is the height of the sand cone increasing when the height is 4 cm?

Ans.

$\begin{array}{l}\mathrm{The}\mathrm{volume}\mathrm{of}\mathrm{a}\mathrm{cone}\mathrm{with}\mathrm{radius}\left(\mathrm{r}\right)\mathrm{and}\mathrm{height}\left(\mathrm{h}\right)\mathrm{is}\mathrm{given}\mathrm{by},\\ \mathrm{V}=\frac{1}{3}{\mathrm{\pi r}}^{2}\mathrm{h}\\ \mathrm{Given},\mathrm{height}\mathrm{of}\mathrm{cone}=\frac{1}{6}\mathrm{r}⇒\mathrm{h}=\frac{1}{6}\mathrm{r}⇒\mathrm{r}=6\mathrm{h}\\ \therefore \mathrm{ }\mathrm{V}=\frac{1}{3}\mathrm{\pi }{\left(6\mathrm{h}\right)}^{2}\mathrm{h}\\ =12{\mathrm{\pi h}}^{3}\\ \mathrm{Differentiating}\mathrm{both}\mathrm{sides}\mathrm{with}\mathrm{respect}\mathrm{to}\mathrm{time}\left(\mathrm{t}\right),\mathrm{we}\mathrm{get}\\ \frac{\mathrm{dV}}{\mathrm{dt}}=\frac{\mathrm{d}}{\mathrm{dt}}\left(12{\mathrm{\pi h}}^{3}\right)\\ =12\mathrm{\pi }\frac{\mathrm{d}}{\mathrm{dt}}{\mathrm{h}}^{3}\\ =36{\mathrm{\pi h}}^{2}\frac{\mathrm{dh}}{\mathrm{dt}}\left[\mathrm{By}\mathrm{chain}\mathrm{rule}\right]\\ \mathrm{ }12=36\mathrm{\pi }{\left(4\right)}^{2}\frac{\mathrm{dh}}{\mathrm{dt}}\left[\because \frac{\mathrm{dV}}{\mathrm{dt}}=12\mathrm{ }\mathrm{and}\mathrm{ }\mathrm{h}=4\mathrm{cm}\right]\\ \frac{\mathrm{dh}}{\mathrm{dt}}=\frac{12}{36\mathrm{\pi }{\left(4\right)}^{2}}=\frac{1}{48\mathrm{\pi }}\\ \mathrm{Hence},\mathrm{when}\mathrm{the}\mathrm{height}\mathrm{of}\mathrm{the}\mathrm{sand}\mathrm{cone}\mathrm{is}4\mathrm{cm},\mathrm{its}\mathrm{height}\mathrm{is}\\ \mathrm{increasing}\mathrm{at}\mathrm{the}\mathrm{rate}\mathrm{of}\frac{1}{48\mathrm{\pi }}\mathrm{cm}/\mathrm{s}\mathrm{.}\end{array}$

Q.15 The total cost C (x) in Rupees associated with the production of x units of an item is given
by C(x) = 0.007x3 – 0.003 x2 + 15x + 4000 Find the marginal cost when 17 units are produced

Ans.

$\begin{array}{l}\mathrm{The}\mathrm{total}\mathrm{cost}\mathrm{C}\left(\mathrm{x}\right)=\mathrm{0}.007{\mathrm{x}}^{\mathrm{3}}-0.003{\mathrm{x}}^{\mathrm{2}}+15\mathrm{x}+ 4000\\ \mathrm{Marginal}\mathrm{cost}\mathrm{is}\mathrm{the}\mathrm{rate}\mathrm{of}\mathrm{change}\mathrm{of}\mathrm{total}\mathrm{cost}\mathrm{with}\mathrm{respect}\\ \mathrm{to}\mathrm{output}\mathrm{.}\\ \therefore \mathrm{Marginal}\mathrm{cost}\left(\mathrm{MC}\right)=\frac{\mathrm{dC}}{\mathrm{dx}}\\ \mathrm{ }=\mathrm{0}.007\left(3{\mathrm{x}}^{\mathrm{2}}\right)-\mathrm{0}.003\left(2\mathrm{x}\right)+15\\ \mathrm{When}\mathrm{x}= 17, \mathrm{ }\mathrm{MC}=\mathrm{0}.021{\left(\mathrm{17}\right)}^{2}-\mathrm{0}.006\left(\mathrm{17}\right)+15\\ =6.069-0.102+15\\ =20.967\\ \mathrm{Hence},\mathrm{when}17\mathrm{units}\mathrm{are}\mathrm{produced},\mathrm{the}\mathrm{marginal}\mathrm{cost}\\ \mathrm{is}\mathrm{Rs}.20.967\mathrm{.}\end{array}$

Q.16 The total revenue in Rupees received from the sale of x units of a product is given by
R(x) = 13x2 + 26x + 15 Find the marginal revenue when x = 7.

Ans.

$\begin{array}{l}\mathrm{The}\mathrm{total}\mathrm{revenue}\mathrm{in}\mathrm{Rupees}\mathrm{received}\mathrm{from}\mathrm{the}\mathrm{sale}\mathrm{of}\mathrm{x}\mathrm{units}\\ \mathrm{of}\mathrm{a}\mathrm{product}\mathrm{is}\\ \mathrm{R}\left(\mathrm{x}\right)=13{\mathrm{x}}^{2}+26\mathrm{x}+15\\ \therefore \mathrm{Marginal}\mathrm{Revenue}\left(\mathrm{MR}\right)=\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{R}\left(\mathrm{x}\right)\\ \mathrm{ }=\frac{\mathrm{d}}{\mathrm{dx}}\left(13{\mathrm{x}}^{2}+26\mathrm{x}+15\right)\\ \mathrm{ }=26\mathrm{x}+26\\ \mathrm{ }=26\left(7\right)+26\left[\mathrm{When}\mathrm{ }\mathrm{x}= 7\right]\\ \mathrm{ }=208\\ \mathrm{Hence},\mathrm{the}\mathrm{required}\mathrm{marginal}\mathrm{revenue}\mathrm{is}\mathrm{Rs}208\mathrm{.}\end{array}$

Q.17 The rate of change of the area of a circle with respect to its radius r at r =6 cmis(A)10π (B)12π (C)8π (D)11π

Ans.

$\begin{array}{l}\mathrm{T}\mathrm{h}\mathrm{e}\phantom{\rule{thickmathspace}{0ex}}\mathrm{a}\mathrm{r}\mathrm{e}\mathrm{a}\phantom{\rule{thickmathspace}{0ex}}\left(\mathrm{A}\right)\phantom{\rule{thickmathspace}{0ex}}\mathrm{o}\mathrm{f}\phantom{\rule{thickmathspace}{0ex}}\mathrm{a}\phantom{\rule{thickmathspace}{0ex}}\mathrm{c}\mathrm{i}\mathrm{r}\mathrm{c}\mathrm{l}\mathrm{e}\phantom{\rule{thickmathspace}{0ex}}\mathrm{w}\mathrm{i}\mathrm{t}\mathrm{h}\phantom{\rule{thickmathspace}{0ex}}\mathrm{r}\mathrm{a}\mathrm{d}\mathrm{i}\mathrm{u}\mathrm{s}\phantom{\rule{thickmathspace}{0ex}}\left(\mathrm{r}\right)\phantom{\rule{thickmathspace}{0ex}}\mathrm{i}\mathrm{s}\phantom{\rule{thickmathspace}{0ex}}\mathrm{g}\mathrm{i}\mathrm{v}\mathrm{e}\mathrm{n}\phantom{\rule{thickmathspace}{0ex}}\mathrm{b}\mathrm{y},\\ \phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\mathrm{A}=\pi {\mathrm{r}}^{2}\\ \mathrm{D}\mathrm{i}\mathrm{f}\mathrm{f}\mathrm{e}\mathrm{r}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{i}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{n}\mathrm{g}\phantom{\rule{thickmathspace}{0ex}}\mathrm{b}\mathrm{o}\mathrm{t}\mathrm{h}\phantom{\rule{thickmathspace}{0ex}}\mathrm{s}\mathrm{i}\mathrm{d}\mathrm{e}\mathrm{s}\phantom{\rule{thickmathspace}{0ex}}\mathrm{w}\mathrm{i}\mathrm{t}\mathrm{h}\phantom{\rule{thickmathspace}{0ex}}\mathrm{r}\mathrm{e}\mathrm{s}\mathrm{p}\mathrm{e}\mathrm{c}\mathrm{t}\phantom{\rule{thickmathspace}{0ex}}\mathrm{t}\mathrm{o}\phantom{\rule{thickmathspace}{0ex}}\mathrm{r},\phantom{\rule{thickmathspace}{0ex}}\mathrm{w}\mathrm{e}\phantom{\rule{thickmathspace}{0ex}}\mathrm{g}\mathrm{e}\mathrm{t}\\ \phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\frac{dA}{dr}=\frac{d}{dr}\pi {\mathrm{r}}^{2}\\ \phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}=2\pi \mathrm{r}\\ \phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}{\left(\frac{dA}{dr}\right)}_{r=6}=2\pi \left(6\right)\\ \phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}=12\pi \phantom{\rule{thinmathspace}{0ex}}\\ \mathrm{H}\mathrm{e}\mathrm{n}\mathrm{c}\mathrm{e},\phantom{\rule{thickmathspace}{0ex}}\mathrm{t}\mathrm{h}\mathrm{e}\phantom{\rule{thickmathspace}{0ex}}\mathrm{r}\mathrm{e}\mathrm{q}\mathrm{u}\mathrm{i}\mathrm{r}\mathrm{e}\mathrm{d}\phantom{\rule{thickmathspace}{0ex}}\mathrm{r}\mathrm{a}\mathrm{t}\mathrm{e}\phantom{\rule{thickmathspace}{0ex}}\mathrm{o}\mathrm{f}\phantom{\rule{thickmathspace}{0ex}}\mathrm{c}\mathrm{h}\mathrm{a}\mathrm{n}\mathrm{g}\mathrm{e}\phantom{\rule{thickmathspace}{0ex}}\mathrm{o}\mathrm{f}\phantom{\rule{thickmathspace}{0ex}}\mathrm{t}\mathrm{h}\mathrm{e}\phantom{\rule{thickmathspace}{0ex}}\mathrm{a}\mathrm{r}\mathrm{e}\mathrm{a}\phantom{\rule{thickmathspace}{0ex}}\mathrm{o}\mathrm{f}\phantom{\rule{thickmathspace}{0ex}}\mathrm{a}\phantom{\rule{thickmathspace}{0ex}}\mathrm{c}\mathrm{i}\mathrm{r}\mathrm{c}\mathrm{l}\mathrm{e}\\ \mathrm{i}\mathrm{s}\phantom{\rule{thickmathspace}{0ex}}12\pi .\\ \mathrm{T}\mathrm{h}\mathrm{e}\phantom{\rule{thickmathspace}{0ex}}\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{r}\mathrm{e}\mathrm{c}\mathrm{t}\phantom{\rule{thickmathspace}{0ex}}\mathrm{a}\mathrm{n}\mathrm{s}\mathrm{w}\mathrm{e}\mathrm{r}\phantom{\rule{thickmathspace}{0ex}}\mathrm{i}\mathrm{s}\phantom{\rule{thickmathspace}{0ex}}\mathrm{B}.\end{array}$

Q.18 The total revenue in Rupees received from the sale of x units of a product is given by  Rx =3x2+36x+5.The marginal revenue, when x = 15 is:(A) 116    (B) 96    (C) 90    (D) 126

Ans.

$\begin{array}{l}\because \mathrm{ }\mathrm{Marginal}\mathrm{revenue}=\mathrm{The}\mathrm{rate}\mathrm{of}\mathrm{change}\mathrm{of}\mathrm{total}\mathrm{revenue}\\ \mathrm{with}\mathrm{respect}\mathrm{to}\mathrm{the}\mathrm{number}\mathrm{of}\\ \mathrm{units}\mathrm{sold}\mathrm{.}\\ \therefore \mathrm{Marginal}\mathrm{Revenue}\left(\mathrm{MR}\right)=\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{R}\left(\mathrm{x}\right)\\ \mathrm{ }=\frac{\mathrm{d}}{\mathrm{dx}}\left(3{\mathrm{x}}^{2}+36\mathrm{x}+5\right)\\ \mathrm{ }=6\mathrm{x}+36\\ \mathrm{Putting}\mathrm{x}=15,\mathrm{we}\mathrm{get}\\ \mathrm{Marginal}\mathrm{Revenue}\left(\mathrm{MR}\right)=6\left(15\right)+36\\ \mathrm{ }=90+36\\ \mathrm{ }=126\\ \mathrm{Hence},\mathrm{the}\mathrm{required}\mathrm{marginal}\mathrm{revenue}\mathrm{is}\mathrm{Rs}126\mathrm{.}\\ \mathrm{The}\mathrm{correct}\mathrm{answer}\mathrm{is}\mathrm{D}\mathrm{.}\end{array}$

Q.19 Show that the function given by f(x) = 3x + 17 is strictly increasing on R.

Ans.

$\begin{array}{l}\mathrm{Function}\mathrm{is}\mathrm{given}\mathrm{by}\mathrm{f}\left(\mathrm{x}\right)=3\mathrm{x}+17\\ \mathrm{Differentiating}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathrm{x},\mathrm{we}\mathrm{get}\\ \mathrm{f}‘\left(\mathrm{x}\right)=3>0\mathrm{ }\mathrm{in}\mathrm{every}\mathrm{interval}\mathrm{of}\mathrm{R}\mathrm{.}\\ \mathrm{Thus},\mathrm{the}\mathrm{function}\mathrm{is}\mathrm{strictly}\mathrm{increasing}\mathrm{on}\mathbf{R}\mathrm{.}\end{array}$

Q.20 Show that the function given by f(x) = e2x is strictly increasing on R

Ans.

$\begin{array}{l}\mathrm{Let}{\mathrm{x}}_{\mathrm{1}}\mathrm{and}{\mathrm{x}}_{\mathrm{2}}\mathrm{be}\mathrm{any}\mathrm{two}\mathrm{numbers}\mathrm{in}\mathbf{R}.\\ \mathrm{Then},\mathrm{ }\mathrm{we}\mathrm{have}:\\ {\mathrm{x}}_{\mathrm{1}}<{\mathrm{x}}_{\mathrm{2}}⇒2{\mathrm{x}}_{\mathrm{1}}< 2{\mathrm{x}}_{\mathrm{2}}\\ ⇒ \mathrm{ }{\mathrm{e}}^{2{\mathrm{x}}_{\mathrm{1}}}<{\mathrm{e}}^{2{\mathrm{x}}_{\mathrm{2}}}\\ ⇒\mathrm{f}\left({\mathrm{x}}_{1}\right)<\mathrm{f}\left({\mathrm{x}}_{2}\right)\\ \mathrm{Hence},\mathrm{f}\mathrm{is}\mathrm{strictly}\mathrm{increasing}\mathrm{on}\mathbf{R}\mathrm{.}\end{array}$

Q.21

$\begin{array}{l}\text{Show that the function given by f(x) = sin x is}\\ \text{(a) strictly increasing in}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\left(0,\frac{\pi }{2}\right),\\ \text{(b) strictly decreasing in}\phantom{\rule{thinmathspace}{0ex}}\left(\frac{\pi }{2},\pi \right),\\ \text{(c) neither increasing nor decreasing in}\phantom{\rule{thinmathspace}{0ex}}\left(0,\pi \right)\end{array}$

Ans.

$\begin{array}{l}\mathrm{The}\mathrm{given}\mathrm{function}\mathrm{is}\mathrm{f}\left(\mathrm{x}\right) =\mathrm{sin}\mathrm{x}\mathrm{.}\\ \therefore \mathrm{f}‘\left(\mathrm{x}\right)=\mathrm{cosx}\\ \left(\mathrm{a}\right)\mathrm{Since}\mathrm{for}\mathrm{each} \mathrm{x}\in \left(0,\frac{\mathrm{\pi }}{2}\right),\mathrm{ }\mathrm{cosx}>0\\ \mathrm{ }\mathrm{we}\mathrm{ }\mathrm{have}\mathrm{f}‘\left(\mathrm{x}\right)>0.\\ \mathrm{Hence},\mathrm{f}\mathrm{is}\mathrm{strictly}\mathrm{increasing}\mathrm{in}\mathrm{ }\left(0,\frac{\mathrm{\pi }}{2}\right).\\ \left(\mathrm{b}\right)\mathrm{Since}\mathrm{for}\mathrm{each} \mathrm{x}\in \left(\frac{\mathrm{\pi }}{2},\mathrm{\pi }\right),\mathrm{ }\mathrm{cosx}<0\\ \mathrm{ }\mathrm{we}\mathrm{ }\mathrm{have}\mathrm{f}‘\left(\mathrm{x}\right)<0.\\ \mathrm{Hence},\mathrm{f}\mathrm{is}\mathrm{strictly}\mathrm{decreasing}\mathrm{in}\mathrm{ }\left(\frac{\mathrm{\pi }}{2},\mathrm{\pi }\right).\\ \left(\mathrm{c}\right)\mathrm{From}\mathrm{the}\mathrm{results}\mathrm{obtained}\mathrm{in}\left(\mathrm{a}\right)\mathrm{and}\left(\mathrm{b}\right),\mathrm{it}\mathrm{is}\mathrm{clear}\mathrm{that}\mathrm{f}\mathrm{is}\\ \mathrm{ }\mathrm{neither}\mathrm{increasing}\mathrm{nor}\mathrm{ }\mathrm{decreasing}\mathrm{in}\left(0,\mathrm{\pi }\right).\end{array}$

Q.22 Find the intervals in which the function f given by f(x) = 2x2 − 3x is

(a) strictly increasing (b) strictly decreasing

Ans.

$\begin{array}{l}\mathrm{The}\mathrm{given}\mathrm{function}\mathrm{is}\mathrm{f}\left(\mathrm{x}\right)=2{\mathrm{x}}^{\mathrm{2}}-3\mathrm{x}\mathrm{.}\\ \mathrm{f}‘\left(\mathrm{x}\right)=4\mathrm{x}-3\\ \mathrm{Therefore},\mathrm{f}‘\left(\mathrm{x}\right)=0⇒4\mathrm{x}-3=0\\ ⇒ \mathrm{ }\mathrm{x}=\frac{3}{4}\\ \mathrm{Now}\mathrm{the}\mathrm{point}\mathrm{x}=\frac{3}{4}\mathrm{divides}\mathrm{the}\mathrm{real}\mathrm{line}\mathrm{into}\mathrm{two}\mathrm{disjoint}\\ \mathrm{intervals}\mathrm{ }\mathrm{namely},\left(-\mathrm{\infty },\mathrm{ }\frac{3}{4}\right)\mathrm{and}\left(\frac{3}{4},\mathrm{\infty }\right).\end{array}$

$\begin{array}{l}\text{In the interval}\left(-\infty ,\text{\hspace{0.17em}}\frac{3}{4}\right),\\ \text{}\text{}\text{f’}\left(x\right)=4x-3<0\\ Therefore,\text{f is strictly deacreasing in this interval}\text{. Also, in}\\ \text{the interval}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(\frac{3}{4},\infty \right)\text{, f’}\left(x\right)>0\text{​}\text{\hspace{0.17em}}\text{and so the function f is strictly}\\ \text{increasing in this interval}\text{.}\end{array}$

Q.23 Find the intervals in which the function f given by f(x) = 2x3 − 3x2 − 36x + 7 is
(a) strictly increasing (b) strictly decreasing

Ans.

$\begin{array}{l}\therefore \mathrm{f}‘\left(\mathrm{x}\right)=0⇒6\mathrm{ }\left(\mathrm{x}-3\right)\mathrm{ }\left(\mathrm{x}+2\right)=0\\ ⇒\mathrm{x}=-2,3\\ \mathrm{The}\mathrm{points}\mathrm{x}=-2\mathrm{and}\mathrm{x}= 3\mathrm{divide}\mathrm{the}\mathrm{real}\mathrm{line}\mathrm{into}\mathrm{three}\\ \mathrm{disjoint}\mathrm{intervals}\mathrm{i}.\mathrm{e}.,\mathrm{ }\left(-\infty ,-2\right),\mathrm{ }\left(-2,3\right)\mathrm{and}\left(3,\infty \right).\end{array}$

$\begin{array}{l}\text{In intervals}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(-\infty ,-2\right)\text{and}\left(3,\infty \right),\text{\hspace{0.17em}}f‘\left(x\right)\text{\hspace{0.17em}}\text{is positive while in interval}\\ \left(-2,3\right),\text{\hspace{0.17em}}f‘\left(x\right)\text{is negative}\text{.}\\ \text{Hence, the given function (f) is strictly increasing in intervals}\\ \left(-\infty ,-2\right)\text{and}\left(3,\infty \right),\text{while function (f) is strictly decreasing in}\\ \text{interval}\text{\hspace{0.17em}}\left(-2,3\right).\end{array}$

Q.24 Find the intervals in which the following functions are strictly increasing or decreasing:

(a) x2 + 2x − 5 (b)10 − 6x − 2x2

(c) −2x3 − 9x2 − 12x + 1 (d) 6 − 9x − x2

(e) (x + 1)3 (x − 3)3

Ans.

