NCERT Solutions Class 12 Maths Chapter 6

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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:

Chapter 6: Exercise 6.1 Solutions – 18 Questions (10 Long, 6 Short, 2 MCQs)
Chapter 6: Exercise 6.2 Solutions – 19 Questions (10 Long, 7 Short, 2 MCQs)
Chapter 6: Exercise 6.3 Solutions – 27 Questions (14 Long, 11 Short, 2 MCQs)
Chapter 6: Exercise 6.4 Solutions – 9 Questions (7 Short, 2 MCQs)
Chapter 6: Exercise 6.5 Solutions – 29 Questions (15 Long, 11 Short, 3 MCQs)

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.

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

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.

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

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.

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

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.

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

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.

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

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.

$\begin{array}{l} Let the circumference of a circle be C with radius (r) is \\ given byC = 2 πr \\ Then, the rate of change of circumference (C) with respect \\ to time (t) is given by, \\ \frac{dC}{dt} = \frac{d}{dt} 2 πr \\ = 2 π \frac{dr}{dt} [By chain rule] \\ \Rightarrow \frac{dC}{dt} = 2 π (0 .7) [∵ \frac{dr}{dt} = 0 .7 cm / s] \\ \Rightarrow \frac{dC}{dt} = 1 .4 π \\ Hence, the rate of increase of the circumference is 1 .4 π cm / s . \end{array}$

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.

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

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} The volume of a sphere (V) with radius (r) is given by, \\ V = \frac{4}{3} {πr}^{3} \\ Differentiating both sides, with respect to t, we get \\ \frac{dv}{dt} = \frac{d}{dt} \frac{4}{3} {πr}^{3} \\ = \frac{4}{3} π \times 3 r^{2} \frac{dr}{dt} \\ 900 = 4 {πr}^{2} \frac{dr}{dt} [∵ \frac{dV}{dt} = 900 {cm}^{3} / \sec] \\ \Rightarrow \frac{dr}{dt} = \frac{900}{4 {πr}^{2}} \\ = \frac{225}{π {(15)}^{2}} [∵ r = 15 ​ cm] \\ = \frac{225}{π (225)} = \frac{1}{π} \\ Hence, the rate at which the radius of the balloon increases \\ when the radius is 15 cm is \frac{1}{π} cm / \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} Volume of sphere having radius r = \frac{4}{3} {πr}^{3} \\ \Rightarrow V = \frac{4}{3} {πr}^{3} \\ Differentiating with respect to r, we get \\ \Rightarrow \frac{dV}{dr} = \frac{d}{dr} (\frac{4}{3} {πr}^{3}) \\ = \frac{4}{3} π \frac{d}{dr} r^{3} \\ = \frac{4}{3} π \times 3 r^{2} \\ = 4 {πr}^{2} \\ = 4 π {(10)}^{2} \\ = 400 π \\ Hence, the volu e of the balloon is increasing at the rate \\ of 400 π^{} . \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.

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

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} The equation of the curve is given as : \\ 6 y = x^{3} + 2 \\ The rate of change of the position of the particle with \\ respect to time (t) is given by, \\ 6 \frac{dy}{dt} = \frac{d}{dt} (x^{3} + 2) \\ 6 \frac{dy}{dt} = 3 x^{2} \frac{dx}{dt} \\ When the y - coordinate of the particle changes 8 times as fast \\ as the x - coordinate i . e ., \frac{dy}{dt} = 8 \frac{dx}{dt}, we have \\ 6 (8 \frac{dx}{dt}) = 3 x^{2} \frac{dx}{dt} \\ \Rightarrow (16 - x^{2}) \frac{dx}{dt} = 0 \\ \Rightarrow 16 - x^{2} = 0 \\ \Rightarrow x = \pm 4 \\ When x = 4, \\ y = \frac{{(4)}^{3} + 2}{6} = 11 \\ When x = - 4, \\ y = \frac{{(- 4)}^{3} + 2}{6} \\ = - \frac{62}{6} = - \frac{31}{3} \\ Hence, the points required on the curve are (4, 11) and (- 4, - \frac{31}{3}) . \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} The air bubble is in the shape of a sphere . \\ So, the volume of an air bubble (V) with radius (r) is given by, \\ V = \frac{4}{3} {πr}^{3} \\ The rate of