$\begin{array}{l}\left(\mathrm{a}\right)\mathrm{We}\mathrm{have},\\ \mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^{\mathrm{2}}+2\mathrm{x}-\mathrm{5}\\ \mathrm{Differentiating}\mathrm{both}\mathrm{sides}\mathrm{with}\mathrm{respect}\mathrm{to}\mathrm{x},\mathrm{we}\mathrm{get}\\ \mathrm{f}‘\left(\mathrm{x}\right)=2\mathrm{x}+2\\ \therefore \mathrm{f}‘\left(\mathrm{x}\right)=0⇒2\mathrm{x}+2=0\\ ⇒ \mathrm{x}=-1\\ \mathrm{Point}\mathrm{x}=-1\mathrm{divides}\mathrm{the}\mathrm{real}\mathrm{line}\mathrm{into}\mathrm{two}\mathrm{disjoint}\\ \mathrm{intervals}\mathrm{i}.\mathrm{e}.,\mathrm{ }\left(-\mathrm{\infty },-1\right)\mathrm{ }\mathrm{and}\mathrm{ }\left(-1,\mathrm{\infty }\right).\\ \mathrm{In}\mathrm{interval}\left(-\mathrm{\infty },-1\right),\mathrm{ }\mathrm{f}‘\left(\mathrm{x}\right)=2\mathrm{x}+2<0\\ \therefore \mathrm{f}\mathrm{is}\mathrm{strictly}\mathrm{decreasing}\mathrm{in}\mathrm{interval}\mathrm{ }\left(-\mathrm{\infty },-1\right).\\ \mathrm{Thus},\mathrm{f}\mathrm{is}\mathrm{strictly}\mathrm{decreasing}\mathrm{for}\mathrm{x}<-1.\\ \mathrm{In}\mathrm{interval} \left(-1,\mathrm{\infty }\right),\mathrm{ }\mathrm{f}‘\left(\mathrm{x}\right)=2\mathrm{x}+2>0\\ \therefore \mathrm{f}\mathrm{is}\mathrm{strictly}\mathrm{increasing}\mathrm{in}\mathrm{interval}\mathrm{ }\left(-1,\mathrm{\infty }\right).\\ \mathrm{Thus},\mathrm{f}\mathrm{is}\mathrm{strictly}\mathrm{increasing}\mathrm{for}\mathrm{x}>-1.\\ \left(\mathrm{b}\right)\mathrm{We}\mathrm{have},\\ \mathrm{f}\left(\mathrm{x}\right)=\mathrm{10}-6\mathrm{x}-2{\mathrm{x}}^{\mathrm{2}}\\ \mathrm{Differentiating}\mathrm{both}\mathrm{sides}\mathrm{with}\mathrm{respect}\mathrm{to}\mathrm{x},\mathrm{we}\mathrm{get}\\ \mathrm{f}‘\left(\mathrm{x}\right)=-\mathrm{ }6-4\mathrm{x}\\ \therefore \mathrm{f}‘\left(\mathrm{x}\right)=0⇒-\mathrm{ }6-4\mathrm{x}=0\\ ⇒ \mathrm{x}=-\frac{3}{2}\\ \mathrm{Point}\mathrm{x}=-\frac{3}{2}\mathrm{divides}\mathrm{the}\mathrm{real}\mathrm{line}\mathrm{into}\mathrm{two}\mathrm{disjoint}\mathrm{intervals}\\ \mathrm{i}.\mathrm{e}.,\mathrm{ }\left(-\mathrm{\infty },-\frac{3}{2}\right)\mathrm{ }\mathrm{and}\mathrm{ }\left(-\frac{3}{2},\mathrm{\infty }\right).\\ \mathrm{In}\mathrm{interval}\left(-\mathrm{\infty },-\frac{3}{2}\right),\mathrm{ }\mathrm{f}‘\left(\mathrm{x}\right)=-\mathrm{ }6-4\mathrm{x}>0\\ \therefore \mathrm{f}\mathrm{is}\mathrm{strictly}\mathrm{increasing}\mathrm{in}\mathrm{interval}\mathrm{ }\left(-\mathrm{\infty },-\frac{3}{2}\right).\\ \mathrm{Thus},\mathrm{f}\mathrm{is}\mathrm{strictly}\mathrm{increasing}\mathrm{for}\mathrm{x}<-\frac{3}{2}\mathrm{.}\\ \mathrm{In}\mathrm{interval} \left(-\frac{3}{2},\mathrm{\infty }\right),\mathrm{ }\mathrm{f}‘\left(\mathrm{x}\right)=-6-4\mathrm{x}<0\\ \therefore \mathrm{f}\mathrm{is}\mathrm{strictly}\mathrm{decreasing}\mathrm{in}\mathrm{interval}\mathrm{ }\left(-\frac{3}{2},\mathrm{\infty }\right).\\ \mathrm{Thus},\mathrm{f}\mathrm{is}\mathrm{strictly}\mathrm{decreasing}\mathrm{for}\mathrm{x} >-\frac{3}{2}\mathrm{.}\\ \left(\mathrm{c}\right)\mathrm{We}\mathrm{have},\\ \mathrm{f}\left(\mathrm{x}\right)=-2{\mathrm{x}}^{\mathrm{3}}-9{\mathrm{x}}^{\mathrm{2}}-12\mathrm{x}+\mathrm{1}\\ \mathrm{Differentiating}\mathrm{both}\mathrm{sides}\mathrm{with}\mathrm{respect}\mathrm{to}\mathrm{x},\mathrm{we}\mathrm{get}\\ \mathrm{f}‘\left(\mathrm{x}\right)=-\mathrm{ }6{\mathrm{x}}^{2}-18\mathrm{x}-12=-6\left({\mathrm{x}}^{2}+3\mathrm{x}+2\right)\\ \therefore \mathrm{f}‘\left(\mathrm{x}\right)=0⇒-6\left({\mathrm{x}}^{2}+3\mathrm{x}+2\right)=0\\ ⇒\left(\mathrm{x}+1\right)\left(\mathrm{x}+2\right)=0\\ ⇒ \mathrm{x}=-1,-2\\ \mathrm{The}\mathrm{points}\mathrm{x}=-1\mathrm{and}\mathrm{x}=-2\mathrm{divide}\mathrm{the}\mathrm{real}\mathrm{line}\mathrm{into}\mathrm{three}\\ \mathrm{disjoint}\mathrm{intervals}\mathrm{i}.\mathrm{e}.,\mathrm{ }\left(-\mathrm{\infty },-2\right),\left(-2,-1\right)\mathrm{and}\left(-1,\mathrm{\infty }\right).\\ \mathrm{In}\mathrm{intervals} \left(-\mathrm{\infty },-2\right)\mathrm{and}\left(-1,\mathrm{\infty }\right),\mathrm{ }\mathrm{f}‘\left(\mathrm{x}\right)\mathrm{ }\mathrm{is}\mathrm{negative}.\\ \mathrm{Hence},\mathrm{the}\mathrm{given}\mathrm{function}\left(\mathrm{f}\right)\mathrm{is}\mathrm{strictly}\mathrm{decreasing}\mathrm{in}\mathrm{intervals}\\ \left(-\mathrm{\infty },-2\right)\mathrm{and}\left(-1,\mathrm{\infty }\right),\\ \mathrm{i}.\mathrm{e}.,\mathrm{ }\mathrm{function}\left(\mathrm{f}\right)\mathrm{is}\mathrm{strictly}\mathrm{decreasing}\mathrm{when}\mathrm{x}<-2\mathrm{and}\mathrm{x}>-1.\end{array}$ $\begin{array}{l}\text{In intervals}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(-2,-1\right),\text{\hspace{0.17em}}f‘\left(x\right)\text{\hspace{0.17em}}\text{is positive}\text{.}\\ \text{So, function (f) is strictly increasing in interval}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(-2,-1\right),\\ i.e.,\text{\hspace{0.17em}}\text{function (f) is strictly increasing}\text{\hspace{0.17em}}when\text{\hspace{0.17em}}-20\\ \therefore \text{f is strictly increasing in interval}\text{\hspace{0.17em}}\left(-\infty ,-\frac{9}{2}\right).\\ \text{Thus, f is strictly increasing for x <}-\frac{9}{2}\text{.}\\ \text{In interval}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(-\frac{9}{2},\infty \right),\text{\hspace{0.17em}}f‘\left(x\right)=-\text{9}-2\text{x}<0\\ \therefore \text{f is strictly decreasing in interval}\text{\hspace{0.17em}}\left(-\frac{9}{2},\infty \right).\\ \text{Thus, f is strictly decreasing for x >}-\frac{9}{2}\text{.}\end{array}$ $\begin{array}{l}\left(e\right)\text{We have,}\\ \text{f(x)}={\text{(x + 1)}}^{\text{3}}\text{(x}-{\text{3)}}^{\text{3}}\\ Differentiating\text{both sides with respect to x, we get}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{f’}\left(x\right)={\left(\text{x + 1}\right)}^{\text{3}}\frac{d}{dx}{\left(\text{x}-\text{3}\right)}^{\text{3}}+{\left(\text{x}-\text{3}\right)}^{\text{3}}\frac{d}{dx}{\left(\text{x + 1}\right)}^{\text{3}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\left[By\text{product rule}\right]\\ ⇒\text{f’}\left(x\right)={\left(\text{x + 1}\right)}^{\text{3}}3{\left(\text{x}-\text{3}\right)}^{\text{2}}\frac{d}{dx}\left(\text{x}-\text{3}\right)+{\left(\text{x}-\text{3}\right)}^{\text{3}}3{\left(\text{x + 1}\right)}^{\text{2}}\frac{d}{dx}\left(\text{x + 1}\right)\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\left[By\text{chain rule}\right]\\ ⇒\text{f’}\left(x\right)=3{\left(\text{x + 1}\right)}^{\text{3}}{\left(\text{x}-\text{3}\right)}^{\text{2}}\left(1-0\right)+3{\left(\text{x}-\text{3}\right)}^{\text{3}}{\left(\text{x + 1}\right)}^{\text{2}}\left(1+0\right)\\ ⇒\text{f’}\left(x\right)=3{\left(\text{x + 1}\right)}^{\text{2}}{\left(\text{x}-\text{3}\right)}^{\text{2}}\left(x+1+x-3\right)\\ ⇒\text{f’}\left(x\right)=3{\left(\text{x + 1}\right)}^{\text{2}}{\left(\text{x}-\text{3}\right)}^{\text{2}}\left(2x-2\right)\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=6{\left(\text{x + 1}\right)}^{\text{2}}{\left(\text{x}-\text{3}\right)}^{\text{2}}\left(x-1\right)\\ Now,\text{f’}\left(x\right)=0⇒x=-1,\text{\hspace{0.17em}}1,\text{\hspace{0.17em}}\text{\hspace{0.17em}}3\\ \text{The points x}=-1,\text{}x=1,\text{}and\text{}x=3\text{divide the real line into}\\ \text{four disjoint intervals i}\text{.e}\text{.,}\left(-\infty ,-1\right),\left(-1,\text{\hspace{0.17em}}1\right),\left(1,3\right)\text{and}\left(3,\infty \right).\\ \text{In intervals}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(-\infty ,-1\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{and (}-\text{1, 1),}\end{array}$ $\begin{array}{l}\text{\hspace{0.17em}}\text{f’}\left(x\right)=6{\left(\text{x + 1}\right)}^{\text{2}}{\left(\text{x}-\text{3}\right)}^{\text{2}}\left(x-1\right)<0\\ \therefore \text{f is strictly decreasing in intervals}\text{\hspace{0.17em}}\left(-\infty ,-1\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{and}\left(-\text{1, 1}\right).\\ \text{In intervals (1, 3) and}\text{\hspace{0.17em}}\left(3,\infty \right),\text{\hspace{0.17em}}\text{f’}\left(x\right)=6{\left(\text{x + 1}\right)}^{\text{2}}{\left(\text{x}-\text{3}\right)}^{\text{2}}\left(x-1\right)>0\\ \therefore \text{f is strictly increasing in intervals}\left(\text{1, 3}\right)\text{and}\text{\hspace{0.17em}}\left(3,\infty \right).\end{array}$

Q.25 Showthat y=log(1+x)2x2+x,x> 1, is an increasing function of x throughout its domain.

Ans.

$\begin{array}{l}\mathrm{We}\mathrm{have}, \mathrm{y}=\mathrm{log}\left(1+\mathrm{x}\right)-\frac{2\mathrm{x}}{2+\mathrm{x}}\\ \mathrm{Differentiating}\mathrm{both}\mathrm{sides}\mathrm{with}\mathrm{respect}\mathrm{to}\mathrm{x},\mathrm{we}\mathrm{get}\\ \mathrm{ }\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{log}\left(1+\mathrm{x}\right)-\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{2\mathrm{x}}{2+\mathrm{x}}\right)\\ =\frac{1}{1+\mathrm{x}}\frac{\mathrm{d}}{\mathrm{dx}}\left(1+\mathrm{x}\right)-\frac{\left(2+\mathrm{x}\right)\frac{\mathrm{d}}{\mathrm{dx}}2\mathrm{x}-2\mathrm{x}\frac{\mathrm{d}}{\mathrm{dx}}\left(2+\mathrm{x}\right)}{{\left(2+\mathrm{x}\right)}^{2}}\\ =\frac{1}{1+\mathrm{x}}×\left(0+1\right)-\frac{2\left(2+\mathrm{x}\right)-2\mathrm{x}\left(0+1\right)}{{\left(2+\mathrm{x}\right)}^{2}}\\ =\frac{1}{1+\mathrm{x}}-\frac{4}{{\left(2+\mathrm{x}\right)}^{2}}\\ =\frac{4+4\mathrm{x}+{\mathrm{x}}^{2}-4-4\mathrm{x}}{\left(1+\mathrm{x}\right){\left(2+\mathrm{x}\right)}^{2}}\\ =\frac{{\mathrm{x}}^{2}}{\left(1+\mathrm{x}\right){\left(2+\mathrm{x}\right)}^{2}}\\ \mathrm{Now}, \mathrm{ }\frac{\mathrm{dy}}{\mathrm{dx}}=0\\ ⇒\mathrm{ }\frac{{\mathrm{x}}^{2}}{\left(1+\mathrm{x}\right){\left(2+\mathrm{x}\right)}^{2}}=0\\ ⇒ \mathrm{ }{\mathrm{x}}^{2}=0\left[\left(1+\mathrm{x}\right)\ne 0\mathrm{and}{\left(2+\mathrm{x}\right)}^{2}\ne 0\right]\\ ⇒ \mathrm{ }\mathrm{x}=0\\ \mathrm{Since}\mathrm{x}>-1,\mathrm{point}\mathrm{x}= 0\mathrm{divides}\mathrm{the}\mathrm{domain}\left(-1,\mathrm{\infty }\right)\mathrm{in}\mathrm{two}\\ \mathrm{disjoint}\mathrm{intervals}\mathrm{i}.\mathrm{e}.,-1<\mathrm{ }\mathrm{x}< 0\mathrm{and}\mathrm{x}> 0\mathrm{.}\\ \mathrm{When}-1<\mathrm{x}< 0,\mathrm{we}\mathrm{have}:\\ \mathrm{x}< 0⇒{\mathrm{x}}^{2}>0\\ \mathrm{x}>-1⇒\mathrm{x}+2>0⇒{\left(\mathrm{x}+2\right)}^{2}>0\\ \therefore \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{{\mathrm{x}}^{2}}{{\left(\mathrm{x}+2\right)}^{2}}>0\\ \mathrm{Also},\mathrm{when}\mathrm{x}> 0:\\ \mathrm{x}>0⇒{\mathrm{x}}^{2}>0, \left(2+{\mathrm{x}}^{2}\right)>0\\ \therefore \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{{\mathrm{x}}^{2}}{{\left(\mathrm{x}+2\right)}^{2}}>0\\ \mathrm{Hence},\mathrm{function}\mathrm{f}\mathrm{is}\mathrm{increasing}\mathrm{throughout}\mathrm{this}\mathrm{domain}\mathrm{.}\end{array}$

Q.26 Find the values of x for whichy=[x(x2)]2is an increasingfunction.

Ans.

$\begin{array}{l}\mathrm{We}\mathrm{have},\\ \mathrm{y}={\left[\mathrm{x}\left(\mathrm{x}-\mathrm{2}\right)\right]}^{2}\\ ={\mathrm{x}}^{2}{\left(\mathrm{x}-2\right)}^{2}\\ \mathrm{Differentiating}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathrm{x},\mathrm{we}\mathrm{get}\\ \mathrm{ }\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}{\mathrm{x}}^{2}{\left(\mathrm{x}-2\right)}^{2}\\ ={\mathrm{x}}^{2}\frac{\mathrm{d}}{\mathrm{dx}}{\left(\mathrm{x}-2\right)}^{2}+{\left(\mathrm{x}-2\right)}^{2}\frac{\mathrm{d}}{\mathrm{dx}}{\mathrm{x}}^{2}\\ ={\mathrm{x}}^{2}.2\left(\mathrm{x}-2\right)\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}-2\right)+{\left(\mathrm{x}-2\right)}^{2}.2\mathrm{x}\\ =2{\mathrm{x}}^{2}\left(\mathrm{x}-2\right)\left(1-0\right)+2\mathrm{x}{\left(\mathrm{x}-2\right)}^{2}\\ =2\mathrm{x}\left(\mathrm{x}-2\right)\left(\mathrm{x}+\mathrm{x}-2\right)\\ =2\mathrm{x}\left(\mathrm{x}-2\right)\left(2\mathrm{x}-2\right)\\ =4\mathrm{x}\left(\mathrm{x}-2\right)\left(\mathrm{x}-1\right)\\ \therefore \frac{\mathrm{dy}}{\mathrm{dx}}=0⇒\mathrm{x}=0,1,2\\ \mathrm{The}\mathrm{points}\mathrm{x}= 0,\mathrm{x}= 1,\mathrm{and}\mathrm{x}= 2\mathrm{divide}\mathrm{the}\mathrm{real}\mathrm{line}\mathrm{into}\mathrm{four}\\ \mathrm{disjoint}\mathrm{intervals}\mathrm{i}.\mathrm{e}.,\left(-\mathrm{\infty },0\right),\left(0,1\right),\left(1,2\right),\left(2,\mathrm{\infty }\right).\\ \mathrm{In}\mathrm{intervals}\left(-\mathrm{\infty },0\right)\mathrm{and} \left(1,2\right),\mathrm{ }\frac{\mathrm{dy}}{\mathrm{dx}}<0\\ \therefore \mathrm{y}\mathrm{is}\mathrm{strictly}\mathrm{decreasing}\mathrm{in}\mathrm{intervals}\mathrm{ }\left(-\mathrm{\infty },0\right)\mathrm{and} \left(1,2\right).\\ \mathrm{However},\mathrm{in}\mathrm{intervals}\left(0, 1\right)\mathrm{and}\left(2,\mathrm{\infty }\right),\mathrm{ }\frac{\mathrm{dy}}{\mathrm{dx}}>0\\ \therefore \mathrm{y}\mathrm{is}\mathrm{strictly}\mathrm{increasing}\mathrm{in}\mathrm{intervals}\left(0, 1\right)\mathrm{and}\left(2,\infty \right)\mathrm{.}\\ \therefore \mathrm{y}\mathrm{is}\mathrm{strictly}\mathrm{increasing}\mathrm{for}0 <\mathrm{x}< 1\mathrm{and}\mathrm{x}> 2\mathrm{.}\end{array}$

Q.27 Prove that y=4sinθ(2+cosθ)θ is an increasing function of θ in [0,π2].

Ans.

$\begin{array}{l}\mathrm{We}\mathrm{have}, \mathrm{y}=\frac{4\mathrm{sin\theta }}{\left(\mathrm{2}+\mathrm{cos\theta }\right)}-\mathrm{\theta }\\ \mathrm{Differentiating}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathrm{\theta },\mathrm{we}\mathrm{get}\\ \frac{\mathrm{dy}}{\mathrm{d\theta }}=\frac{\mathrm{d}}{\mathrm{d\theta }}\left\{\frac{4\mathrm{sin\theta }}{\left(\mathrm{2}+\mathrm{cos\theta }\right)}-\mathrm{\theta }\right\}\\ =\frac{\left(2+\mathrm{cos\theta }\right)\frac{\mathrm{d}}{\mathrm{d\theta }}\left(4\mathrm{sin\theta }\right)-4\mathrm{sin\theta }\frac{\mathrm{d}}{\mathrm{d\theta }}\left(2+\mathrm{cos\theta }\right)}{{\left(2+\mathrm{cos\theta }\right)}^{2}}-\frac{\mathrm{d}}{\mathrm{d\theta }}\mathrm{\theta }\\ =\frac{4\left(2+\mathrm{cos\theta }\right)\mathrm{cos\theta }-4\mathrm{sin\theta }\left(0-\mathrm{sin\theta }\right)}{{\left(2+\mathrm{cos\theta }\right)}^{2}}-1\\ =\frac{4\left(2+\mathrm{cos\theta }\right)\mathrm{cos\theta }+4{\mathrm{sin}}^{2}\mathrm{\theta }}{{\left(2+\mathrm{cos\theta }\right)}^{2}}-1\\ =\frac{8\mathrm{cos\theta }+4\left({\mathrm{cos}}^{2}\mathrm{\theta }+{\mathrm{sin}}^{2}\mathrm{\theta }\right)}{{\left(2+\mathrm{cos\theta }\right)}^{2}}-1\\ =\frac{8\mathrm{cos\theta }+4}{{\left(2+\mathrm{cos\theta }\right)}^{2}}-1\\ \mathrm{Now},\mathrm{}\frac{\mathrm{dy}}{\mathrm{d\theta }}=0\\ ⇒\frac{8\mathrm{cos\theta }+4}{{\left(2+\mathrm{cos\theta }\right)}^{2}}-1=0⇒\frac{8\mathrm{cos\theta }+4}{{\left(2+\mathrm{cos\theta }\right)}^{2}}=1\\ ⇒8\mathrm{cos\theta }+4={\left(2+\mathrm{cos\theta }\right)}^{2}\\ ⇒8\mathrm{cos\theta }+\overline{)4}=\overline{)4}+4\mathrm{cos\theta }+{\mathrm{cos}}^{2}\mathrm{\theta }\\ ⇒{\mathrm{cos}}^{2}\mathrm{\theta }-4\mathrm{cos\theta }=0\\ ⇒\mathrm{cos\theta }\left(\mathrm{cos\theta }-4\right)=0\\ ⇒\mathrm{cos\theta }=0 \mathrm{or}\mathrm{ }\mathrm{cos\theta }=4\mathrm{ }\\ \mathrm{Since}\mathrm{cos}\mathrm{\theta }\ne \mathrm{4}\\ \therefore \mathrm{cos\theta }=0⇒\mathrm{\theta }=\frac{\mathrm{\pi }}{2}\\ \mathrm{Now}, \mathrm{ }\frac{\mathrm{dy}}{\mathrm{d\theta }}=\frac{8\mathrm{cos\theta }+4}{{\left(2+\mathrm{cos\theta }\right)}^{2}}-1\\ =\frac{8\mathrm{cos\theta }+4-{\left(2+\mathrm{cos\theta }\right)}^{2}}{{\left(2+\mathrm{cos\theta }\right)}^{2}}\\ =\frac{8\mathrm{cos\theta }+4-4-4\mathrm{cos\theta }-{\mathrm{cos}}^{2}\mathrm{\theta }}{{\left(2+\mathrm{cos\theta }\right)}^{2}}\\ =\frac{\mathrm{cos\theta }\left(4-\mathrm{cos\theta }\right)}{{\left(2+\mathrm{cos\theta }\right)}^{2}}\\ \mathrm{In}\mathrm{interval} \left(0,\frac{\mathrm{\pi }}{2}\right), \mathrm{we}\mathrm{have}\mathrm{cos\theta }> 0.\mathrm{Also}, 4 >\mathrm{cos\theta }\\ ⇒\mathrm{4}-\mathrm{cos\theta }\mathrm{ }>0\mathrm{.}\\ \therefore \mathrm{cos\theta }\left(4-\mathrm{cos\theta }\right)>0\mathrm{and}\mathrm{also}{\left(2+\mathrm{cos\theta }\right)}^{2}>0\\ ⇒\frac{\mathrm{cos\theta }\left(4-\mathrm{cos\theta }\right)}{{\left(2+\mathrm{cos\theta }\right)}^{2}}>0⇒\frac{\mathrm{dy}}{\mathrm{d\theta }}>0\\ \mathrm{Therefore},\mathrm{y}\mathrm{is}\mathrm{strictly}\mathrm{increasing}\mathrm{in}\mathrm{interval} \left(0,\frac{\mathrm{\pi }}{2}\right).\\ \mathrm{Also},\mathrm{the}\mathrm{given}\mathrm{function}\mathrm{is}\mathrm{continuous}\mathrm{at} \mathrm{x}=0\mathrm{and}\mathrm{x}=\frac{\mathrm{\pi }}{2}.\\ \mathrm{Hence},\mathrm{y}\mathrm{is}\mathrm{increasing}\mathrm{in}\mathrm{interval}\mathrm{ }\left(0,\frac{\mathrm{\pi }}{2}\right).\end{array}$

Q.28 Provethatthelogarithmicfunctionisstrictly increasingon (0, ).

Ans.

$\begin{array}{l}\mathrm{The}\mathrm{given}\mathrm{function}\mathrm{if}\mathrm{f}\left(\mathrm{x}\right)=\mathrm{logx}\\ \therefore \mathrm{f}‘\left(\mathrm{x}\right)=\frac{1}{\mathrm{x}}\\ \mathrm{It}\mathrm{is}\mathrm{clear}\mathrm{that}\mathrm{for}\mathrm{x}> 0,\mathrm{f}‘\left(\mathrm{x}\right)=\frac{1}{\mathrm{x}}>0\\ \mathrm{Hence},\mathrm{f}\left(\mathrm{x}\right) =\mathrm{log}\mathrm{x}\mathrm{is}\mathrm{strictly}\mathrm{increasing}\mathrm{in}\mathrm{interval}\left(0,\infty \right)\mathrm{.}\\ \end{array}$

Q.29 Prove that the function f given by f(x)= x2 − x + 1 is neither strictly increasing
nor strictly decreasing on (−1, 1).

Ans.

$\begin{array}{l}{\text{The given function is f(x) = x}}^{\text{2}}-\text{x + 1}\text{.}\\ Differentiating\text{with respect to x, we get}\\ \text{f’}\left(x\right)=2x-1\\ Now,\text{\hspace{0.17em}}f‘\left(x\right)=0⇒2x-1=0⇒x=\frac{1}{2}\\ \text{The point}\frac{1}{2}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{divides the interval}\left(-\text{1, 1}\right)\text{into two disjoint}\\ \text{intervals i}\text{.e}\text{.,}\left(-1,\text{}\frac{1}{2}\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}and\text{\hspace{0.17em}}\left(\frac{1}{2}\text{\hspace{0.17em}}\text{, 1}\right)\text{.}\\ \text{Now, for interval}\left(-1,\text{}\frac{1}{2}\right)\text{\hspace{0.17em}},\text{\hspace{0.17em}}f‘\left(x\right)=2x-1<0\\ \text{Therefore, f is strictly decreasing in interval}\text{\hspace{0.17em}}\left(-1,\text{}\frac{1}{2}\right).\\ \text{Now, for interval}\left(\frac{1}{2}\text{\hspace{0.17em}}\text{, 1}\right)\text{\hspace{0.17em}},\text{\hspace{0.17em}}f‘\left(x\right)=2x-1>0\\ \text{So, f is strictly increasing in interval}\left(\frac{1}{2}\text{\hspace{0.17em}}\text{, 1}\right)\text{.}\\ \text{Hence, f is neither strictly increasing nor decreasing in}\\ \text{interval}\left(-\text{1, 1}\right).\end{array}$

Q.30 Whichofthefollowingfunctionsarestrictly decreasingon(0,π2)?(A)cosx (B)cos2x (C)cos3x (D)tanx

Ans.

$\begin{array}{l}\left(A\right)\text{\hspace{0.17em}}Let\text{\hspace{0.17em}}{f}_{1}\left(x\right)=\mathrm{cos}x\\ \text{Differentiating w}\text{.r}\text{.t}\text{. x, we get}\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\text{f’}}_{\text{1}}\left(x\right)=\frac{d}{dx}\mathrm{cos}x\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=-\mathrm{sin}x\\ In\text{interval}\left(0,\frac{\pi }{2}\right),\text{sinx is positive in first quadrant}\text{.}\\ \text{So,}\text{\hspace{0.17em}}\text{}\text{}{\text{f’}}_{\text{1}}\left(x\right)<0\\ Therefore,\text{\hspace{0.17em}}{f}_{1}\left(x\right)=\mathrm{cos}x\text{\hspace{0.17em}}\text{is strictly decreasing in interval}\text{\hspace{0.17em}}\left(0,\frac{\pi }{2}\right).\\ \left(B\right)Let\text{\hspace{0.17em}}{f}_{2}\left(x\right)=\mathrm{cos}2x\\ \text{Differentiating w}\text{.r}\text{.t}\text{. x, we get}\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\text{f’}}_{\text{2}}\left(x\right)=\frac{d}{dx}\mathrm{cos}2x\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=-2\mathrm{sin}2x\\ In\text{interval}\left(0,\frac{\pi }{2}\right),\text{sin2x is positive in first quadrant}\text{.}\\ \text{So,}\text{\hspace{0.17em}}\text{}\text{}{\text{f’}}_{\text{2}}\left(x\right)<0\\ Therefore,\text{\hspace{0.17em}}{f}_{2}\left(x\right)=\mathrm{cos}2x\text{\hspace{0.17em}}\text{is strictly decreasing in interval}\text{\hspace{0.17em}}\left(0,\frac{\pi }{2}\right).\\ \left(C\right)Let\text{\hspace{0.17em}}{f}_{3}\left(x\right)=\mathrm{cos}3x\\ \text{Differentiating w}\text{.r}\text{.t}\text{. x, we get}\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\text{f’}}_{\text{3}}\left(x\right)=\frac{d}{dx}\mathrm{cos}3x\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=-3\mathrm{sin}3x\\ Now,\text{\hspace{0.17em}}{\text{f’}}_{\text{3}}\left(x\right)=0⇒-3\mathrm{sin}3x=0\\ ⇒\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}3x=\pi ⇒x=\frac{\pi }{3}\\ \text{The point}x=\frac{\pi }{3}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{divides the interval}\left(0,\frac{\pi }{2}\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{into two disjoint}\\ \text{intervals i}\text{.e}\text{.,}\text{\hspace{0.17em}}\left(0,\frac{\pi }{3}\right)\text{and}\left(\frac{\pi }{3},\frac{\pi }{2}\right).\\ Now,\text{​}\text{in interval}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(0,\frac{\pi }{3}\right),\\ {\text{f’}}_{\text{3}}\left(x\right)=-3\mathrm{sin}3x<0\text{}\text{}\left[\because 00\\ Therefore,\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\text{f}}_{\text{3}}\text{is strictly increasing in interval}\text{\hspace{0.17em}}\left(\frac{\pi }{3},\frac{\pi }{2}\right).\\ {\text{Hence, f}}_{\text{3}}\text{is neither increasing nor decreasing in interval}\text{\hspace{0.17em}}\left(0,\frac{\pi }{2}\right).\\ \left(D\right)Let\text{\hspace{0.17em}}{f}_{4}\left(x\right)=\mathrm{tan}x\\ Differentiating\text{w}\text{.r}\text{.t}\text{. x, we get}\\ \text{}{\text{f’}}_{\text{4}}\left(x\right)=\frac{d}{dx}\mathrm{tan}x\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}={\mathrm{sec}}^{2}x\\ In\text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathrm{int}erval\text{}\left(0,\frac{\pi }{2}\right),\text{​}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathrm{sec}x>0⇒{\mathrm{sec}}^{2}x>0\\ \therefore {f}_{4}\left(x\right)>0\\ \therefore \text{f4 is strictly increasing in interval}\text{\hspace{0.17em}}\left(0,\frac{\pi }{2}\right).\\ \text{Therefore, functions cos x and cos 2x are strictly decreasing}\\ \text{in}\text{\hspace{0.17em}}\left(0,\frac{\pi }{2}\right).\\ \text{Hence, the correct answers are A and B}\text{.}\end{array}$

Q.31 On which of the following intervals is the function f given byf(x)=x100 +sinx– 1strictly decreasing?(A)(0, 1)(B)(π2, π)(C)(0, π2)(D)None of these

Ans.

$\begin{array}{l}\mathrm{We} \mathrm{have},\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^{100}+\mathrm{sinx}-1\\ \mathrm{Differentiating}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathrm{x},\mathrm{we}\mathrm{get}\\ \mathrm{f}‘\left(\mathrm{x}\right)=100{\mathrm{x}}^{99}+\mathrm{cosx}\\ \mathrm{f}‘\left(\mathrm{x}\right)>0\left[\because 100{\mathrm{x}}^{99}>0\mathrm{and}\mathrm{cosx}>0\mathrm{in}\mathrm{interval}\left(0,1\right)\right]\\ \mathrm{Thus},\mathrm{function}\mathrm{f}\mathrm{is}\mathrm{strictly}\mathrm{increasing}\mathrm{in}\mathrm{interval}\left(0, 1\right)\mathrm{.}\\ \mathrm{In}\mathrm{interval}\left(\frac{\mathrm{\pi }}{2},\mathrm{\pi }\right),\\ \mathrm{cosx}<0\mathrm{and}100{\mathrm{x}}^{\mathrm{99}}>0\\ \mathrm{So}, \mathrm{ }\mathrm{f}‘\left(\mathrm{x}\right)>0\\ \mathrm{Thus},\mathrm{function}\mathrm{f}\mathrm{is}\mathrm{strictly}\mathrm{increasing}\mathrm{in}\mathrm{interval}\left(\frac{\mathrm{\pi }}{2},\mathrm{\pi }\right).\\ \mathrm{In}\mathrm{interval}\left(0,\frac{\mathrm{\pi }}{2}\right),\\ \mathrm{cosx}>0\mathrm{and}100{\mathrm{x}}^{\mathrm{99}}>0\\ ⇒100{\mathrm{x}}^{99}+\mathrm{cosx}>0\\ \mathrm{So}, \mathrm{ }\mathrm{f}‘\left(\mathrm{x}\right)>0\\ \mathrm{Thus},\mathrm{function}\mathrm{f}\mathrm{is}\mathrm{strictly}\mathrm{increasing}\mathrm{in}\mathrm{interval}\left(0,\frac{\mathrm{\pi }}{2}\right).\\ \mathrm{Hence},\mathrm{function}\mathrm{f}\mathrm{is}\mathrm{strictly}\mathrm{decreasing}\mathrm{in}\mathrm{none}\mathrm{of}\mathrm{the}\mathrm{intervals}\mathrm{.}\\ \mathrm{The}\mathrm{correct}\mathrm{answer}\mathrm{is}\mathrm{D}\mathrm{.}\end{array}$

Q.32 Find the least value of a such that the function f given f(x) =x2+ax+1 is strictly increasing on (1, 2).

Ans.

$\begin{array}{l}\mathrm{We}\mathrm{have},\\ \mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^{2}+\mathrm{ax}+1\\ \mathrm{Differentiating}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathrm{x},\mathrm{we}\mathrm{get}\\ \mathrm{f}‘\left(\mathrm{x}\right)=2\mathrm{x}+\mathrm{a}\\ \mathrm{Now},\mathrm{ }\mathrm{function}\mathrm{f}\mathrm{will}\mathrm{be}\mathrm{increasing}\mathrm{in}\mathrm{ }\left(1,2\right),\mathrm{if}\\ \mathrm{ }\mathrm{f}‘\left(\mathrm{x}\right)>0⇒2\mathrm{x}+\mathrm{a}>0\\ \mathrm{ }\mathrm{x}>-\frac{\mathrm{a}}{2}\\ \mathrm{Therefore},\mathrm{we}\mathrm{have}\mathrm{to}\mathrm{find}\mathrm{the}\mathrm{least}\mathrm{value}\mathrm{of}\mathrm{a}\mathrm{such}\mathrm{that}\\ \mathrm{ }\mathrm{x}>-\frac{\mathrm{a}}{2}, \mathrm{ }\mathrm{when}\mathrm{ }\mathrm{x}\in \left(1,2\right)\\ ⇒\mathrm{ }\mathrm{x}>-\frac{\mathrm{a}}{2}, \mathrm{ }\left(\mathrm{when}\mathrm{ }1<\mathrm{x}<2\right)\\ \mathrm{Thus},\mathrm{the}\mathrm{least}\mathrm{value}\mathrm{of}\mathrm{a}\mathrm{for}\mathrm{f}\mathrm{to}\mathrm{be}\mathrm{increasing}\mathrm{on}\left(1, 2\right)\mathrm{is}\\ \mathrm{given}\mathrm{by},\\ \mathrm{–}\frac{\mathrm{a}}{2}=1⇒\mathrm{a}=-2\\ \mathrm{Hence},\mathrm{the}\mathrm{required}\mathrm{value}\mathrm{of}\mathrm{a}\mathrm{is}-2.\end{array}$

Q.33 Let I be any interval disjoint from (1,1).Prove that the function fgiven byf(x)=x+1x is strictly increasing onI.

Ans.

$\begin{array}{l}\mathrm{we}\mathrm{}\mathrm{have},\\ \mathrm{f}\left(\mathrm{x}\right)=\mathrm{x}+\frac{1}{\mathrm{x}}\\ \mathrm{Differentiating}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathrm{x},\mathrm{we}\mathrm{get}\\ \mathrm{f}‘\left(\mathrm{x}\right)=1-\frac{1}{{\mathrm{x}}^{2}}\\ \mathrm{Now},\mathrm{ }\mathrm{f}‘\left(\mathrm{x}\right)=0⇒1-\frac{1}{{\mathrm{x}}^{2}}=0\\ ⇒1=\frac{1}{{\mathrm{x}}^{2}}⇒\mathrm{x}=±1\\ \mathrm{The}\mathrm{points}\mathrm{x}=1\mathrm{and}\mathrm{x}=-1\mathrm{divide}\mathrm{the}\mathrm{real}\mathrm{line}\mathrm{in}\mathrm{three}\mathrm{disjoint}\\ \mathrm{intervals}\mathrm{i}.\mathrm{e}.,\mathrm{ }\left(-\mathrm{\infty },-1\right),\left(-1,1\right) \mathrm{and} \left(1,\mathrm{\infty }\right).\\ \mathrm{In}\mathrm{interval}\left(-1, 1\right),\mathrm{it}\mathrm{is}\mathrm{noticed}\mathrm{that}:\\ -1<\mathrm{x}<1⇒{\mathrm{x}}^{2}<1\\ ⇒1<\frac{1}{{\mathrm{x}}^{2}}, \mathrm{x}\ne 0\\ ⇒1-\frac{1}{{\mathrm{x}}^{2}}<0, \mathrm{x}\ne 0\\ \mathrm{f}‘\left(\mathrm{x}\right)=1-\frac{1}{{\mathrm{x}}^{2}}<0 \mathrm{on}\mathrm{}\left(-1,1\right)~\left\{0\right\}\\ \therefore \mathrm{f}\mathrm{is}\mathrm{strictly}\mathrm{decreasing}\mathrm{on} \left(-1,1\right)~\left\{0\right\}\\ \mathrm{In}\mathrm{intervals}\mathrm{ }\left(-\mathrm{\infty },-1\right) \mathrm{and} \left(1,\mathrm{\infty }\right),\mathrm{it}\mathrm{is}\mathrm{seen}\mathrm{that}\\ \mathrm{x}<-1 \mathrm{or}\mathrm{x}>1\\ ⇒{\mathrm{x}}^{\mathrm{2}}>\mathrm{1}⇒1>\frac{1}{{\mathrm{x}}^{2}}\\ ⇒1-\frac{1}{{\mathrm{x}}^{2}}>0\\ \therefore \mathrm{f}‘\left(\mathrm{x}\right)=1-\frac{1}{{\mathrm{x}}^{2}}>0\mathrm{on}\mathrm{ }\left(-\mathrm{\infty },-1\right) \mathrm{and} \left(1,\mathrm{\infty }\right).\\ \therefore \mathrm{f}\mathrm{is}\mathrm{strictly}\mathrm{increasing}\mathrm{on}\mathrm{ }\left(-\mathrm{\infty },-1\right) \mathrm{and} \left(1,\mathrm{\infty }\right).\\ \mathrm{Hence},\mathrm{function}\mathrm{f}\mathrm{is}\mathrm{strictly}\mathrm{increasing}\mathrm{in}\mathrm{interval}\mathrm{I}\mathrm{disjoint}\mathrm{from}\left(-1, 1\right)\mathrm{.}\end{array}$

Q.34 Provethatthefunctionfgivenbyf(x) =logsinxisstrictly increasingon (0,π2)andstrictlydecreasingon(π2,π).

Ans.

$\begin{array}{l}\mathrm{We}\mathrm{have},\\ \mathrm{f}\left(\mathrm{x}\right)=\mathrm{logsinx}\\ \mathrm{Differentiating}\mathrm{both}\mathrm{sides}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathrm{x},\mathrm{ }\mathrm{we}\mathrm{get}\\ \mathrm{f}‘\left(\mathrm{x}\right)=\frac{1}{\mathrm{sinx}}\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{sinx}\left[\mathrm{By}\mathrm{ }\mathrm{chain}\mathrm{rule}\right]\\ \mathrm{ }=\frac{1}{\mathrm{sinx}}\mathrm{cosx}=\mathrm{cotx}\\ \mathrm{f}‘\left(\mathrm{x}\right)=\mathrm{cotx}>0 \mathrm{ }\mathrm{in}\mathrm{interval}\left(0,\frac{\mathrm{\pi }}{2}\right)\\ \therefore \mathrm{f}\left(\mathrm{x}\right)\mathrm{is}\mathrm{strictly}\mathrm{increasing}\mathrm{function}\mathrm{in}\mathrm{interval}\left(0,\frac{\mathrm{\pi }}{2}\right).\\ \mathrm{f}‘\left(\mathrm{x}\right)=\mathrm{cotx}<0 \mathrm{ }\mathrm{in}\mathrm{interval}\left(\frac{\mathrm{\pi }}{2},\mathrm{\pi }\right)\\ \therefore \mathrm{f}\left(\mathrm{x}\right)\mathrm{is}\mathrm{strictly}\mathrm{decreasing}\mathrm{function}\mathrm{in}\mathrm{interval}\left(\frac{\mathrm{\pi }}{2},\mathrm{\pi }\right).\end{array}$

Q.35

$\begin{array}{l}\text{Prove that the function f given by f(x) = logcos x is strictly}\\ \text{decreasing on}\left(0,\frac{\pi }{2}\right)\phantom{\rule{thinmathspace}{0ex}}\text{and strictly increasing on}\phantom{\rule{thinmathspace}{0ex}}\left(\frac{\pi }{2},\pi \right).\end{array}$

Ans.

$\begin{array}{l}\mathrm{We}\mathrm{have},\\ \mathrm{f}\left(\mathrm{x}\right)=\mathrm{logcosx}\\ \mathrm{Differentiating}\mathrm{both}\mathrm{sides}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathrm{x},\mathrm{ }\mathrm{we}\mathrm{get}\\ \mathrm{f}‘\left(\mathrm{x}\right)=\frac{1}{\mathrm{cosx}}\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{cosx}\left[\mathrm{By}\mathrm{ }\mathrm{chain}\mathrm{rule}\right]\\ \mathrm{ }=\frac{1}{\mathrm{cosx}}×-\mathrm{sinx}=-\mathrm{tanx}\\ \mathrm{f}‘\left(\mathrm{x}\right)=-\mathrm{tanx}<0 \mathrm{ }\mathrm{in}\mathrm{interval}\left(0,\frac{\mathrm{\pi }}{2}\right)\\ \therefore \mathrm{f}\left(\mathrm{x}\right)\mathrm{is}\mathrm{strictly}\mathrm{decreasing}\mathrm{function}\mathrm{in}\mathrm{interval}\left(0,\frac{\mathrm{\pi }}{2}\right).\\ \mathrm{f}‘\left(\mathrm{x}\right)=-\mathrm{tanx}>0 \mathrm{ }\mathrm{in}\mathrm{interval}\left(\frac{\mathrm{\pi }}{2},\mathrm{\pi }\right)\\ \therefore \mathrm{f}\left(\mathrm{x}\right)\mathrm{is}\mathrm{strictly}\mathrm{increasing}\mathrm{function}\mathrm{in}\mathrm{interval}\left(\frac{\mathrm{\pi }}{2},\mathrm{\pi }\right).\end{array}$

Q.36 Prove that the function f given byf(x) = x23x2+3x100is increasinginR.

Ans.

$\begin{array}{l}\mathrm{We}\mathrm{have},\\ \mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^{3}-3{\mathrm{x}}^{2}+3\mathrm{x}-100\\ \mathrm{Differentiating}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathrm{x},\mathrm{we}\mathrm{get}\\ \mathrm{f}‘\left(\mathrm{x}\right)=3{\mathrm{x}}^{2}-6\mathrm{x}+3\\ =3\left({\mathrm{x}}^{2}-2\mathrm{x}+1\right)\\ =3{\left(\mathrm{x}-1\right)}^{2}\\ \mathrm{For}\mathrm{any}\mathrm{x}\in \mathbf{R}, {\left(\mathrm{x}-1\right)}^{2}>0.\\ \mathrm{Thus},\mathrm{f}‘\left(\mathrm{x}\right) \mathrm{is}\mathrm{always}\mathrm{positive}\mathrm{in}\mathbf{R}\mathrm{.}\\ \mathrm{Hence},\mathrm{the}\mathrm{given}\mathrm{function}\mathrm{f}\left(\mathrm{x}\right)\mathrm{is}\mathrm{increasing}\mathrm{in}\mathbf{R}\mathrm{.}\end{array}$

Q.37 The interval in which y=x2exis increasing is(A) (, ) (B )(2, 0) (C) (2, ) (D) (0, 2)

Ans.

$\begin{array}{l}\mathrm{We}\mathrm{have},\\ \mathrm{y}={\mathrm{x}}^{2}{\mathrm{e}}^{-\mathrm{x}}\\ \mathrm{Differentiating}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathrm{x},\mathrm{we}\mathrm{get}\\ \frac{\mathrm{dy}}{\mathrm{dx}}={\mathrm{x}}^{2}\frac{\mathrm{d}}{\mathrm{dx}}{\mathrm{e}}^{-\mathrm{x}}+{\mathrm{e}}^{-\mathrm{x}}\frac{\mathrm{d}}{\mathrm{dx}}{\mathrm{x}}^{2}\\ \mathrm{ }=-{\mathrm{x}}^{2}{\mathrm{e}}^{-\mathrm{x}}+{\mathrm{e}}^{-\mathrm{x}}\left(2\mathrm{x}\right)\\ \mathrm{ }={\mathrm{e}}^{-\mathrm{x}}\left(-{\mathrm{x}}^{2}+2\mathrm{x}\right)\\ \mathrm{Now},\mathrm{ }\frac{\mathrm{dy}}{\mathrm{dx}}=0\\ ⇒{\mathrm{e}}^{-\mathrm{x}}\left(-{\mathrm{x}}^{2}+2\mathrm{x}\right)=0\\ ⇒\mathrm{x}=0,2\left[\because {\mathrm{e}}^{-\mathrm{x}}\ne 0\right]\\ \mathrm{The}\mathrm{points}\mathrm{x}=0\mathrm{and}\mathrm{x}=2\mathrm{divide}\mathrm{the}\mathrm{real}\mathrm{line}\mathrm{into}\mathrm{three}\mathrm{disjoint}\\ \mathrm{intervals}\mathrm{i}.\mathrm{e}.,\left(-\mathrm{\infty },0\right),\left(0,2\right)\mathrm{ }\mathrm{and} \left(2,\mathrm{\infty }\right).\\ \mathrm{f}‘\left(\mathrm{x}\right)<0\mathrm{in}\mathrm{intervals}\left(-\mathrm{\infty },0\right)\mathrm{ }\mathrm{and}\mathrm{ }\left(2,\mathrm{\infty }\right).\\ \mathrm{So},\mathrm{ }\mathrm{f}\left(\mathrm{x}\right)\mathrm{}\mathrm{is}\mathrm{decreasing}\mathrm{in}\mathrm{intervals}\left(-\mathrm{\infty },0\right)\mathrm{ }\mathrm{and}\mathrm{ }\left(2,\mathrm{\infty }\right).\\ \mathrm{Now},\mathrm{ }\mathrm{f}‘\left(\mathrm{x}\right)>0\mathrm{in}\mathrm{interval}\left(0,2\right)\\ \mathrm{So},\mathrm{ }\mathrm{f}\left(\mathrm{x}\right)\mathrm{}\mathrm{is}\mathrm{increasing}\mathrm{in}\mathrm{intervals}\left(0,2\right)\mathrm{ }\\ \mathrm{Hence},\mathrm{f}\mathrm{is}\mathrm{strictly}\mathrm{increasing}\mathrm{in}\mathrm{interval}\left(0, 2\right)\mathrm{.}\\ \mathrm{The}\mathrm{correct}\mathrm{answer}\mathrm{is}\mathrm{D}\mathrm{.}\end{array}$

Q.38 Find the slope of the tangent to the curve y = 3x4 − 4x at x = 4

Ans.

$\begin{array}{l}\mathrm{The}\mathrm{given}\mathrm{curve}\mathrm{is}\mathrm{y}=3{\mathrm{x}}^{\mathrm{4}}-4\mathrm{x}\mathrm{.}\\ \mathrm{Differentiating}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathrm{x},\mathrm{we}\mathrm{get}\\ \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\left(3{\mathrm{x}}^{\mathrm{4}}-4\mathrm{x}\right)\\ \mathrm{ }=12{\mathrm{x}}^{3}-4\\ \mathrm{Then},\mathrm{the}\mathrm{slope}\mathrm{of}\mathrm{the}\mathrm{tangent}\mathrm{to}\mathrm{the}\mathrm{given}\mathrm{curve}\mathrm{at}\mathrm{x}=4\mathrm{is}\\ \mathrm{given}\mathrm{by},\\ {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}_{\mathrm{x}=4}={\left(12{\mathrm{x}}^{3}-4\right)}_{\mathrm{x}=4}\\ \mathrm{ }=12{\left(4\right)}^{3}-4\\ \mathrm{ }=768-4\\ \mathrm{ }=764\\ \mathrm{Thus}, \mathrm{}\mathrm{the}\mathrm{slope}\mathrm{of}\mathrm{tangent}\mathrm{to}\mathrm{the}\mathrm{given}\mathrm{curve}\mathrm{is}764\mathrm{.}\end{array}$

Q.39

$\begin{array}{l}\text{Find the slope of the tangent to the curve y =}\frac{\mathrm{x}-1}{\mathrm{x}-2},\phantom{\rule{thinmathspace}{0ex}}\mathrm{x}\ne 2\\ \mathrm{a}\mathrm{t}\mathrm{x}=10.\end{array}$

Ans.

$\begin{array}{l}\mathrm{The}\mathrm{given}\mathrm{curve}\mathrm{is}\mathrm{ }\mathrm{y}=\frac{\mathrm{x}-1}{\mathrm{x}-2}\\ \mathrm{Differentiating}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathrm{x},\mathrm{we}\mathrm{get}\\ \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\mathrm{x}-1}{\mathrm{x}-2}\right)\\ \mathrm{ }=\frac{\left(\mathrm{x}-2\right)\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}-1\right)-\left(\mathrm{x}-1\right)\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}-2\right)}{{\left(\mathrm{x}-2\right)}^{2}}\\ \mathrm{ }=\frac{\left(\mathrm{x}-2\right)\left(1-0\right)-\left(\mathrm{x}-1\right)\left(1-0\right)}{{\left(\mathrm{x}-2\right)}^{2}}\\ \mathrm{ }=\frac{\mathrm{x}-2-\mathrm{x}+1}{{\left(\mathrm{x}-2\right)}^{2}}\\ \mathrm{ }=\frac{-1}{{\left(\mathrm{x}-2\right)}^{2}}\\ \mathrm{Thus},\mathrm{the}\mathrm{slope}\mathrm{of}\mathrm{the}\mathrm{tangent}\mathrm{at}\mathrm{x}= 10\mathrm{is}\mathrm{given}\mathrm{by},\\ {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}_{\mathrm{x}=10}={\left(\frac{-1}{{\left(\mathrm{x}-2\right)}^{2}}\right)}_{\mathrm{x}=10}\\ \mathrm{ }=\frac{-1}{{\left(10-2\right)}^{2}}\\ \mathrm{ }=\frac{-1}{64}\\ \mathrm{Thus}, \mathrm{}\mathrm{the}\mathrm{slope}\mathrm{of}\mathrm{tangent}\mathrm{to}\mathrm{the}\mathrm{given}\mathrm{curve}\mathrm{is}\frac{-1}{64}\mathrm{.}\end{array}$

Q.40 Find the slope of the tangent to curve y = x3−x+1 at the point whose x-coordinate is 2.

Ans.

$\begin{array}{l}\mathrm{The}\phantom{\rule{0ex}{0ex}}\mathrm{given}\phantom{\rule{0ex}{0ex}}\mathrm{curve}\phantom{\rule{0ex}{0ex}}\mathrm{is}\\ \phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{y}={\mathrm{x}}^{3}-\mathrm{x}+1\\ \mathrm{Differentiating}\phantom{\rule{0ex}{0ex}}\mathrm{w}.\mathrm{r}.\mathrm{t}.\phantom{\rule{0ex}{0ex}}\mathrm{x},\phantom{\rule{0ex}{0ex}}\mathrm{we}\phantom{\rule{0ex}{0ex}}\mathrm{get}\\ \phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\left({\mathrm{x}}^{3}-\mathrm{x}+1\right)\\ \phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=3{\mathrm{x}}^{2}-1\\ \mathrm{The}\phantom{\rule{0ex}{0ex}}\mathrm{slope}\mathrm{of}\phantom{\rule{0ex}{0ex}}\mathrm{tangent}\phantom{\rule{0ex}{0ex}}\mathrm{to}\phantom{\rule{0ex}{0ex}}\mathrm{the}\phantom{\rule{0ex}{0ex}}\mathrm{curve}\phantom{\rule{0ex}{0ex}}\mathrm{is}\\ {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}_{\mathrm{x}=2}=3{\left(2\right)}^{2}-1\\ \phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=12-1=11\\ \mathrm{Thus},\phantom{\rule{0ex}{0ex}}\mathrm{the}\phantom{\rule{0ex}{0ex}}\mathrm{slope}\phantom{\rule{0ex}{0ex}}\mathrm{of}\phantom{\rule{0ex}{0ex}}\mathrm{the}\phantom{\rule{0ex}{0ex}}\mathrm{tangent}\phantom{\rule{0ex}{0ex}}\mathrm{at}\phantom{\rule{0ex}{0ex}}\mathrm{the}\phantom{\rule{0ex}{0ex}}\mathrm{point}\mathrm{where}\phantom{\rule{0ex}{0ex}}\mathrm{the}\\ \mathrm{x}-\mathrm{coordinate}\phantom{\rule{0ex}{0ex}}\mathrm{is}\phantom{\rule{0ex}{0ex}}2\phantom{\rule{0ex}{0ex}}\mathrm{is}\phantom{\rule{0ex}{0ex}}11.\end{array}$

Q.41 Find the slope of the tangent to curve y = x3−3x+2 at the point whose x-coordinate
is 3.

Ans.

$\begin{array}{l}\mathrm{The}\mathrm{given}\mathrm{curve}\mathrm{is}\\ \mathrm{y}={\mathrm{x}}^{3}-3\mathrm{x}+2\\ \mathrm{Differentiating}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathrm{x},\mathrm{we}\mathrm{get}\\ \mathrm{ }\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\left({\mathrm{x}}^{3}-3\mathrm{x}+2\right)\\ =3{\mathrm{x}}^{2}-3\\ \mathrm{The}\mathrm{slope}\mathrm{of}\mathrm{tangent}\mathrm{to}\mathrm{the}\mathrm{curve}\mathrm{is}\\ {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}_{\mathrm{x}=3}=3{\left(3\right)}^{\mathrm{2}}-\mathrm{3}\\ \mathrm{ }=\mathrm{27}-\mathrm{3}=\mathrm{24}\\ \mathrm{Thus},\mathrm{the}\mathrm{slope}\mathrm{of}\mathrm{the}\mathrm{tangent}\mathrm{at}\mathrm{the}\mathrm{point}\mathrm{where}\mathrm{the}\\ \mathrm{x}–\mathrm{coordinate}\mathrm{is}3\mathrm{is}24\mathrm{.}\end{array}$

Q.42 Find the slope of the normal to the curve x =acos3θ, y=asin3θat θ=π4.

Ans.

$\begin{array}{l}\mathrm{Given}\mathrm{that}\mathrm{x}={\mathrm{acos}}^{\mathrm{3}}\mathrm{\theta }\mathrm{and}\mathrm{y}={\mathrm{asin}}^{\mathrm{3}}\mathrm{\theta }\\ \mathrm{Differentiating}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathrm{\theta },\mathrm{we}\mathrm{get}\\ \frac{\mathrm{dx}}{\mathrm{d\theta }}=\mathrm{a}\frac{\mathrm{d}}{\mathrm{d\theta }}{\mathrm{cos}}^{\mathrm{3}}\mathrm{\theta }\\ =3{\mathrm{acos}}^{2}\mathrm{\theta }\frac{\mathrm{d}}{\mathrm{d\theta }}\mathrm{cos\theta }\mathrm{ }=-3{\mathrm{acos}}^{2}\mathrm{\theta sin\theta }\\ \frac{\mathrm{dy}}{\mathrm{d\theta }}=\mathrm{a}\frac{\mathrm{d}}{\mathrm{d\theta }}{\mathrm{sin}}^{\mathrm{3}}\mathrm{\theta }\\ =3{\mathrm{asin}}^{2}\mathrm{\theta }\frac{\mathrm{d}}{\mathrm{d\theta }}\mathrm{sin\theta }\\ =3{\mathrm{asin}}^{2}\mathrm{\theta cos\theta }\\ \therefore \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\left(\frac{\mathrm{dy}}{\mathrm{d\theta }}\right)}{\left(\frac{\mathrm{dx}}{\mathrm{d\theta }}\right)}\\ =\frac{3{\mathrm{asin}}^{2}\mathrm{\theta cos\theta }}{-3{\mathrm{acos}}^{2}\mathrm{\theta sin\theta }}\mathrm{ }=-\mathrm{tan\theta }\\ \mathrm{Therefore},\mathrm{the}\mathrm{slope}\mathrm{of}\mathrm{the}\mathrm{tangent}\mathrm{at} \mathrm{\theta }=\frac{\mathrm{\pi }}{4},\mathrm{ }\mathrm{is}\\ {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}_{\left(\mathrm{\theta }=\frac{\mathrm{\pi }}{4}\right)}={\left(-\mathrm{tan\theta }\right)}_{\left(\mathrm{\theta }=\frac{\mathrm{\pi }}{4}\right)}\\ =-\mathrm{tan}\frac{\mathrm{\pi }}{4}\\ =-1\\ \mathrm{Hence},\mathrm{the}\mathrm{slope}\mathrm{of}\mathrm{the}\mathrm{normal}\mathrm{at}\mathrm{ }\mathrm{\theta }=\frac{\mathrm{\pi }}{4},\mathrm{ }\mathrm{is}\\ \mathrm{slope}\mathrm{of}\mathrm{normal}\left(\mathrm{M}\right)=-\frac{1}{\mathrm{slope}\mathrm{of}\mathrm{tangent}\left(\mathrm{m}\right)}\\ \mathrm{ }=-\frac{1}{-1}\\ =1\end{array}$

Q.43

$\begin{array}{l}\text{Find the slope of the normal to the curve x = 1}-\mathrm{a}\phantom{\rule{thickmathspace}{0ex}}\mathrm{s}\mathrm{i}\mathrm{n}\theta ,\\ \mathrm{y}=\mathrm{b}\mathrm{c}\mathrm{o}{\mathrm{s}}^{2}\theta \phantom{\rule{thickmathspace}{0ex}}\mathrm{a}\mathrm{t}\phantom{\rule{thickmathspace}{0ex}}\theta =\frac{\pi }{2}\end{array}$

Ans.

$\begin{array}{l}\mathrm{Given}\mathrm{that}:\\ \mathrm{x}=1-\mathrm{asin\theta }\mathrm{and}\mathrm{y}={\mathrm{bcos}}^{2}\mathrm{\theta }\\ \mathrm{Differentiating}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathrm{\theta },\mathrm{we}\mathrm{get}\\ \frac{\mathrm{dx}}{\mathrm{d\theta }}=\frac{\mathrm{d}}{\mathrm{d\theta }}\left(1-\mathrm{asin\theta }\right)\\ \mathrm{ }=0-\mathrm{acos\theta }\\ \mathrm{ }=-\mathrm{acos\theta }\\ \mathrm{ }\frac{\mathrm{dy}}{\mathrm{d\theta }}=\frac{\mathrm{d}}{\mathrm{d\theta }}\left({\mathrm{bcos}}^{2}\mathrm{\theta }\right)\\ =2\mathrm{bcos\theta }\frac{\mathrm{d}}{\mathrm{d\theta }}\mathrm{cos\theta }\\ =-2\mathrm{bcos\theta sin\theta }\\ \therefore \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\left(\frac{\mathrm{dy}}{\mathrm{d\theta }}\right)}{\left(\frac{\mathrm{dx}}{\mathrm{d\theta }}\right)}\\ =\frac{-2\mathrm{bcos\theta sin\theta }}{-\mathrm{acos\theta }}\\ =\frac{2\mathrm{b}}{\mathrm{a}}\mathrm{sin\theta }\\ \mathrm{Slope}\mathrm{of}\mathrm{tangent}\mathrm{at}\mathrm{\theta }=\frac{\mathrm{\pi }}{2},\mathrm{ }\mathrm{is}\mathrm{given}\mathrm{by}\\ \mathrm{ }\mathrm{m}={\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}_{\mathrm{\theta }=\frac{\mathrm{\pi }}{2}}\\ =\frac{2\mathrm{b}}{\mathrm{a}}\mathrm{sin}\frac{\mathrm{\pi }}{2}=\frac{2\mathrm{b}}{\mathrm{a}}\\ \mathrm{slope}\mathrm{of}\mathrm{normal}\\ \left(\mathrm{M}\right)=-\frac{1}{\mathrm{m}}\\ =-\frac{1}{\left(\frac{2\mathrm{b}}{\mathrm{a}}\right)}\\ =-\frac{\mathrm{a}}{2\mathrm{b}}\\ \mathrm{Thus},\mathrm{slope}\mathrm{of}\mathrm{normal}\mathrm{is}-\frac{\mathrm{a}}{2\mathrm{b}}.\end{array}$

Q.44 Find points at which the tangent to the curve
y = x3 − 3x2 − 9x + 7 is parallel to the x axis.

Ans.

$\begin{array}{l}\text{The equation of the given curve is}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}y={\text{x}}^{\text{3}}-{\text{3x}}^{\text{2}}-\text{9x + 7}\\ \text{Differentiating w}\text{.r}\text{.t}\text{. x, we get}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{dy}{dx}=\frac{d}{dx}\left({\text{x}}^{\text{3}}-{\text{3x}}^{\text{2}}-\text{9x + 7}\right)\\ \text{}=3{x}^{2}-6x-9\\ \text{Now, the tangent is parallel to the x-axis if the slope of the}\\ \text{tangent is zero}\text{.}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}3{x}^{2}-6x-9=0⇒3\left({x}^{2}-2x-3\right)=0\\ ⇒3\left(x-3\right)\left(x+1\right)=0\\ ⇒x=3,-1\\ When\text{x}=\text{3, then}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{y}={\left(3\right)}^{\text{3}}-\text{3}{\left(3\right)}^{\text{2}}-\text{9}\left(3\right)\text{+ 7}\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=27-27-27+7\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=-20\\ When\text{x}=-\text{1, then}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{y}={\left(-\text{1}\right)}^{\text{3}}-\text{3}{\left(-\text{1}\right)}^{\text{2}}-\text{9}\left(-\text{1}\right)\text{+ 7}\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=-\text{1}-3+9+7\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=12\\ \text{Hence, the points at which the tangent is parallel to the x-axis}\\ \text{are}\left(\text{3,}-\text{20}\right)\text{and}\left(-\text{1, 12}\right)\text{.}\end{array}$

Q.45 Find a point on the curve y = (x − 2)2 at which the tangent is parallel to the chord joining the
points (2, 0) and (4, 4).

Ans.

$\begin{array}{l}Given\text{curve is}\\ \text{}\text{}\text{y}={\left(x-2\right)}^{2}\\ Differentiating\text{both sides, w}\text{.r}\text{.t}\text{. x, we get}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{dy}{dx}=\frac{d}{dx}{\left(x-2\right)}^{2}\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=2\left(x-2\right)\frac{d}{dx}\left(x-2\right)\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=2\left(x-2\right)\\ \text{The slope of the chord}\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{4-0}{4-2}\text{}\text{}\text{}\text{}\left[\because m=\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}\right]\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=2\\ \text{Since,}\\ \text{the slope of the tangent}=\text{slope of the chord, we have:}\\ \text{}\text{}\text{}\text{}\text{2}\left(x-2\right)=2\\ ⇒\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}x=3\\ When\text{x}=\text{3, y}={\left(3-2\right)}^{2}=1\\ \text{Hence, the required point is}\left(\text{3, 1}\right)\text{.}\end{array}$

Q.46 Find the point on the curve y = x3 − 11x + 5
at which the tangent is y = x − 11.

Ans.

$\begin{array}{l}\text{The equation of the given curve is}\\ \text{}y={x}^{3}-11x+5\\ \text{The equation of the tangent to the given curve is}\\ \text{y = x}-\text{11}\\ \text{comparing it with y = mx + c, we get}\\ \text{m}=\text{1}\\ \therefore \text{slope of the tangent}\left(m\right)=1\\ \text{Now, the slope of the tangent to the given curve at the}\\ \text{point (x, y) is,}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{dy}{dx}=\frac{d}{dx}\left({x}^{3}-11x+5\right)\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=3{x}^{2}-11\\ \because 3{x}^{2}-11=1\text{}\text{}\text{}\text{}\text{}\text{}\text{}\left[Given\right]\\ ⇒\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{x}^{2}=\frac{12}{3}=4\\ ⇒\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}x=±2\\ \text{When}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{x}=\text{2,}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{y}={\text{(2)}}^{\text{3}}-\text{11(2) + 5}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\text{8}-\text{22 + 5}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=-\text{9}\text{.}\\ \text{When x}=-\text{2,}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{y}=\text{(}-\text{2)3}-\text{11 (}-\text{2) + 5}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=-\text{8 + 22 + 5}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\text{19}\text{.}\\ \text{Hence, the required points are}\left(\text{2,}-\text{9}\right)\text{and}\left(-\text{2, 19}\right)\text{.}\end{array}$

Q.47

$\begin{array}{l}\mathrm{Find}\phantom{\rule{0ex}{0ex}}\mathrm{the}\phantom{\rule{0ex}{0ex}}\mathrm{equation}\phantom{\rule{0ex}{0ex}}\mathrm{of}\phantom{\rule{0ex}{0ex}}\mathrm{all}\phantom{\rule{0ex}{0ex}}\mathrm{lines}\phantom{\rule{0ex}{0ex}}\mathrm{having}\phantom{\rule{0ex}{0ex}}\mathrm{slope}\phantom{\rule{0ex}{0ex}}-1\phantom{\rule{0ex}{0ex}}\mathrm{that}\phantom{\rule{0ex}{0ex}}\mathrm{are}\\ \mathrm{tangents}\phantom{\rule{0ex}{0ex}}\mathrm{to}\phantom{\rule{0ex}{0ex}}\mathrm{the}\phantom{\rule{0ex}{0ex}}\mathrm{curve}\phantom{\rule{0ex}{0ex}}\mathrm{y}\phantom{\rule{0ex}{0ex}}=\frac{1}{\mathrm{x}-1},\phantom{\rule{0ex}{0ex}}\mathrm{x}\phantom{\rule{0ex}{0ex}}\ne \phantom{\rule{0ex}{0ex}}1.\end{array}$

Ans.

$\begin{array}{l}\mathrm{Equation}\mathrm{of}\mathrm{given}\mathrm{curve}\mathrm{is}\mathrm{y}=\frac{1}{\mathrm{x}-1},\mathrm{x}\ne 1\\ \mathrm{Differentiating}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathrm{x},\mathrm{we}\mathrm{get}\\ \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\frac{1}{\mathrm{x}-1}\\ \mathrm{ }=-\frac{1}{{\left(\mathrm{x}-1\right)}^{2}}\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}-1\right)\\ \mathrm{ }=-\frac{1}{{\left(\mathrm{x}-1\right)}^{2}}\\ \mathrm{So},\mathrm{ }\mathrm{m}={\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}_{\left(\mathrm{x},\mathrm{y}\right)}\\ \mathrm{ }=-\frac{1}{{\left(\mathrm{x}-1\right)}^{2}}\\ \mathrm{According} \mathrm{to}\mathrm{question},\\ -\frac{1}{{\left(\mathrm{x}-1\right)}^{2}}=-1⇒{\left(\mathrm{x}-1\right)}^{2}=1\\ ⇒\mathrm{x}-1=±1\\ ⇒\mathrm{x}-1=1 \mathrm{or} \mathrm{x}-1=-1\\ ⇒\mathrm{x}=2,0\\ \mathrm{When}\mathrm{x}=0,\mathrm{y}=-\mathrm{1}\\ \mathrm{When}\mathrm{ }\mathrm{x}=2, \mathrm{y}=1\\ \mathrm{Thus},\mathrm{there}\mathrm{are}\mathrm{two}\mathrm{tangents}\mathrm{to}\mathrm{the}\mathrm{given}\mathrm{curve}\mathrm{having}\mathrm{slope}\\ -\mathrm{1}.\mathrm{These}\mathrm{are}\mathrm{passing}\mathrm{through}\mathrm{the}\mathrm{points}\left(0, -1\right)\mathrm{and}\left(2, 1\right)\mathrm{.}\end{array}$ $\begin{array}{l}\text{And equation of two tangents are}\\ \mathrm{y}-1=-1\left(\mathrm{x}-2\right)\\ \mathrm{y}-1=-\mathrm{x}+2\\ \mathrm{x}+\mathrm{y}-3=0\\ \mathrm{A}\mathrm{n}\mathrm{d},\mathrm{y}+1=-1\left(\mathrm{x}-0\right)\\ \mathrm{y}+1=-\mathrm{x}\\ \mathrm{x}+\mathrm{y}+1=0\end{array}$

Q.48

$\begin{array}{l}\text{Find the equation of all lines having slope 2 which are}\\ \text{tangents to the curve y =}\frac{1}{\mathrm{x}-3},\phantom{\rule{thinmathspace}{0ex}}\mathrm{x}\ne 3.\end{array}$

Ans.

$\begin{array}{l}\mathrm{Equation}\mathrm{of}\mathrm{given}\mathrm{curve}\mathrm{is}\mathrm{y}=\frac{1}{\mathrm{x}-3},\mathrm{x}\ne 3\\ \mathrm{Differentiating}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathrm{x},\mathrm{we}\mathrm{get}\\ \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{1}{\mathrm{x}-3}\right)\\ \mathrm{ }=-\frac{1}{{\left(\mathrm{x}-3\right)}^{2}}\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}-3\right)\\ \mathrm{ }=-\frac{1}{{\left(\mathrm{x}-3\right)}^{2}}\\ \mathrm{So},\mathrm{ }\mathrm{m}={\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}_{\left(\mathrm{x},\mathrm{y}\right)}\\ \mathrm{ }=-\frac{1}{{\left(\mathrm{x}-3\right)}^{2}}\\ \mathrm{According} \mathrm{to}\mathrm{question},\\ \mathrm{ }-\frac{1}{{\left(\mathrm{x}-3\right)}^{2}}=2\\ ⇒{\left(\mathrm{x}-3\right)}^{2}=-2\\ \mathrm{ }\mathrm{This}\mathrm{is}\mathrm{impossible}\mathrm{that}\mathrm{square}\mathrm{of}\mathrm{a}\mathrm{number}\mathrm{is}\mathrm{negative}\mathrm{.}\\ \mathrm{Hence},\mathrm{there}\mathrm{is}\mathrm{no}\mathrm{tangent}\mathrm{to}\mathrm{the}\mathrm{given}\mathrm{curve}\mathrm{having}\mathrm{slope}2\mathrm{.}\end{array}$

Q.49

$\begin{array}{l}\text{Find the equations of all lines having slope 0 which are tangent}\\ \text{to the curve y}\phantom{\rule{thickmathspace}{0ex}}=\phantom{\rule{thickmathspace}{0ex}}\frac{1}{{\mathrm{x}}^{2}-2\mathrm{x}+3}\end{array}$

Ans.

$\begin{array}{l}\text{The equation of the given curve is}\text{\hspace{0.17em}}\text{\hspace{0.17em}}y=\frac{1}{{x}^{2}-2x+3}\\ \text{The slope of the tangent to the given curve at any point}\left(\text{x, y}\right)\\ {\left(\frac{dy}{dx}\right)}_{\left(x,y\right)}=\frac{d}{dx}\frac{1}{{x}^{2}-2x+3}\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=-\frac{1}{{\left({x}^{2}-2x+3\right)}^{2}}\frac{d}{dx}\left({x}^{2}-2x+3\right)\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=-\frac{\left(2x-2\right)}{{\left({x}^{2}-2x+3\right)}^{2}}\\ \text{Since, the slope of the tangent is 0, then we have:}\\ -\frac{\left(2x-2\right)}{{\left({x}^{2}-2x+3\right)}^{2}}=0⇒\left(2x-2\right)=0\\ ⇒x=1\\ When\text{\hspace{0.17em}}\text{x}=\text{1, y}=\frac{1}{1-2+3}=\frac{1}{2}\\ The\text{point on the curve is}\left(1,\frac{1}{2}\right).\\ \therefore \text{The equation of the tangent through}\text{\hspace{0.17em}}\left(1,\frac{1}{2}\right)\text{\hspace{0.17em}}is\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}y-\frac{1}{2}=0\left(x-1\right)\\ \text{}y=\frac{1}{2}\\ \text{Hence, the equation of the required line is}\text{\hspace{0.17em}}y=\frac{1}{2}.\end{array}$

Q.50

$\begin{array}{l}\mathrm{Findpointsonthe}curve\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\frac{{\mathrm{x}}^{2}}{9}+\frac{{\mathrm{y}}^{2}}{16}= 1\mathrm{atwhichthe}tangents\\ \mathrm{are}\phantom{\rule{0ex}{0ex}}\left(\mathrm{i}\right)\phantom{\rule{0ex}{0ex}}\mathrm{paralleltox}–\mathrm{axis}\phantom{\rule{0ex}{0ex}}\left(\mathrm{ii}\right)\mathrm{paralleltoy}–\mathrm{axis}.\end{array}$

Ans.

$\begin{array}{l}\mathrm{The}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{curve}\mathrm{is}\mathrm{ }\frac{{\mathrm{x}}^{2}}{9}+\frac{{\mathrm{y}}^{2}}{16}=1\dots \left(\mathrm{i}\right)\\ \mathrm{On}\mathrm{differentiating}\mathrm{both}\mathrm{sides}\mathrm{with}\mathrm{respect}\mathrm{to}\mathrm{x},\mathrm{we}\mathrm{have}:\\ \frac{2\mathrm{x}}{9}+\frac{2\mathrm{y}}{16}\frac{\mathrm{dy}}{\mathrm{dx}}=0\\ ⇒\frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{2\mathrm{x}}{9}×\frac{16}{2\mathrm{y}}=-\frac{16}{9}\frac{\mathrm{x}}{\mathrm{y}}\\ \left(\mathrm{i}\right)\mathrm{When}\mathrm{the}\mathrm{tangent}\mathrm{is}\mathrm{parallel}\mathrm{to}\mathrm{x}–\mathrm{axis},\mathrm{then}\\ \mathrm{slope}\mathrm{of}\mathrm{tangent}\left(\mathrm{m}\right)=0\\ \frac{\mathrm{dy}}{\mathrm{dx}}=0⇒-\frac{16}{9}\frac{\mathrm{x}}{\mathrm{y}}=0⇒\mathrm{x}=0\\ \mathrm{Putting} \mathrm{ }\mathrm{value}\mathrm{of}\mathrm{x}\mathrm{in}\mathrm{equation}\left(\mathrm{i}\right),\mathrm{ }\mathrm{we} \mathrm{have}\\ \frac{{\left(0\right)}^{2}}{9}+\frac{{\mathrm{y}}^{2}}{16}=1⇒{\mathrm{y}}^{2}=16⇒ \mathrm{y}=±4\\ \mathrm{Hence},\mathrm{the}\mathrm{points}\mathrm{at}\mathrm{which}\mathrm{the}\mathrm{tangents}\mathrm{are}\mathrm{parallel}\mathrm{to}\mathrm{the}\\ \mathrm{x}–\mathrm{axis}\mathrm{are}\left(0, 4\right)\mathrm{and}\left(0,-4\right)\mathrm{.}\\ \left(\mathrm{b}\right)\mathrm{The}\mathrm{tangent}\mathrm{is}\mathrm{parallel}\mathrm{to}\mathrm{the}\mathrm{y}–\mathrm{axis}\mathrm{if}\mathrm{the}\mathrm{slope}\mathrm{of}\mathrm{the}\\ \mathrm{normal}\left(\mathrm{M}\right)\mathrm{is}0,\mathrm{i}.\mathrm{e}.,\mathrm{M}=0\\ \mathrm{So}, \mathrm{ }\frac{1}{-\frac{16}{9}\frac{\mathrm{x}}{\mathrm{y}}}=0 \left[\because \mathrm{M}=\frac{1}{\mathrm{m}}\right]\\ ⇒\frac{\mathrm{y}}{\mathrm{x}}=0⇒\mathrm{y}=0\\ \mathrm{Putting} \mathrm{ }\mathrm{value}\mathrm{of}\mathrm{y}\mathrm{in}\mathrm{equation}\left(\mathrm{i}\right),\mathrm{ }\mathrm{we} \mathrm{have}\\ \frac{{\mathrm{x}}^{2}}{9}+\frac{{\left(0\right)}^{2}}{16}=1⇒{\mathrm{x}}^{2}=9⇒\mathrm{ }\mathrm{x}=±3\\ \mathrm{Hence},\mathrm{the}\mathrm{points}\mathrm{at}\mathrm{which}\mathrm{the}\mathrm{tangents}\mathrm{are}\mathrm{parallel}\mathrm{to}\mathrm{the}\mathrm{y}–\mathrm{axis}\\ \mathrm{are}\left(3, 0\right)\mathrm{and}\left(- 3, 0\right)\mathrm{.}\end{array}$

Q.51 Find the equations of the tangent and normal to the given curves at the indicated points:

(i) y = x4 − 6x3+13x2−10x + 5 at (0, 5)

(ii) y = x4 − 6x3+13x2−10x + 5 at (1, 3)

(iii) y = x3 at (1, 1)

(iv) y = x2 at (0, 0)

$\mathrm{\left(v\right) x}=cost,y=sintat\text{}\text{\hspace{0.17em}}t=\frac{\pi }{4}.$

Ans.

$\begin{array}{l}\text{(i) The equation of the curve is y =}{\mathrm{x}}^{4}-6{\mathrm{x}}^{3}+13{\mathrm{x}}^{2}-10\mathrm{x}+5.\\ \text{On differentiating}\phantom{\rule{thinmathspace}{0ex}}\text{with respect to x, we get:}\\ \phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}=\frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}\left({\mathrm{x}}^{4}-6{\mathrm{x}}^{3}+13{\mathrm{x}}^{2}-10\mathrm{x}+5\right)\end{array}$ $\begin{array}{l}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=4{\mathrm{x}}^{3}-18{\mathrm{x}}^{2}+26\mathrm{x}-10\\ {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}_{\left(0,5\right)}\phantom{\rule{0ex}{0ex}}=4{\left(0\right)}^{3}-18{\left(0\right)}^{2}+26\left(0\right)-10\\ \phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=-10\\ \mathrm{Thus},\phantom{\rule{0ex}{0ex}}\mathrm{the}\phantom{\rule{0ex}{0ex}}\mathrm{slope}\phantom{\rule{0ex}{0ex}}\mathrm{of}\phantom{\rule{0ex}{0ex}}\mathrm{the}\phantom{\rule{0ex}{0ex}}\mathrm{tangent}\phantom{\rule{0ex}{0ex}}\mathrm{at}\phantom{\rule{0ex}{0ex}}\left(0,5\right)\phantom{\rule{0ex}{0ex}}\mathrm{is}-10.\\ \mathrm{The}\phantom{\rule{0ex}{0ex}}\mathrm{equation}\phantom{\rule{0ex}{0ex}}\mathrm{of}\phantom{\rule{0ex}{0ex}}\mathrm{the}\phantom{\rule{0ex}{0ex}}\mathrm{tangent}\phantom{\rule{0ex}{0ex}}\mathrm{is}\phantom{\rule{0ex}{0ex}}\mathrm{given}\phantom{\rule{0ex}{0ex}}\mathrm{as}:\\ \phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{y}-5=-10\left(\mathrm{x}-0\right)\\ ⇒\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{y}=-10\mathrm{x}+5\\ ⇒10\mathrm{x}+\mathrm{y}=5\\ \mathrm{The}\phantom{\rule{0ex}{0ex}}\mathrm{slope}\phantom{\rule{0ex}{0ex}}\mathrm{of}\phantom{\rule{0ex}{0ex}}\mathrm{the}\phantom{\rule{0ex}{0ex}}\mathrm{normal}\phantom{\rule{0ex}{0ex}}\left(\mathrm{M}\right)\phantom{\rule{0ex}{0ex}}\mathrm{at}\phantom{\rule{0ex}{0ex}}\left(0,5\right)=-\frac{1}{-10}=\frac{1}{10}\\ \mathrm{There}\mathrm{fore},\phantom{\rule{0ex}{0ex}}\mathrm{the}\phantom{\rule{0ex}{0ex}}\mathrm{e}\mathrm{quation}\phantom{\rule{0ex}{0ex}}\mathrm{of}\phantom{\rule{0ex}{0ex}}\mathrm{the}\phantom{\rule{0ex}{0ex}}\mathrm{normal}\phantom{\rule{0ex}{0ex}}\mathrm{at}\phantom{\rule{0ex}{0ex}}\left(0,5\right)\phantom{\rule{0ex}{0ex}}\mathrm{is}\phantom{\rule{0ex}{0ex}}\mathrm{given}\phantom{\rule{0ex}{0ex}}\mathrm{as}:\\ \phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{y}-5=\frac{1}{10}\left(\mathrm{x}-0\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\left[\because \phantom{\rule{0ex}{0ex}}\mathrm{y}-{\mathrm{y}}_{1}=\mathrm{M}\left(\mathrm{x}-{\mathrm{x}}_{1}\right)\right]\\ ⇒\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}10\mathrm{y}-50=\mathrm{x}\\ ⇒\mathrm{x}-10\mathrm{y}+50=0\\ \left(\mathrm{ii}\right)\phantom{\rule{0ex}{0ex}}\mathrm{The}\phantom{\rule{0ex}{0ex}}\mathrm{equation}\phantom{\rule{0ex}{0ex}}\mathrm{of}\phantom{\rule{0ex}{0ex}}\mathrm{the}\phantom{\rule{0ex}{0ex}}\mathrm{curve}\phantom{\rule{0ex}{0ex}}\mathrm{is}\phantom{\rule{0ex}{0ex}}\mathrm{y}={\mathrm{x}}^{4}-6{\mathrm{x}}^{3}+13{\mathrm{x}}^{2}-10\mathrm{x}+5.\\ \mathrm{On}\phantom{\rule{0ex}{0ex}}\mathrm{differentiating}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{with}\phantom{\rule{0ex}{0ex}}\mathrm{respect}\phantom{\rule{0ex}{0ex}}\mathrm{to}\phantom{\rule{0ex}{0ex}}\mathrm{x},\phantom{\rule{0ex}{0ex}}\mathrm{we}\phantom{\rule{0ex}{0ex}}\mathrm{get}:\\ \phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\left({\mathrm{x}}^{4}-6{\mathrm{x}}^{3}+13{\mathrm{x}}^{2}-10\mathrm{x}+5\right)\\ \phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=4{\mathrm{x}}^{3}-18{\mathrm{x}}^{2}+26\mathrm{x}-10\\ {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}_{\left(1,3\right)}=4{\left(1\right)}^{3}-18{\left(1\right)}^{2}+26\left(1\right)-10\\ \phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=2\\ \mathrm{Thus},\phantom{\rule{0ex}{0ex}}\mathrm{the}\phantom{\rule{0ex}{0ex}}\mathrm{slope}\phantom{\rule{0ex}{0ex}}\mathrm{of}\phantom{\rule{0ex}{0ex}}\mathrm{the}\phantom{\rule{0ex}{0ex}}\mathrm{tangent}\phantom{\rule{0ex}{0ex}}\mathrm{at}\phantom{\rule{0ex}{0ex}}\left(1,3\right)\phantom{\rule{0ex}{0ex}}\mathrm{is}\phantom{\rule{0ex}{0ex}}2.\\ \mathrm{The}\phantom{\rule{0ex}{0ex}}\mathrm{equation}\phantom{\rule{0ex}{0ex}}\mathrm{of}\phantom{\rule{0ex}{0ex}}\mathrm{the}\phantom{\rule{0ex}{0ex}}\mathrm{tangent}\phantom{\rule{0ex}{0ex}}\mathrm{is}\phantom{\rule{0ex}{0ex}}\mathrm{given}\phantom{\rule{0ex}{0ex}}\mathrm{as}:\end{array}$

$\begin{array}{l}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\mathrm{y}-3=2\left(\mathrm{x}-1\right)\\ ⇒\phantom{\rule{thinmathspace}{0ex}}\mathrm{y}-3=2x-2\\ ⇒\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}y=2x+1\\ \mathrm{T}\mathrm{h}\mathrm{e}\phantom{\rule{thickmathspace}{0ex}}\mathrm{s}\mathrm{l}\mathrm{o}\mathrm{p}\mathrm{e}\phantom{\rule{thickmathspace}{0ex}}\mathrm{o}\mathrm{f}\phantom{\rule{thickmathspace}{0ex}}\mathrm{t}\mathrm{h}\mathrm{e}\phantom{\rule{thickmathspace}{0ex}}\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}\mathrm{a}\mathrm{l}\phantom{\rule{thickmathspace}{0ex}}\left(\mathrm{M}\right)\phantom{\rule{thickmathspace}{0ex}}\mathrm{a}\mathrm{t}\phantom{\rule{thickmathspace}{0ex}}\left(1,3\right)=-\frac{1}{2}\\ \mathrm{T}\mathrm{h}\mathrm{e}\mathrm{r}\mathrm{e}\mathrm{f}\mathrm{o}\mathrm{r}\mathrm{e},\phantom{\rule{thickmathspace}{0ex}}\mathrm{t}\mathrm{h}\mathrm{e}\phantom{\rule{thickmathspace}{0ex}}\mathrm{e}\mathrm{q}\mathrm{u}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}\phantom{\rule{thickmathspace}{0ex}}\mathrm{o}\mathrm{f}\phantom{\rule{thickmathspace}{0ex}}\mathrm{t}\mathrm{h}\mathrm{e}\phantom{\rule{thickmathspace}{0ex}}\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}\mathrm{a}\mathrm{l}\phantom{\rule{thickmathspace}{0ex}}\mathrm{a}\mathrm{t}\phantom{\rule{thickmathspace}{0ex}}\left(1,3\right)\phantom{\rule{thickmathspace}{0ex}}\mathrm{i}\mathrm{s}\phantom{\rule{thickmathspace}{0ex}}\mathrm{g}\mathrm{i}\mathrm{v}\mathrm{e}\mathrm{n}\phantom{\rule{thickmathspace}{0ex}}\mathrm{a}\mathrm{s}:\\ \mathrm{y}-3=-\frac{1}{2}\left(x-1\right)\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\left[\because \phantom{\rule{thickmathspace}{0ex}}y-{y}_{1}=M\left(x-{x}_{1}\right)\right]\\ ⇒\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}x+2y-7=0\\ \left(\mathrm{i}\mathrm{i}\right)\phantom{\rule{thickmathspace}{0ex}}\mathrm{T}\mathrm{h}\mathrm{e}\phantom{\rule{thickmathspace}{0ex}}\mathrm{e}\mathrm{q}\mathrm{u}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}\phantom{\rule{thickmathspace}{0ex}}\mathrm{o}\mathrm{f}\phantom{\rule{thickmathspace}{0ex}}\mathrm{t}\mathrm{h}\mathrm{e}\phantom{\rule{thickmathspace}{0ex}}\mathrm{c}\mathrm{u}\mathrm{r}\mathrm{v}\mathrm{e}\phantom{\rule{thickmathspace}{0ex}}\mathrm{i}\mathrm{s}\phantom{\rule{thickmathspace}{0ex}}\mathrm{y}\phantom{\rule{thickmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}{\mathrm{x}}^{3}\\ \mathrm{O}\mathrm{n}\phantom{\rule{thickmathspace}{0ex}}\mathrm{d}\mathrm{i}\mathrm{f}\mathrm{f}\mathrm{e}\mathrm{r}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{i}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{n}\mathrm{g}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\mathrm{w}\mathrm{i}\mathrm{t}\mathrm{h}\phantom{\rule{thickmathspace}{0ex}}\mathrm{r}\mathrm{e}\mathrm{s}\mathrm{p}\mathrm{e}\mathrm{c}\mathrm{t}\phantom{\rule{thickmathspace}{0ex}}\mathrm{t}\mathrm{o}\phantom{\rule{thickmathspace}{0ex}}\mathrm{x},\phantom{\rule{thickmathspace}{0ex}}\mathrm{w}\mathrm{e}\phantom{\rule{thickmathspace}{0ex}}\mathrm{g}\mathrm{e}\mathrm{t}:\\ \phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\frac{dy}{dx}=\frac{d}{dx}\left({\mathrm{x}}^{3}\right)=3{x}^{2}\\ {\left(\frac{dy}{dx}\right)}_{\left(1,1\right)}=3{\left(1\right)}^{2}=3\\ \mathrm{T}\mathrm{h}\mathrm{u}\mathrm{s},\phantom{\rule{thickmathspace}{0ex}}\mathrm{t}\mathrm{h}\mathrm{e}\phantom{\rule{thickmathspace}{0ex}}\mathrm{s}\mathrm{l}\mathrm{o}\mathrm{p}\mathrm{e}\phantom{\rule{thickmathspace}{0ex}}\mathrm{o}\mathrm{f}\phantom{\rule{thickmathspace}{0ex}}\mathrm{t}\mathrm{h}\mathrm{e}\phantom{\rule{thickmathspace}{0ex}}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{g}\mathrm{e}\mathrm{n}\mathrm{t}\phantom{\rule{thickmathspace}{0ex}}\mathrm{a}\mathrm{t}\phantom{\rule{thickmathspace}{0ex}}\left(1,1\right)\phantom{\rule{thickmathspace}{0ex}}\mathrm{i}\mathrm{s}\phantom{\rule{thickmathspace}{0ex}}3.\\ \mathrm{E}\mathrm{q}\mathrm{u}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}\phantom{\rule{thickmathspace}{0ex}}\mathrm{o}\mathrm{f}\phantom{\rule{thickmathspace}{0ex}}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{g}\mathrm{e}\mathrm{n}\mathrm{t}\phantom{\rule{thickmathspace}{0ex}}\mathrm{a}\mathrm{t}\phantom{\rule{thickmathspace}{0ex}}\left(1,1\right)\\ \mathrm{o}\mathrm{f}\phantom{\rule{thickmathspace}{0ex}}\mathrm{t}\mathrm{h}\mathrm{e}\phantom{\rule{thickmathspace}{0ex}}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{g}\mathrm{e}\mathrm{n}\mathrm{t}\phantom{\rule{thickmathspace}{0ex}}\mathrm{i}\mathrm{s}:\phantom{\rule{thickmathspace}{0ex}}y-1=3\left(x-1\right)\phantom{\rule{thickmathspace}{0ex}}⇒y=3x-2\\ \mathrm{T}\mathrm{h}\mathrm{e}\phantom{\rule{thickmathspace}{0ex}}\mathrm{s}\mathrm{l}\mathrm{o}\mathrm{p}\mathrm{e}\phantom{\rule{thickmathspace}{0ex}}\mathrm{o}\mathrm{f}\phantom{\rule{thickmathspace}{0ex}}\mathrm{t}\mathrm{h}\mathrm{e}\phantom{\rule{thickmathspace}{0ex}}\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}\mathrm{a}\mathrm{l}\phantom{\rule{thickmathspace}{0ex}}\mathrm{a}\mathrm{t}\phantom{\rule{thickmathspace}{0ex}}\left(1,1\right)=-\frac{1}{3}\\ \mathrm{T}\mathrm{h}\mathrm{e}\mathrm{r}\mathrm{e}\mathrm{f}\mathrm{o}\mathrm{r}\mathrm{e},\phantom{\rule{thickmathspace}{0ex}}\mathrm{t}\mathrm{h}\mathrm{e}\phantom{\rule{thickmathspace}{0ex}}\mathrm{e}\mathrm{q}\mathrm{u}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}\phantom{\rule{thickmathspace}{0ex}}\mathrm{o}\mathrm{f}\phantom{\rule{thickmathspace}{0ex}}\mathrm{t}\mathrm{h}\mathrm{e}\phantom{\rule{thickmathspace}{0ex}}\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}\mathrm{a}\mathrm{l}\phantom{\rule{thickmathspace}{0ex}}\mathrm{a}\mathrm{t}\phantom{\rule{thickmathspace}{0ex}}\left(1,1\right)\phantom{\rule{thickmathspace}{0ex}}\mathrm{i}\mathrm{s}:\\ \phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\mathrm{y}-1=-\frac{1}{3}\left(x-1\right)\\ ⇒x+3y-4=0\\ \left(\mathrm{i}\mathrm{v}\right)\mathrm{T}\mathrm{h}\mathrm{e}\phantom{\rule{thickmathspace}{0ex}}\mathrm{e}\mathrm{q}\mathrm{u}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}\phantom{\rule{thickmathspace}{0ex}}\mathrm{o}\mathrm{f}\phantom{\rule{thickmathspace}{0ex}}\mathrm{t}\mathrm{h}\mathrm{e}\phantom{\rule{thickmathspace}{0ex}}\mathrm{c}\mathrm{u}\mathrm{r}\mathrm{v}\mathrm{e}\phantom{\rule{thickmathspace}{0ex}}\mathrm{i}\mathrm{s}\phantom{\rule{thickmathspace}{0ex}}\mathrm{y}\phantom{\rule{thickmathspace}{0ex}}=\phantom{\rule{thickmathspace}{0ex}}{\mathrm{x}}^{2}\\ \mathrm{O}\mathrm{n}\phantom{\rule{thickmathspace}{0ex}}\mathrm{d}\mathrm{i}\mathrm{f}\mathrm{f}\mathrm{e}\mathrm{r}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{i}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{n}\mathrm{g}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\mathrm{w}\mathrm{i}\mathrm{t}\mathrm{h}\phantom{\rule{thickmathspace}{0ex}}\mathrm{r}\mathrm{e}\mathrm{s}\mathrm{p}\mathrm{e}\mathrm{c}\mathrm{t}\phantom{\rule{thickmathspace}{0ex}}\mathrm{t}\mathrm{o}\phantom{\rule{thickmathspace}{0ex}}\mathrm{x},\phantom{\rule{thickmathspace}{0ex}}\mathrm{w}\mathrm{e}\phantom{\rule{thickmathspace}{0ex}}\mathrm{g}\mathrm{e}\mathrm{t}:\end{array}$ $\begin{array}{l}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\frac{dy}{dx}=\frac{d}{dx}\left({\mathrm{x}}^{2}\right)=2x\\ {\left(\frac{dy}{dx}\right)}_{\left(0,0\right)}=2\left(0\right)=0\\ \mathrm{E}\mathrm{q}\mathrm{u}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}\phantom{\rule{thickmathspace}{0ex}}\mathrm{o}\mathrm{f}\phantom{\rule{thickmathspace}{0ex}}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{g}\mathrm{e}\mathrm{n}\mathrm{t}\phantom{\rule{thickmathspace}{0ex}}\mathrm{a}\mathrm{t}\phantom{\rule{thickmathspace}{0ex}}\left(0,0\right):\\ y-0=0\left(x-0\right)\\ ⇒\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}y=0\\ \mathrm{T}\mathrm{h}\mathrm{e}\phantom{\rule{thickmathspace}{0ex}}\mathrm{s}\mathrm{l}\mathrm{o}\mathrm{p}\mathrm{e}\phantom{\rule{thickmathspace}{0ex}}\mathrm{o}\mathrm{f}\phantom{\rule{thickmathspace}{0ex}}\mathrm{t}\mathrm{h}\mathrm{e}\phantom{\rule{thickmathspace}{0ex}}\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}\mathrm{a}\mathrm{l}\phantom{\rule{thickmathspace}{0ex}}\mathrm{a}\mathrm{t}\phantom{\rule{thickmathspace}{0ex}}\left(0,0\right)=-\frac{1}{0}=\frac{1}{0}\\ \mathrm{T}\mathrm{h}\mathrm{e}\mathrm{r}\mathrm{e}\mathrm{f}\mathrm{o}\mathrm{r}\mathrm{e},\phantom{\rule{thickmathspace}{0ex}}\mathrm{t}\mathrm{h}\mathrm{e}\phantom{\rule{thickmathspace}{0ex}}\mathrm{e}\mathrm{q}\mathrm{u}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}\phantom{\rule{thickmathspace}{0ex}}\mathrm{o}\mathrm{f}\mathrm{t}\mathrm{h}\mathrm{e}\phantom{\rule{thickmathspace}{0ex}}\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}\mathrm{a}\mathrm{l}\phantom{\rule{thickmathspace}{0ex}}\mathrm{a}\mathrm{t}\phantom{\rule{thickmathspace}{0ex}}\left(0,0\right)\phantom{\rule{thickmathspace}{0ex}}\mathrm{i}\mathrm{s}:\\ \mathrm{y}-0=\frac{1}{0}\left(x-0\right)⇒x=0\\ \left(v\right)\phantom{\rule{thickmathspace}{0ex}}\mathrm{T}\mathrm{h}\mathrm{e}\phantom{\rule{thickmathspace}{0ex}}\mathrm{e}\mathrm{q}\mathrm{u}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}\phantom{\rule{thickmathspace}{0ex}}\mathrm{o}\mathrm{f}\phantom{\rule{thickmathspace}{0ex}}\mathrm{t}\mathrm{h}\mathrm{e}\phantom{\rule{thickmathspace}{0ex}}\mathrm{c}\mathrm{u}\mathrm{r}\mathrm{v}\mathrm{e}\phantom{\rule{thickmathspace}{0ex}}\mathrm{i}\mathrm{s}\phantom{\rule{thickmathspace}{0ex}}\mathrm{x}\phantom{\rule{thickmathspace}{0ex}}=\phantom{\rule{thickmathspace}{0ex}}\mathrm{c}\mathrm{o}\mathrm{s}\phantom{\rule{thickmathspace}{0ex}}\mathrm{t},\phantom{\rule{thickmathspace}{0ex}}\mathrm{y}\phantom{\rule{thickmathspace}{0ex}}=\phantom{\rule{thickmathspace}{0ex}}\mathrm{s}\mathrm{i}\mathrm{n}\phantom{\rule{thickmathspace}{0ex}}\mathrm{t}.\\ \mathrm{D}\mathrm{i}\mathrm{f}\mathrm{f}\mathrm{e}\mathrm{r}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{i}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{n}\mathrm{g}\phantom{\rule{thickmathspace}{0ex}}\mathrm{w}.\mathrm{r}.\mathrm{t}.\phantom{\rule{thickmathspace}{0ex}}\mathrm{t},\phantom{\rule{thickmathspace}{0ex}}\mathrm{w}\mathrm{e}\phantom{\rule{thickmathspace}{0ex}}\mathrm{g}\mathrm{e}\mathrm{t}\\ \phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\frac{dx}{dt}=-\mathrm{sin}t\phantom{\rule{thickmathspace}{0ex}}\mathrm{a}\mathrm{n}\mathrm{d}\phantom{\rule{thickmathspace}{0ex}}\frac{dy}{dt}\phantom{\rule{thickmathspace}{0ex}}=\phantom{\rule{thickmathspace}{0ex}}\mathrm{cos}t\\ \therefore \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\\ \phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}=\frac{\mathrm{cos}t}{-\mathrm{sin}t}=-\mathrm{cot}t\\ {\left(\frac{dy}{dx}\right)}_{x=\frac{\pi }{4}}=-\mathrm{cot}\left(\frac{\pi }{4}\right)\\ \phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}=-1\\ \therefore \phantom{\rule{thickmathspace}{0ex}}Equation\phantom{\rule{thickmathspace}{0ex}}\mathrm{o}\mathrm{f}\phantom{\rule{thickmathspace}{0ex}}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{g}\mathrm{e}\mathrm{n}\mathrm{t}\phantom{\rule{thickmathspace}{0ex}}\mathrm{a}\mathrm{t}\phantom{\rule{thickmathspace}{0ex}}\left(\mathrm{cos}\frac{\pi }{4},\mathrm{sin}\frac{\pi }{4}\right)\phantom{\rule{thickmathspace}{0ex}}i.e.,\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right):\end{array}$ $\begin{array}{l}y-\frac{1}{\sqrt{2}}=-1\left(x-\frac{1}{\sqrt{2}}\right)\\ ⇒x+y=\sqrt{2}\\ \mathrm{T}\mathrm{h}\mathrm{e}\phantom{\rule{thickmathspace}{0ex}}\mathrm{s}\mathrm{l}\mathrm{o}\mathrm{p}\mathrm{e}\phantom{\rule{thickmathspace}{0ex}}\mathrm{o}\mathrm{f}\phantom{\rule{thickmathspace}{0ex}}\mathrm{t}\mathrm{h}\mathrm{e}\phantom{\rule{thickmathspace}{0ex}}\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}\mathrm{a}\mathrm{l}\phantom{\rule{thickmathspace}{0ex}}\mathrm{a}\mathrm{t}\phantom{\rule{thickmathspace}{0ex}}\mathrm{t}=\frac{\pi }{4}:\\ M=-\frac{1}{{\left(\frac{dy}{dx}\right)}_{x=\frac{\pi }{4}}}\\ \phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}=-\frac{1}{-1}=1\\ \mathrm{T}\mathrm{h}\mathrm{e}\mathrm{r}\mathrm{e}\mathrm{f}\mathrm{o}\mathrm{r}\mathrm{e},\phantom{\rule{thickmathspace}{0ex}}\mathrm{t}\mathrm{h}\mathrm{e}\phantom{\rule{thickmathspace}{0ex}}\mathrm{e}\mathrm{q}\mathrm{u}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}\phantom{\rule{thickmathspace}{0ex}}\mathrm{o}\mathrm{f}\phantom{\rule{thickmathspace}{0ex}}\mathrm{t}\mathrm{h}\mathrm{e}\phantom{\rule{thickmathspace}{0ex}}\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}\mathrm{a}\mathrm{l}\phantom{\rule{thickmathspace}{0ex}}\mathrm{a}\mathrm{t}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right):\\ \phantom{\rule{thinmathspace}{0ex}}y-\frac{1}{\sqrt{2}}=1\left(x-\frac{1}{\sqrt{2}}\right)⇒y=x\end{array}$

Q.52 Find the equation of the tangent line to the curve y = x2 − 2x + 7 which is

(a) parallel to the line 2x – y + 9 = 0

(b) perpendicular to the line 5y −15x = 13.

Ans.

$\begin{array}{l}\mathrm{The}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{curve}\mathrm{is}\\ \mathrm{y}={\mathrm{x}}^{\mathrm{2}}-2\mathrm{x}+7\mathrm{}...\left(\mathrm{i}\right)\\ \mathrm{On}\mathrm{differentiating}\mathrm{with}\mathrm{respect}\mathrm{to}\mathrm{x},\mathrm{we}\mathrm{get}:\\ \mathrm{ }\frac{\mathrm{dy}}{\mathrm{dx}}=2\mathrm{x}-2\\ \mathrm{slope}\mathrm{of}\mathrm{tangent}\mathrm{to}\mathrm{curve}={\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}_{\left(\mathrm{x},\mathrm{y}\right)}=2\mathrm{x}-2\\ \left(\mathrm{a}\right)\mathrm{The}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{line}\mathrm{is}2\mathrm{x}-\mathrm{y}+ 9 = 0\\ ⇒\mathrm{y}=2\mathrm{x}+9\\ \mathrm{comparing}\mathrm{with}\mathrm{y}=\mathrm{mx}+\mathrm{c},\mathrm{we}\mathrm{get}\\ \mathrm{ }\mathrm{m}=\mathrm{2}\\ \mathrm{When}\mathrm{tangent}\mathrm{is}\mathrm{parallel}\mathrm{to}\mathrm{given}\mathrm{line}\mathrm{than},\\ {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}_{\left(\mathrm{x},\mathrm{y}\right)}=\mathrm{m}\\ 2\mathrm{x}-2=2⇒\mathrm{x}=2\\ \mathrm{Putting}\mathrm{x}=2\mathrm{in}\mathrm{equation}\left(\mathrm{i}\right),\mathrm{we}\mathrm{get}\\ \mathrm{y}={\left(2\right)}^{\mathrm{2}}-2\left(2\right)+7=7\\ \mathrm{Thus},\mathrm{the}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{tangent}\mathrm{passing}\mathrm{through}\left(2, 7\right)\mathrm{is}:\\ \mathrm{y}-7=2\left(\mathrm{x}-2\right)\\ \mathrm{ }\mathrm{y}-7=2\mathrm{x}-4\\ \mathrm{y}-2\mathrm{x}-3=0\\ \mathrm{Thus},\mathrm{the}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{tangent}\mathrm{line}\mathrm{to}\mathrm{the}\mathrm{given}\mathrm{curve}\left(\mathrm{which}\\ \mathrm{is}\mathrm{parallel}\mathrm{to}\mathrm{line}2\mathrm{x}-\mathrm{y}+ 9 = 0\right)\mathrm{is}\mathrm{y}-2\mathrm{x}-3=0.\\ \left(\mathrm{b}\right)\mathrm{The}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{line}\mathrm{is}5\mathrm{y}-15\mathrm{x}=13.\\ \mathrm{This}\mathrm{equation}\mathrm{can}\mathrm{be}\mathrm{written}\mathrm{as}\\ \mathrm{y}=3\mathrm{x}+\frac{13}{5}\mathrm{which}\mathrm{in}\mathrm{the}\mathrm{form}\mathrm{of}\mathrm{y}=\mathrm{mx}+\mathrm{c},\\ \mathrm{ }\mathrm{So},\mathrm{m}=3\\ \mathrm{If}\mathrm{a}\mathrm{tangent}\mathrm{is}\mathrm{perpendicular}\mathrm{to}\mathrm{the}\mathrm{line}5\mathrm{y}-\mathrm{15}=13,\\ \mathrm{then}\mathrm{the}\mathrm{slope}\mathrm{of}\mathrm{the}\mathrm{tangent}\mathrm{is}\\ \mathrm{ }\frac{-1}{{\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}_{\left(\mathrm{x},\mathrm{y}\right)}}=\frac{-1}{3}\\ ⇒2\mathrm{x}-2=-\frac{1}{3}\\ ⇒\mathrm{x}=\frac{5}{6}\\ \mathrm{When}\mathrm{ }\mathrm{x}=\frac{5}{6},\mathrm{y}={\left(\frac{5}{6}\right)}^{\mathrm{2}}-\mathrm{2}\left(\frac{5}{6}\right)+ 7=\frac{217}{36}\\ \mathrm{Thus},\mathrm{the}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{tangent}\mathrm{passing}\mathrm{through}\left(\frac{5}{6},\frac{217}{36}\right)\mathrm{is}\\ \mathrm{ }\mathrm{y}-\frac{217}{36}=\frac{-1}{3}\left(\mathrm{x}-\frac{5}{6}\right)\\ ⇒12\mathrm{x}+36\mathrm{y}-227=0\\ \mathrm{Hence},\mathrm{the}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{required}\mathrm{tangent}\mathrm{line}\mathrm{is}\\ 12\mathrm{x}+36\mathrm{y}-227=0.\end{array}$

Q.53 Show that the tangents to the curve y = 7x3 + 11 at the points where x = 2 and x = −2 are parallel.

Ans.

$\begin{array}{l}We\text{have,}\\ y={\text{7x}}^{\text{3}}\text{+ 11}\\ \text{Differentiating w}\text{.r}\text{.t}\text{. x, we get}\\ \frac{dy}{dx}=21{x}^{2}\\ Slope\text{of tangent at x=2,}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\text{m}}_{\text{1}}={\left(\frac{dy}{dx}\right)}_{x=2}\\ \text{}=21{\left(2\right)}^{2}\\ \text{}=84\\ Slope\text{of tangent at x=}-\text{2,}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\text{m}}_{\text{2}}={\left(\frac{dy}{dx}\right)}_{x=-\text{2}}\\ \text{}=21{\left(-\text{2}\right)}^{2}\\ \text{}=84\\ Since,\text{\hspace{0.17em}}\text{\hspace{0.17em}}{m}_{1}={m}_{2}\\ \text{Hence, the two tangents are parallel}.\end{array}$

Q.54 Find the points on the curve y = x3 at which the slope of the tangent is equal to the y coordinate of the point.

Ans.

$\begin{array}{l}\mathrm{The}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{curve}\mathrm{is}\mathrm{y}={\mathrm{x}}^{\mathrm{3}}\mathrm{.}\\ \mathrm{Diffferentiating}\mathrm{both}\mathrm{sides},\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathrm{x},\mathrm{we}\mathrm{get}\\ \mathrm{ }\frac{\mathrm{dy}}{\mathrm{dx}}=3{\mathrm{x}}^{2}\\ \mathrm{Slope}\mathrm{}\mathrm{of}\mathrm{tangent}\mathrm{at}\mathrm{point}\left(\mathrm{x},\mathrm{y}\right)\mathrm{is}\mathrm{given}\mathrm{by},\\ {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}_{\left(\mathrm{x},\mathrm{y}\right)}=3{\mathrm{x}}^{2}\\ \because {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}_{\left(\mathrm{x},\mathrm{y}\right)}=\mathrm{y}\left[\mathrm{Given}\right]\\ ⇒ \mathrm{ }3{\mathrm{x}}^{\mathrm{2}}=\mathrm{y}\\ \mathrm{But}\mathrm{ }\mathrm{y}={\mathrm{x}}^{\mathrm{3}}\left[\mathrm{Given}\right]\\ ⇒ \mathrm{ }{\mathrm{x}}^{\mathrm{3}}=3{\mathrm{x}}^{\mathrm{2}}\\ {\mathrm{x}}^{\mathrm{3}}-3{\mathrm{x}}^{2}=0\\ ⇒ \mathrm{ }{\mathrm{x}}^{2}\left(\mathrm{x}-3\right)=0\\ ⇒\mathrm{x}=0\mathrm{and}3\\ \mathrm{When}\mathrm{x}= 0,\mathrm{then}\mathrm{y}= 0\mathrm{and}\mathrm{when}\mathrm{x}= 3,\mathrm{then}\mathrm{y}= 3{\left(\mathrm{3}\right)}^{\mathrm{2}}=27\mathrm{.}\\ \mathrm{Hence},\mathrm{the}\mathrm{required}\mathrm{points}\mathrm{are}\left(0, 0\right)\mathrm{and}\left(3, 27\right)\mathrm{.}\end{array}$

Q.55 For the curve y = 4x3 − 2x5, find all the points at which the tangents

passes through the origin.

Ans.

$\begin{array}{l}\mathrm{The}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{curve}\mathrm{is}\\ \mathrm{y}=4{\mathrm{x}}^{\mathrm{3}}-2{\mathrm{x}}^{\mathrm{5}}...\left(\mathrm{i}\right)\\ \mathrm{Differentiating}\mathrm{with}\mathrm{respect}\mathrm{to}\mathrm{x},\mathrm{we}\mathrm{get}\\ \frac{\mathrm{dy}}{\mathrm{dx}}=12{\mathrm{x}}^{2}-10{\mathrm{x}}^{4}\\ \mathrm{slope}\mathrm{of}\mathrm{tangent}\mathrm{at}\mathrm{the}\mathrm{point}\left({\mathrm{x}}_{\mathrm{1}},{\mathrm{y}}_{\mathrm{1}}\right)\mathrm{is}\\ {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}_{\left({\mathrm{x}}_{\mathrm{1}},{\mathrm{y}}_{\mathrm{1}}\right)}=12{{\mathrm{x}}_{1}}^{2}-10{{\mathrm{x}}_{1}}^{4}\\ \mathrm{The}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{tangent}\mathrm{at}\left({\mathrm{x}}_{\mathrm{1}},{\mathrm{y}}_{\mathrm{1}}\right)\mathrm{is}\mathrm{given}\mathrm{by}\\ \mathrm{y}-{\mathrm{y}}_{1}={\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}_{\left({\mathrm{x}}_{\mathrm{1}},{\mathrm{y}}_{\mathrm{1}}\right)}\left(\mathrm{x}-{\mathrm{x}}_{1}\right)\\ \mathrm{y}-{\mathrm{y}}_{1}=\left(12{{\mathrm{x}}_{1}}^{2}-10{{\mathrm{x}}_{1}}^{4}\right)\left(\mathrm{x}-{\mathrm{x}}_{1}\right)\\ \mathrm{When}\mathrm{the}\mathrm{tangent}\mathrm{passes}\mathrm{through}\mathrm{the}\mathrm{origin}\left(0, 0\right),\\ \mathrm{then}\mathrm{x}=\mathrm{y}=0.\\ 0-{\mathrm{y}}_{1}=\left(12{{\mathrm{x}}_{1}}^{2}-10{{\mathrm{x}}_{1}}^{4}\right)\left(0-{\mathrm{x}}_{1}\right)\\ ⇒ \mathrm{ }{\mathrm{y}}_{1}=\left(12{{\mathrm{x}}_{1}}^{2}-10{{\mathrm{x}}_{1}}^{4}\right){\mathrm{x}}_{1}\\ ⇒ \mathrm{ }{\mathrm{y}}_{1}=12{{\mathrm{x}}_{1}}^{3}-10{{\mathrm{x}}_{1}}^{5} ...\left(\mathrm{ii}\right)\\ \mathrm{Point}\mathrm{}\left({\mathrm{x}}_{1},{\mathrm{y}}_{1}\right)\mathrm{lies}\mathrm{on}\mathrm{curve}\left(\mathrm{i}\right),\mathrm{ }\mathrm{then}\\ \mathrm{ }{\mathrm{y}}_{\mathrm{1}}=4{{\mathrm{x}}_{\mathrm{1}}}^{\mathrm{3}}-2{{\mathrm{x}}_{\mathrm{1}}}^{\mathrm{5}} .\dots \left(\mathrm{iii}\right)\\ \mathrm{From}\mathrm{ }\mathrm{equation}\left(\mathrm{ii}\right)\mathrm{and}\mathrm{equation}\left(\mathrm{iii}\right),\mathrm{ }\mathrm{we}\mathrm{ }\mathrm{have}\\ \mathrm{ }12{{\mathrm{x}}_{1}}^{3}-10{{\mathrm{x}}_{1}}^{5}\mathrm{}=4{{\mathrm{x}}_{\mathrm{1}}}^{\mathrm{3}}-2{{\mathrm{x}}_{\mathrm{1}}}^{\mathrm{5}}\\ ⇒ 8{{\mathrm{x}}_{1}}^{3}-8{{\mathrm{x}}_{1}}^{5}\mathrm{}=0\\ ⇒\mathrm{ }8{{\mathrm{x}}_{1}}^{3}\left(1-{{\mathrm{x}}_{1}}^{2}\right)=0\\ ⇒{\mathrm{x}}_{1}=0,\mathrm{ }±1\\ \mathrm{When}\mathrm{x}=0,\mathrm{y}=4\mathrm{}{\left(0\right)}^{3}-2\mathrm{}{\left(0\right)}^{5}=\mathrm{}0.\\ \mathrm{When}\mathrm{x}=1,\mathrm{y}=4\mathrm{}{\left(1\right)}^{3}-2{\left(1\right)}^{5}=\mathrm{}2.\\ \mathrm{When}\mathrm{x}=-1,\mathrm{}\mathrm{y}=4\mathrm{}{\left(-1\right)}^{3}-2{\left(-1\right)}^{5}=-2.\\ \mathrm{Hence},\mathrm{the}\mathrm{required}\mathrm{points}\mathrm{are}\left(0, 0\right),\left(1, 2\right)\mathrm{and}\left(-1,-\mathrm{2}\right)\mathrm{.}\end{array}$

Q.56 Find the points on the curve x2 + y2 − 2x − 3 = 0 at which the tangents are
parallel to the x-axis.

Ans.

$\begin{array}{l}\text{The equation of the given curve is}\\ {\text{x}}^{\text{2}}{\text{+ y}}^{\text{2}}-\text{2x}-\text{3}=\text{0}\text{}\dots \left(i\right)\\ \text{Differentiating both sides, w}\text{.r}\text{.t}\text{. x, we get}\\ \text{2x}+\text{2y}\frac{dy}{dx}-2=0\\ ⇒\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{dy}{dx}=\frac{2-2x}{2y}\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{1-x}{y}\\ Slope\text{of tangent of the given curve at point}\left({x}_{1},{y}_{1}\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}is\\ ⇒\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\left(\frac{dy}{dx}\right)}_{\left({x}_{1},{y}_{1}\right)}=\frac{1-{x}_{1}}{{y}_{1}}\\ It\text{​}\text{\hspace{0.17em}}\text{is given that}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\left(\frac{dy}{dx}\right)}_{\left({x}_{1},{y}_{1}\right)}=0\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{1-{x}_{1}}{{y}_{1}}=0⇒1-{x}_{1}=0\\ ⇒\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{x}_{1}=1\\ Since,\text{point}\left({x}_{1},{y}_{1}\right)\text{lies on curve}\left(i\right),\text{we get}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\text{x}}_{\text{1}}{}^{\text{2}}+{\text{y}}_{\text{1}}{}^{\text{2}}-{\text{2x}}_{\text{1}}-\text{3}=\text{0}\\ {\left(1\right)}^{2}+{\text{y}}_{\text{1}}{}^{\text{2}}-\text{2}\left(1\right)-\text{3}=\text{0}\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\text{y}}_{\text{1}}{}^{\text{2}}=4⇒{\text{y}}_{\text{1}}=±2\\ \text{Hence, the points at which the tangents are parallel to the}\\ \text{x-axis are}\left(\text{1, 2}\right)\text{and}\left(\text{1,}-\text{2}\right)\text{.}\end{array}$

Q.57 Find the equation of the normal at the point (am2, am3) for the curve ay2 = x3

Ans.

$\begin{array}{l}\mathrm{The}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{curve}\mathrm{is}{\mathrm{ay}}^{\mathrm{2}}={\mathrm{x}}^{\mathrm{3}}\mathrm{.}\\ \mathrm{On}\mathrm{differentiating}\mathrm{with}\mathrm{respect}\mathrm{to}\mathrm{x},\mathrm{we}\mathrm{have}:\\ 2\mathrm{ay}\frac{\mathrm{dy}}{\mathrm{dx}}=3{\mathrm{x}}^{2}\\ ⇒ \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{3{\mathrm{x}}^{2}}{2\mathrm{ay}}\\ \mathrm{Slope}\mathrm{at}\mathrm{the}\mathrm{point}\left({\mathrm{am}}^{\mathrm{2}},{\mathrm{am}}^{\mathrm{3}}\right)\\ ={\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}_{\left({\mathrm{am}}^{\mathrm{2}},{\mathrm{am}}^{\mathrm{3}}\right)}\\ =\frac{3{\left({\mathrm{am}}^{\mathrm{2}}\right)}^{2}}{2\mathrm{a}\left({\mathrm{am}}^{\mathrm{3}}\right)}\\ =\frac{3{\mathrm{a}}^{2}{\mathrm{m}}^{4}}{2{\mathrm{a}}^{2}{\mathrm{m}}^{3}}\\ =\frac{3}{2}\mathrm{m}\\ \mathrm{Slope}\mathrm{of}\mathrm{normal}=-\frac{1}{\left(\frac{3}{2}\mathrm{m}\right)}\\ =-\frac{2}{3\mathrm{m}}\\ \mathrm{Hence},\mathrm{the}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{normal}\mathrm{at}\left({\mathrm{am}}^{\mathrm{2}},{\mathrm{am}}^{\mathrm{3}}\right)\mathrm{is}\mathrm{given}\mathrm{by},\\ \mathrm{y}-{\mathrm{am}}^{\mathrm{3}}=-\frac{2}{3\mathrm{m}}\left(\mathrm{x}-{\mathrm{am}}^{\mathrm{2}}\right)\\ ⇒ \mathrm{ }3\mathrm{my}-3{\mathrm{am}}^{4}=-2\mathrm{x}+2{\mathrm{am}}^{2}\\ ⇒2\mathrm{x}+3\mathrm{my}-{\mathrm{am}}^{2}\left(2-3{\mathrm{m}}^{2}\right)=0\end{array}$

Q.58 Find the equation of the normals to the curve y = x3 + 2x + 6 which are parallel to the line
x + 14y + 4 = 0.

Ans.

$\begin{array}{l}\mathrm{The}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{curve}\mathrm{is}\mathrm{y}={\mathrm{x}}^{\mathrm{3}}+ 2\mathrm{x}+ 6\mathrm{.}\\ \mathrm{Differentiating}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathrm{x},\mathrm{we}\mathrm{get}\\ \frac{\mathrm{dy}}{\mathrm{dx}}=3{\mathrm{x}}^{2}+2\\ \mathrm{The}\mathrm{slope}\mathrm{of}\mathrm{the}\mathrm{tangent}\mathrm{to}\mathrm{the}\mathrm{given}\mathrm{curve}\mathrm{at}\mathrm{any}\mathrm{point}\left({\mathrm{x}}_{\mathrm{1}},{\mathrm{y}}_{\mathrm{1}}\right)\mathrm{}\\ {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}_{\left({\mathrm{x}}_{\mathrm{1}},{\mathrm{y}}_{\mathrm{1}}\right)}=3{{\mathrm{x}}_{1}}^{2}+2\\ \mathrm{Slope}\mathrm{of}\mathrm{normal}\mathrm{to}\mathrm{the}\mathrm{given}\mathrm{curve}\mathrm{at}\mathrm{any}\mathrm{point}\left({\mathrm{x}}_{\mathrm{1}},{\mathrm{y}}_{\mathrm{1}}\right)\\ =-\frac{1}{{\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}_{\left({\mathrm{x}}_{\mathrm{1}},{\mathrm{y}}_{\mathrm{1}}\right)}}\\ =-\frac{1}{3{{\mathrm{x}}_{1}}^{2}+2}\\ \mathrm{The}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{line}\mathrm{is}\\ \mathrm{ }\mathrm{x}+ 14\mathrm{y}+ 4=\mathrm{0}\\ \mathrm{Differentiating}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathrm{x},\mathrm{we}\mathrm{get}\\ \frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{1}{14}\\ \mathrm{Slope}\mathrm{of}\mathrm{line}\mathrm{ }\mathrm{at}\mathrm{any}\mathrm{point}\left({\mathrm{x}}_{\mathrm{1}},{\mathrm{y}}_{\mathrm{1}}\right)\\ ={\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}_{\left({\mathrm{x}}_{\mathrm{1}},{\mathrm{y}}_{\mathrm{1}}\right)}\\ =-\frac{1}{14}\\ \mathrm{Since}\mathrm{normal}\mathrm{is}\mathrm{parallel}\mathrm{to}\mathrm{line},\mathrm{so}\\ \mathrm{ }\mathrm{slope}\mathrm{of}\mathrm{normal}=\mathrm{slope}\mathrm{of}\mathrm{line}\\ -\frac{1}{3{\mathrm{x}}^{2}+2}=-\frac{1}{14}\\ ⇒ \mathrm{ }14=3{\mathrm{x}}^{2}+2\\ ⇒ {\mathrm{x}}^{2}=\frac{12}{3}=4\\ ⇒ \mathrm{x}=±2\\ \mathrm{When}\mathrm{x}=2,\mathrm{}\mathrm{y}=8+4+6=18.\\ \mathrm{When}\mathrm{}\mathrm{x}=-2,\mathrm{y}=-\mathrm{ }8-4+6=-\mathrm{ }6.\\ \mathrm{Therefore},\mathrm{the}\mathrm{equations}\mathrm{of}\mathrm{normals}\mathrm{passing}\mathrm{through}\\ \mathrm{the}\mathrm{points}\left(2,18\right)\mathrm{and}\left(-2,-6\right).\\ \mathrm{ }\mathrm{y}-18=-\frac{1}{14}\left(\mathrm{x}-2\right)⇒\mathrm{x}+14\mathrm{y}-254=0\\ \mathrm{and} \mathrm{y}+6=-\frac{1}{14}\left(\mathrm{x}+2\right) ⇒\mathrm{x}+14\mathrm{y}+86=0\\ \mathrm{Hence},\mathrm{the}\mathrm{equations}\mathrm{of}\mathrm{the}\mathrm{normals}\mathrm{to}\mathrm{the}\mathrm{given}\mathrm{curve}\\ \left(\mathrm{which}\mathrm{are}\mathrm{parallel}\mathrm{to}\mathrm{the}\mathrm{given}\mathrm{line}\right)\mathrm{are}\\ \mathrm{x}+14\mathrm{y}-254=0,\mathrm{ }\mathrm{x}+14\mathrm{y}+86=0. \end{array}$

Q.59 Find the equations of the tangent and normal to the parabola y2= 4ax at the point (at2, 2at).

Ans.

$\begin{array}{l}\mathrm{The}\mathrm{equation}\mathrm{of}\mathrm{parabola}\mathrm{is}{\mathrm{y}}^{\mathrm{2}}=4\mathrm{ax}\\ \mathrm{Differentiating}\mathrm{both}\mathrm{sides}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathrm{x},\mathrm{we}\mathrm{get}\\ \mathrm{ }2\mathrm{y}\frac{\mathrm{dy}}{\mathrm{dx}}=4\mathrm{a}\\ ⇒\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{4\mathrm{a}}{2\mathrm{y}}=\frac{2\mathrm{a}}{\mathrm{y}}\\ \mathrm{Slope}\mathrm{of}\mathrm{tangent}\mathrm{to}\mathrm{parabola}\mathrm{at}\mathrm{point}\left({\mathrm{at}}^{\mathrm{2}},2\mathrm{at}\right)\\ ⇒ \mathrm{ }{\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}_{\left({\mathrm{at}}^{\mathrm{2}},2\mathrm{at}\right)}=\frac{2\mathrm{a}}{2\mathrm{at}}=\frac{1}{\mathrm{t}}\\ \mathrm{Slope}\mathrm{of}\mathrm{normal}\mathrm{to}\mathrm{parabola}\mathrm{at}\mathrm{point}\left({\mathrm{at}}^{\mathrm{2}},2\mathrm{at}\right)\\ =-\frac{1}{{\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}_{\left({\mathrm{at}}^{\mathrm{2}},2\mathrm{at}\right)}}\\ =-\frac{1}{\left(\frac{1}{\mathrm{t}}\right)}=-\mathrm{t}\\ \mathrm{Then},\mathrm{the}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{tangent}\mathrm{at}\mathrm{ }\left({\mathrm{at}}^{\mathrm{2}},2\mathrm{at}\right)\\ \mathrm{y}-2\mathrm{at}=\frac{1}{\mathrm{t}}\left(\mathrm{x}-{\mathrm{at}}^{2}\right)\\ ⇒ \mathrm{ }\mathrm{ty}=\mathrm{x}+{\mathrm{at}}^{2}\\ \mathrm{And},\mathrm{the}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{normal}\mathrm{at}\mathrm{ }\left({\mathrm{at}}^{\mathrm{2}},2\mathrm{at}\right)\\ \mathrm{y}-2\mathrm{at}=-\mathrm{t}\left(\mathrm{x}-{\mathrm{at}}^{2}\right)\\ ⇒ \mathrm{tx}+ \mathrm{y}=2\mathrm{at}+{\mathrm{at}}^{3}\\ \mathrm{Thus},\mathrm{the}\mathrm{required}\mathrm{equations}\mathrm{of}\mathrm{tangent}\mathrm{and}\mathrm{normal}\mathrm{are}\\ \mathrm{ty}=\mathrm{x}+{\mathrm{at}}^{\mathrm{2}}\mathrm{and}\mathrm{tx}+ \mathrm{y}=2\mathrm{at}+{\mathrm{at}}^{3}.\end{array}$

Q.60 Prove that the curves x = y2 and xy = k cut at right angles if 8k2 = 1.

Ans.

$\begin{array}{l}\mathrm{Given}\mathrm{equations}\mathrm{of}\mathrm{the}\mathrm{curves}\mathrm{are}\mathrm{x}={\mathrm{y}}^{\mathrm{2}}\mathrm{and}\mathrm{xy}=\mathrm{k}\mathrm{.}\\ \mathrm{Putting}\mathrm{x}={\mathrm{y}}^{\mathrm{2}}\mathrm{in}\mathrm{xy}=\mathrm{k},\mathrm{we}\mathrm{get}\\ \mathrm{ }{\mathrm{y}}^{\mathrm{2}}.\mathrm{y}=\mathrm{k}⇒\mathrm{y}={\mathrm{k}}^{\frac{1}{3}}\\ \mathrm{Putting}\mathrm{value}\mathrm{of}\mathrm{y}\mathrm{in}\mathrm{xy}=\mathrm{k},\mathrm{we}\mathrm{get}\\ \mathrm{ }{\mathrm{k}}^{\frac{1}{3}}\mathrm{x}=\mathrm{k}⇒\mathrm{x}={\mathrm{k}}^{\frac{2}{3}}\\ \mathrm{Thus},\mathrm{the}\mathrm{point}\mathrm{of}\mathrm{intersection}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{curves}\mathrm{is}\left({\mathrm{k}}^{\frac{2}{3}},{\mathrm{k}}^{\frac{1}{3}}\right).\\ \mathrm{Differentiating}\mathrm{}\mathrm{the}\mathrm{curves}\mathrm{x}={\mathrm{y}}^{\mathrm{2}}\mathrm{and}\mathrm{xy}=\mathrm{k}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathrm{x},\mathrm{we}\mathrm{get}\\ \frac{\mathrm{d}}{\mathrm{dx}}\mathrm{x}=\frac{\mathrm{d}}{\mathrm{dx}}{\mathrm{y}}^{2}⇒1=2\mathrm{y}\frac{\mathrm{dy}}{\mathrm{dx}}⇒\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{2\mathrm{y}}\\ \mathrm{Slope}\mathrm{of}\mathrm{tangent}\mathrm{to}\mathrm{the}\mathrm{curves}\mathrm{ }\mathrm{x}={\mathrm{y}}^{\mathrm{2}}\mathrm{ }\mathrm{at}\mathrm{point}\left({\mathrm{k}}^{\frac{2}{3}},{\mathrm{k}}^{\frac{1}{3}}\right)\\ \mathrm{ }{\mathrm{m}}_{1}=\mathrm{ }{\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}_{\left({\mathrm{k}}^{\frac{2}{3}},{\mathrm{k}}^{\frac{1}{3}}\right)}\\ =\frac{1}{2{\mathrm{k}}^{\frac{1}{3}}}\\ \mathrm{and}\mathrm{}\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{xy}\right)=\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{k}⇒\mathrm{x}\frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{y}\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{x}=0⇒\frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{\mathrm{y}}{\mathrm{x}}\\ \mathrm{Slope}\mathrm{of}\mathrm{tangent}\mathrm{to}\mathrm{the}\mathrm{curve}\mathrm{xy}=\mathrm{k}\mathrm{at}\mathrm{point}\left({\mathrm{k}}^{\frac{2}{3}},{\mathrm{k}}^{\frac{1}{3}}\right)\\ \mathrm{ }{\mathrm{m}}_{2}={\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}_{\left({\mathrm{k}}^{\frac{2}{3}},{\mathrm{k}}^{\frac{1}{3}}\right)}\\ =-\frac{{\mathrm{k}}^{\frac{1}{3}}}{{\mathrm{k}}^{\frac{2}{3}}}\\ =-\frac{1}{{\mathrm{k}}^{\frac{1}{3}}}\\ \mathrm{Since},\mathrm{two}\mathrm{curves}\mathrm{intersect}\mathrm{at}\mathrm{right}\mathrm{angles}\mathrm{if}\\ \mathrm{ }{\mathrm{m}}_{\mathrm{1}}.{\mathrm{m}}_{\mathrm{2}}=-\mathrm{1}\\ \frac{1}{2{\mathrm{k}}^{\frac{1}{3}}}×-\frac{1}{{\mathrm{k}}^{\frac{1}{3}}}=-1\\ 1=2{\mathrm{k}}^{\frac{2}{3}}\\ \mathrm{Cubing}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get}\\ \mathrm{1}=8{\mathrm{k}}^{\mathrm{2}}.\\ \mathrm{Hence}\mathrm{ },\mathrm{the}\mathrm{given}\mathrm{two}\mathrm{curves}\mathrm{cut}\mathrm{at}\mathrm{right}\mathrm{angels}\mathrm{if}8{\mathrm{k}}^{\mathrm{2}}=1.\end{array}$

Q.61

$\begin{array}{l}\text{Find the equations of the tangent and normal to the}\\ \text{hyperbola}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{{\text{x}}^{\text{2}}}{{\text{a}}^{\text{2}}}-\frac{{\text{y}}^{\text{2}}}{{\text{b}}^{\text{2}}}\text{=1 at the point}\left({\text{x}}_{\text{0}}{\text{,y}}_{\text{0}}\right)\text{.}\end{array}$

Ans.

$\begin{array}{l}\text{Find the equations of the tangent and normal to the}\\ \text{hyperbola}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{{\text{x}}^{\text{2}}}{{\text{a}}^{\text{2}}}-\frac{{\text{y}}^{\text{2}}}{{\text{b}}^{\text{2}}}\text{=1 at the point}\left({\text{x}}_{\text{0}}{\text{,y}}_{\text{0}}\right)\text{.}\end{array}$

Q.62

$\begin{array}{l}\text{Find the equation of the tangent to the curve y=}\sqrt{\text{3x}-\text{2}}\text{}\\ \text{which is parallel to the line 4x}-\text{2y + 5 = 0}\text{.}\end{array}$

Ans.

$\begin{array}{l}\mathrm{The}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{curve}\mathrm{is}\mathrm{y}=\sqrt{3\mathrm{x}-2}\\ \mathrm{Differentiating}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathrm{x},\mathrm{we}\mathrm{get}\\ \mathrm{ }\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}{\left(3\mathrm{x}-2\right)}^{\frac{1}{2}}\\ =\frac{1}{2}{\left(3\mathrm{x}-2\right)}^{-\frac{1}{2}}\frac{\mathrm{d}}{\mathrm{dx}}\left(3\mathrm{x}-2\right)\left[\mathrm{By}\mathrm{chain}\mathrm{rule}\right]\\ =\frac{1}{2\sqrt{3\mathrm{x}-2}}×3\\ =\frac{3}{2\sqrt{3\mathrm{x}-2}}\\ \mathrm{The}\mathrm{slope}\mathrm{of}\mathrm{the}\mathrm{tangent}\mathrm{to}\mathrm{the}\mathrm{given}\mathrm{curve}\mathrm{at}\mathrm{any}\mathrm{point}\left({\mathrm{x}}_{\mathrm{1}},{\mathrm{y}}_{\mathrm{1}}\right),\\ \mathrm{ }{\mathrm{m}}_{1}={\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}_{\left({\mathrm{x}}_{\mathrm{1}},{\mathrm{y}}_{\mathrm{1}}\right)}\\ =\frac{3}{2\sqrt{3{\mathrm{x}}_{1}-2}}\\ \mathrm{The}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{line}\mathrm{is}4\mathrm{x}-2\mathrm{y}+ 5 = 0\mathrm{.}\\ \mathrm{Differentiating}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathrm{x},\mathrm{we}\mathrm{get}\\ 4-2\frac{\mathrm{dy}}{\mathrm{dx}}=0⇒\frac{\mathrm{dy}}{\mathrm{dx}}=2\\ \mathrm{The}\mathrm{slope}\mathrm{of}\mathrm{the}\mathrm{tangent}\mathrm{to}\mathrm{the}\mathrm{given}\mathrm{line}\mathrm{at}\mathrm{any}\mathrm{point}\left({\mathrm{x}}_{\mathrm{1}},{\mathrm{y}}_{\mathrm{1}}\right),\\ \mathrm{ }{\mathrm{m}}_{2}={\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}_{\left({\mathrm{x}}_{\mathrm{1}},{\mathrm{y}}_{\mathrm{1}}\right)}=2\\ \mathrm{Since}, \mathrm{ }\mathrm{tangent}\mathrm{is}\mathrm{parallel}\mathrm{to}\mathrm{given}\mathrm{line}.\mathrm{So},\\ \mathrm{ }{\mathrm{m}}_{\mathrm{1}}={\mathrm{m}}_{\mathrm{2}}\\ ⇒\frac{3}{2\sqrt{3{\mathrm{x}}_{1}-2}}=2\\ ⇒ \sqrt{3{\mathrm{x}}_{1}-2}=\frac{3}{4}\\ ⇒ \mathrm{ }3{\mathrm{x}}_{1}-2=\frac{9}{16}\\ ⇒3{\mathrm{x}}_{1}=\frac{9}{16}+2\\ ⇒3{\mathrm{x}}_{1}=\frac{41}{16}⇒{\mathrm{x}}_{1}=\frac{41}{48}\\ \mathrm{Since},\mathrm{point}\left({\mathrm{x}}_{1},{\mathrm{y}}_{1}\right)\mathrm{}\mathrm{lies}\mathrm{on}\mathrm{given}\mathrm{curve} \mathrm{y}=\sqrt{3\mathrm{x}-2}\\ \mathrm{so}, {\mathrm{y}}_{\mathrm{1}}=\sqrt{3{\mathrm{x}}_{1}-2}=\frac{3}{4}\\ \therefore \mathrm{Equation}\mathrm{of}\mathrm{the}\mathrm{tangent}\mathrm{passing}\mathrm{through}\mathrm{the}\mathrm{point}\mathrm{ }\left(\frac{41}{48},\frac{3}{4}\right)\mathrm{ }\mathrm{is}\\ \mathrm{y}-\frac{3}{4}=2\left(\mathrm{x}-\frac{41}{48}\right)\\ ⇒ \mathrm{ }48\mathrm{x}-24\mathrm{y}=23\\ \mathrm{Hence},\mathrm{the}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{required}\mathrm{tangent}\mathrm{is} 48\mathrm{x}-24\mathrm{y}=23.\end{array}$

Q.63 The slope of the normal to the curve y = 2x2 + 3 sin x at x = 0 is
(A) 3
(B) 1/3
(C) – 3
(D) –1/3

Ans.

$\begin{array}{l}The\text{equation of given curve is y}={\text{2x}}^{\text{2}}\text{+ 3sinx}\\ \text{Differentiating w}\text{.r}\text{.t}\text{. x, we get}\\ \text{\hspace{0.17em}}\frac{dy}{dx}=\frac{d}{dx}\left({\text{2x}}^{\text{2}}\text{+ 3sinx}\right)\\ \text{}=4x+3\mathrm{cos}x\\ Putting\text{x= 0, we get}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{m}={\left(\frac{dy}{dx}\right)}_{\text{x= 0}}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{}=4\left(0\right)+3\mathrm{cos}0\\ \text{}=3\\ Slope\text{of normal to the given curve is}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{M}=-\frac{1}{m}\\ \text{}=-\frac{1}{3}\\ \text{The correct answer is D}\text{.}\end{array}$

Q.64 The line y = x + 1 is a tangent to the curve y2 = 4x at the point
(A) (1, 2) (B) (2, 1) (C) (1, −2) (D) (−1, 2)

Ans.

$\begin{array}{l}\mathrm{The}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{curve}\mathrm{is} \\ {\mathrm{y}}^{\mathrm{2}}=4\mathrm{x}\\ \mathrm{Differentiating}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathrm{x},\mathrm{we}\mathrm{get}\\ 2\mathrm{y}\frac{\mathrm{dy}}{\mathrm{dx}}=4⇒\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{4}{2\mathrm{y}}=\frac{2}{\mathrm{y}}\\ \mathrm{Therefore},\mathrm{the}\mathrm{slope}\mathrm{of}\mathrm{the}\mathrm{tangent}\mathrm{to}\mathrm{the}\mathrm{given}\mathrm{curve}\mathrm{at}\mathrm{any}\\ \mathrm{point}\left({\mathrm{x}}_{\mathrm{1}},{\mathrm{y}}_{1}\right)={\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}_{\left({\mathrm{x}}_{\mathrm{1}},{\mathrm{y}}_{1}\right)}\\ =\frac{2}{{\mathrm{y}}_{1}}\\ \mathrm{The}\mathrm{given}\mathrm{line}\mathrm{is}\mathrm{y}=\mathrm{x}+ 1\mathrm{which}\mathrm{is}\mathrm{in}\mathrm{the}\mathrm{form}\mathrm{of}\mathrm{y}=\mathrm{mx}+\mathrm{c},\\ \mathrm{m}=1\\ \mathrm{Since},\mathrm{tangent}\mathrm{and}\mathrm{line}\mathrm{are}\mathrm{parallel}.\mathrm{So}\\ {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}_{\left({\mathrm{x}}_{\mathrm{1}},{\mathrm{y}}_{1}\right)}=\mathrm{m}⇒\frac{2}{{\mathrm{y}}_{1}}=1\\ \mathrm{ }{\mathrm{y}}_{1}=2\\ \mathrm{Since},\mathrm{ }\mathrm{y}=\mathrm{x}+1,\mathrm{ }\mathrm{so}\\ \mathrm{x}=\mathrm{y}-1\\ =2-1=1\\ \mathrm{Hence},\mathrm{the}\mathrm{line}\mathrm{y}=\mathrm{x}+ 1\mathrm{is}\mathrm{a}\mathrm{tangent}\mathrm{to}\mathrm{the}\mathrm{given}\mathrm{curve}\mathrm{at}\mathrm{the}\\ \mathrm{point}\left(1, 2\right)\mathrm{.}\\ \mathrm{The}\mathrm{correct}\mathrm{answer}\mathrm{is}\mathrm{A}\mathrm{.}\end{array}$

Q.65

$\begin{array}{l}\text{Using differentials, find the approximate value}\phantom{\rule{thinmathspace}{0ex}}\mathrm{o}\mathrm{f}\mathrm{e}\mathrm{a}\mathrm{c}\mathrm{h}\mathrm{o}\mathrm{f}\\ \text{the following up to 3 places of}\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}\mathrm{e}\mathrm{c}\mathrm{i}\mathrm{m}\mathrm{a}\mathrm{l}:\\ \left(\mathrm{i}\right)\phantom{\rule{thickmathspace}{0ex}}\sqrt{25.3}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\left(\mathrm{i}\mathrm{i}\right)\phantom{\rule{thickmathspace}{0ex}}\sqrt{49.5}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\left(\mathrm{i}\mathrm{i}\mathrm{i}\right)\phantom{\rule{thickmathspace}{0ex}}\sqrt{0.6}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\left(\mathrm{i}\mathrm{v}\right)\phantom{\rule{thickmathspace}{0ex}}{\left(0.009\right)}^{\frac{1}{3}}& \phantom{\rule{thinmathspace}{0ex}}\\ \left(\mathrm{v}\right)\phantom{\rule{thickmathspace}{0ex}}{\left(0.999\right)}^{\frac{1}{10}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\left(\mathrm{v}\mathrm{i}\right)\phantom{\rule{thickmathspace}{0ex}}{\left(15\right)}^{\frac{1}{4}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\left(\mathrm{v}\mathrm{i}\mathrm{i}\right)\phantom{\rule{thickmathspace}{0ex}}{\left(26\right)}^{\frac{1}{3}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\left(\mathrm{v}\mathrm{i}\mathrm{i}\mathrm{i}\right)\phantom{\rule{thickmathspace}{0ex}}{\left(255\right)}^{\frac{1}{4}}& \\ \left(\mathrm{i}\mathrm{x}\right)\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{\left(82\right)}^{\frac{1}{4}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\left(\mathrm{x}\right)\phantom{\rule{thickmathspace}{0ex}}{\left(401\right)}^{\frac{1}{2}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\left(\mathrm{x}\mathrm{i}\right)\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}{\left(0.0037\right)}^{\frac{1}{2}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\left(\mathrm{x}\mathrm{i}\mathrm{i}\right)\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}{\left(26.57\right)}^{\frac{1}{3}}& \\ \left(\mathrm{x}\mathrm{i}\mathrm{i}\mathrm{i}\right)\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{\left(81.5\right)}^{\frac{1}{4}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\left(\mathrm{x}\mathrm{i}\mathrm{v}\right)\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{\left(3.968\right)}^{\frac{3}{2}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\left(\mathrm{x}\mathrm{v}\right)\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{\left(32.15\right)}^{\frac{1}{5}}\end{array}$

Ans.

$\begin{array}{l}\left(\mathrm{i}\right)\sqrt{25.3}\\ \mathrm{Let}\mathrm{ }\mathrm{y}=\sqrt{\mathrm{x}}, \mathrm{ }\mathrm{x}=25\mathrm{and}\mathrm{\Delta x}=\mathrm{0}.3\\ ⇒\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{2\sqrt{\mathrm{x}}}\\ \mathrm{Then},\\ \mathrm{ }\mathrm{\Delta y}=\sqrt{\mathrm{x}+\mathrm{\Delta x}}-\sqrt{\mathrm{x}}\\ \mathrm{ }=\sqrt{25.3}-\sqrt{25}\\ \mathrm{ }=\sqrt{25.3}-5\\ \sqrt{25.3}=\mathrm{\Delta y}+5\\ \mathrm{Now},\mathrm{dy}\mathrm{is}\mathrm{approximately}\mathrm{equal}\mathrm{to}\mathrm{\Delta y}\mathrm{and}\mathrm{is}\mathrm{given}\mathrm{by},\\ \mathrm{ }\mathrm{dy}=\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)\mathrm{\Delta x}\\ \mathrm{ }=\frac{1}{2\sqrt{\mathrm{x}}}\left(0.3\right)\\ \mathrm{ }=\frac{1}{2\sqrt{25}}\left(0.3\right)\\ \mathrm{ }=\frac{1}{2×5}×0.3\\ \mathrm{ }=0.03\\ \mathrm{Hence},\mathrm{the}\mathrm{approximate}\mathrm{value}\mathrm{of}\mathrm{ }\sqrt{25.3}\mathrm{ }\mathrm{is}5+\mathrm{0}.03=\mathrm{5}.03\mathrm{.}\\ \left(\mathrm{ii}\right)\sqrt{49.5}\\ \mathrm{Let}\mathrm{ }\mathrm{y}=\sqrt{\mathrm{x}}, \mathrm{ }\mathrm{x}=49\mathrm{and}\mathrm{\Delta x}=\mathrm{0}.5\\ ⇒\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{2\sqrt{\mathrm{x}}}\\ \mathrm{Then},\\ \mathrm{ }\mathrm{\Delta y}=\sqrt{\mathrm{x}+\mathrm{\Delta x}}-\sqrt{\mathrm{x}}\\ \mathrm{ }=\sqrt{49.5}-\sqrt{49}\\ \mathrm{ }=\sqrt{49.5}-7\\ \sqrt{49.5}=\mathrm{\Delta y}+7\\ \mathrm{Now},\mathrm{dy}\mathrm{is}\mathrm{approximately}\mathrm{equal}\mathrm{to}\mathrm{\Delta y}\mathrm{and}\mathrm{is}\mathrm{given}\mathrm{by},\\ \mathrm{ }\mathrm{dy}=\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)\mathrm{\Delta x}\\ \mathrm{ }=\frac{1}{2\sqrt{\mathrm{x}}}\left(0.5\right)\\ \mathrm{ }=\frac{1}{2\sqrt{49}}\left(0.5\right)\\ \mathrm{ }=\frac{1}{2×7}×0.5\\ \mathrm{ }=0.035\\ \mathrm{Hence},\mathrm{the}\mathrm{approximate}\mathrm{value}\mathrm{of}\mathrm{ }\sqrt{49.5}\mathrm{ }\mathrm{is}7+0.035=\mathrm{7}.035\mathrm{.}\\ \left(\mathrm{iii}\right)\sqrt{0.6}\\ \mathrm{Let}\mathrm{ }\mathrm{y}=\sqrt{\mathrm{x}}, \mathrm{ }\mathrm{x}=1\mathrm{and}\mathrm{\Delta x}=- 0.4\\ ⇒\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{2\sqrt{\mathrm{x}}}\\ \mathrm{Then},\\ \mathrm{ }\mathrm{\Delta y}=\sqrt{\mathrm{x}+\mathrm{\Delta x}}-\sqrt{\mathrm{x}}\\ \mathrm{ }=\sqrt{0.6}-1\\ \mathrm{ }\sqrt{0.6}=\mathrm{\Delta y}+1\\ \mathrm{Now},\mathrm{dy}\mathrm{is}\mathrm{approximately}\mathrm{equal}\mathrm{to}\mathrm{\Delta y}\mathrm{and}\mathrm{is}\mathrm{given}\mathrm{by},\\ \mathrm{ }\mathrm{dy}=\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)\mathrm{\Delta x}\\ \mathrm{ }=\frac{1}{2\sqrt{\mathrm{x}}}\left(- 0.4\right)\\ \mathrm{ }=\frac{1}{2\sqrt{1}}\left(- 0.4\right)\\ \mathrm{ }=- 0.2\\ \mathrm{Hence},\mathrm{the}\mathrm{approximate}\mathrm{value}\mathrm{of}\mathrm{ }\sqrt{0.6}\mathrm{ }\mathrm{is}1- 0.2=\mathrm{0}.8\\ \left(\mathrm{iv}\right)\mathrm{ }{\left(0.009\right)}^{\frac{1}{3}}\\ \mathrm{Let}\mathrm{ }\mathrm{y}={\mathrm{x}}^{\frac{1}{3}}, \mathrm{ }\mathrm{x}=0.008\mathrm{and}\mathrm{\Delta x}=0.001\\ ⇒\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{3{\mathrm{x}}^{\frac{2}{3}}}\\ \mathrm{Then},\\ \mathrm{\Delta y}={\left(\mathrm{x}+\mathrm{\Delta x}\right)}^{\frac{1}{3}}-{\mathrm{x}}^{\frac{1}{3}}\\ ={\left(0.009\right)}^{\frac{1}{3}}-0.2\\ {\left(0.009\right)}^{\frac{1}{3}}=\mathrm{\Delta y}+0.2\\ \mathrm{Now},\mathrm{dy}\mathrm{is}\mathrm{approximately}\mathrm{equal}\mathrm{to}\mathrm{\Delta y}\mathrm{and}\mathrm{is}\mathrm{given}\mathrm{by},\\ ⇒ \mathrm{ }\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{3{\mathrm{x}}^{\frac{2}{3}}}\left(\mathrm{\Delta x}\right)\\ \mathrm{ }=\frac{1}{3{\left(0.008\right)}^{\frac{2}{3}}}\left(0.001\right)\\ \mathrm{ }=\frac{1}{3\left(0.04\right)}\left(0.001\right)=\frac{0.001}{0.12}\\ \mathrm{ }=0.008\\ \mathrm{Hence},\mathrm{the}\mathrm{approximate}\mathrm{value}\mathrm{of}\mathrm{ }{\left(0.009\right)}^{\frac{2}{3}}\mathrm{ }\mathrm{is}0.2+ 0.008=\mathrm{0}.208\end{array}$ $\begin{array}{l}\left(v\right){\left(0.999\right)}^{\frac{1}{10}}\\ Let\text{​}\text{y}={x}^{\frac{1}{10}},\text{\hspace{0.17em}}x=1\text{and}\Delta \text{x}=-0.001\\ ⇒\frac{dy}{dx}=\frac{1}{10{x}^{\frac{9}{10}}}\\ Then,\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\Delta y={\left(x+\Delta x\right)}^{\frac{1}{10}}-{\left(x\right)}^{\frac{1}{10}}\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}={\left(0.999\right)}^{\frac{1}{10}}-{\left(1\right)}^{\frac{1}{10}}\\ {\left(0.999\right)}^{\frac{1}{10}}=\Delta y+1\\ \text{Now, dy is approximately equal to}\Delta \text{y and is given by,}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{dy}=\left(\frac{dy}{dx}\right)\Delta x\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{1}{10{x}^{\frac{9}{10}}}×-0.001\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{1}{10{\left(1\right)}^{\frac{9}{10}}}×-0.001\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=-\frac{0.001}{10}\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=-0.0001\\ \text{Hence, the approximate value of}\text{\hspace{0.17em}}{\left(0.999\right)}^{\frac{1}{10}}\text{\hspace{0.17em}}is\\ \text{}1+\left(-0.001\right)=0.9999\\ \left(vi\right)\text{}{\left(15\right)}^{\frac{1}{4}}\\ Let\text{\hspace{0.17em}}y={x}^{\frac{1}{4}}\text{,}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{x}=16\text{and}\Delta \text{x}=-1\\ ⇒\frac{dy}{dx}=\frac{1}{4{x}^{\frac{3}{4}}}\\ \text{Then,}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\Delta y={\left(x+\Delta x\right)}^{\frac{1}{4}}-{x}^{\frac{1}{4}}\\ \text{}\text{}={\left(15\right)}^{\frac{1}{4}}-{\left(16\right)}^{\frac{1}{4}}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\left(15\right)}^{\frac{1}{4}}=\Delta y+2\\ \text{Now, dy is approximately equal to}\Delta \text{y and is given by,}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{dy}=\left(\frac{dy}{dx}\right)\Delta x\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{1}{4{x}^{\frac{3}{4}}}×-1\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{1}{4{\left(16\right)}^{\frac{3}{4}}}×-1\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{1}{4×8}×-1=-0.03125\\ \text{Hence, the approximate value of}\text{\hspace{0.17em}}{\left(15\right)}^{\frac{1}{4}}\text{\hspace{0.17em}}is\text{\hspace{0.17em}}\text{}\\ \text{}\text{2}+\left(-0.03125\right)=1.96875\\ \left(vii\right)\text{}{\left(26\right)}^{\frac{1}{3}}\\ Let\text{\hspace{0.17em}}y={x}^{\frac{1}{3}}\text{,}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{x}=27\text{and}\Delta \text{x}=-1\\ ⇒\frac{dy}{dx}=\frac{1}{3{x}^{\frac{2}{3}}}\\ \text{Then,}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\Delta y={\left(x+\Delta x\right)}^{\frac{1}{3}}-{x}^{\frac{1}{3}}\\ \text{}\text{}={\left(26\right)}^{\frac{1}{3}}-{\left(27\right)}^{\frac{1}{3}}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\left(26\right)}^{\frac{1}{3}}=\Delta y+3\\ \text{Now, dy is approximately equal to}\Delta \text{y and is given by,}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{dy}=\left(\frac{dy}{dx}\right)\Delta x\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{1}{3{x}^{\frac{2}{3}}}×-1\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{1}{3{\left(27\right)}^{\frac{2}{3}}}×-1\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{1}{3×9}×-1\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=-0.0\overline{370}\\ \text{Hence, the approximate value of}\text{\hspace{0.17em}}{\left(26\right)}^{\frac{1}{3}}\text{\hspace{0.17em}}is\text{\hspace{0.17em}}\text{}\\ \text{}\text{3}+\left(-0.0370\right)=\text{2}\text{.9629}\\ \left(viii\right){\left(255\right)}^{\frac{1}{4}}\\ Let\text{\hspace{0.17em}}y={x}^{\frac{1}{4}}\text{,}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{x}=256\text{and}\Delta \text{x}=-1\\ ⇒\frac{dy}{dx}=\frac{1}{4{x}^{\frac{3}{4}}}\\ \text{Then,}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\Delta y={\left(x+\Delta x\right)}^{\frac{1}{4}}-{x}^{\frac{1}{4}}\\ \text{}\text{}={\left(255\right)}^{\frac{1}{4}}-{\left(256\right)}^{\frac{1}{4}}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\left(255\right)}^{\frac{1}{4}}=\Delta y+4\\ \text{Now, dy is approximately equal to}\Delta \text{y and is given by,}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{dy}=\left(\frac{dy}{dx}\right)\Delta x\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{1}{4{x}^{\frac{3}{4}}}×\left(-1\right)\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{1}{4{\left(256\right)}^{\frac{3}{4}}}×\left(-1\right)\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{-1}{4×{4}^{3}}=-0.0039\\ \text{Hence, the approximate value of}\text{\hspace{0.17em}}{\left(255\right)}^{\frac{1}{4}}\text{\hspace{0.17em}}is\\ \text{}\text{4}+\text{(}-\text{0}\text{.0039)}=\text{3}\text{.9961}\end{array}$ $\begin{array}{l}\left(ix\right)\text{\hspace{0.17em}}{\left(82\right)}^{\frac{1}{4}}\\ Let\text{\hspace{0.17em}}y={x}^{\frac{1}{4}}\text{,}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{x}=81\text{and}\Delta \text{x}=1\\ ⇒\frac{dy}{dx}=\frac{1}{4{x}^{\frac{3}{4}}}\\ \text{Then,}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\Delta y={\left(x+\Delta x\right)}^{\frac{1}{4}}-{x}^{\frac{1}{4}}\\ \text{}\text{}={\left(82\right)}^{\frac{1}{4}}-{\left(81\right)}^{\frac{1}{4}}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\left(82\right)}^{\frac{1}{4}}=\Delta y+3\\ \text{Now, dy is approximately equal to}\Delta \text{y and is given by,}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{dy}=\left(\frac{dy}{dx}\right)\Delta x\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{1}{4{x}^{\frac{3}{4}}}×\left(1\right)\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{1}{4{\left(81\right)}^{\frac{3}{4}}}×\left(1\right)\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{1}{4×{3}^{3}}=0.009\\ \text{Hence, the approximate value of}\text{\hspace{0.17em}}{\left(82\right)}^{\frac{1}{4}}\text{\hspace{0.17em}}is\\ \text{}\text{3}+\text{(0}\text{.009)}=\text{3}\text{.009}\\ \left(x\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\left(401\right)}^{\frac{1}{2}}\\ Let\text{\hspace{0.17em}}y={x}^{\frac{1}{2}}\text{,}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{x}=\text{400 and}\Delta \text{x}=\text{1}\\ ⇒\frac{dy}{dx}=\frac{1}{2\sqrt{x}}\\ \text{Then,}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\Delta y=\sqrt{x+\Delta x}-\sqrt{x}\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\sqrt{401}-\sqrt{400}\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\sqrt{401}-20\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\sqrt{401}=\Delta y+20\\ \text{Now, dy is approximately equal to}\Delta \text{y and is given by,}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{dy}=\left(\frac{dy}{dx}\right)\Delta x\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{1}{2\sqrt{x}}\left(1\right)\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{1}{2\sqrt{400}}\left(1\right)\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{1}{2×20}×\left(1\right)\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=0.025\\ \text{Hence, the approximate value of}\text{\hspace{0.17em}}\sqrt{401}\text{\hspace{0.17em}}\text{is 20}+\text{0}\text{.025}=\text{20}\text{.025}\text{.}\\ \left(xi\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\left(0.0037\right)}^{\frac{1}{2}}\\ Let\text{\hspace{0.17em}}y={x}^{\frac{1}{2}}\text{,}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{x}=0.0036\text{and}\Delta \text{x}=0.0001\\ ⇒\frac{dy}{dx}=\frac{1}{2\sqrt{x}}\\ \text{Then,}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\Delta y=\sqrt{x+\Delta x}-\sqrt{x}\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\sqrt{0.0037}-\sqrt{0.0036}\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\sqrt{0.0037}-0.06\\ \text{\hspace{0.17em}}\sqrt{0.0037}=\Delta y+0.06\\ \text{Now, dy is approximately equal to}\Delta \text{y and is given by,}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{dy}=\left(\frac{dy}{dx}\right)\Delta x\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{1}{2\sqrt{x}}\left(0.0001\right)\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{1}{2\sqrt{0.0036}}\left(0.0001\right)\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{1}{2×0.06}×\left(0.0001\right)\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=0.00083\\ \text{Hence, the approximate value of}\text{\hspace{0.17em}}\sqrt{0.0037}\text{\hspace{0.17em}}\text{is}\\ \text{}\text{}0.06+0.00083=\text{0}\text{.06083}\\ \left(xii\right)\text{\hspace{0.17em}}{\left(26.57\right)}^{\frac{1}{3}}\\ Let\text{\hspace{0.17em}}y={x}^{\frac{1}{3}}\text{,}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{x}=27\text{and}\Delta \text{x}=-0.43\\ ⇒\frac{dy}{dx}=\frac{1}{3{x}^{\frac{2}{3}}}\\ \text{Then,}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\Delta y={\left(x+\Delta x\right)}^{\frac{1}{3}}-{x}^{\frac{1}{3}}\\ \text{}\text{}={\left(26.57\right)}^{\frac{1}{3}}-{\left(27\right)}^{\frac{1}{3}}\\ \text{\hspace{0.17em}}{\left(26.57\right)}^{\frac{1}{3}}=\Delta y+3\\ \text{Now, dy is approximately equal to}\Delta \text{y and is given by,}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{dy}=\left(\frac{dy}{dx}\right)\Delta x\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{1}{3{x}^{\frac{2}{3}}}×-0.43\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{1}{3{\left(27\right)}^{\frac{2}{3}}}×-0.43\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{1}{3\left(9\right)}×-0.43=-0.015\\ \text{Hence, the approximate value of}\text{\hspace{0.17em}}{\left(26.57\right)}^{\frac{1}{3}}\text{\hspace{0.17em}}\text{is}\\ \text{}\text{}3-0.015=\text{2}\text{.984}\\ \left(xiii\right)\text{\hspace{0.17em}}{\left(81.5\right)}^{\frac{1}{4}}\\ Let\text{\hspace{0.17em}}y={x}^{\frac{1}{4}}\text{,}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{x}=81\text{and}\Delta \text{x}=0.5\\ ⇒\frac{dy}{dx}=\frac{1}{4{x}^{\frac{3}{4}}}\\ \text{Then,}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\Delta y={\left(x+\Delta x\right)}^{\frac{1}{4}}-{x}^{\frac{1}{4}}\end{array}$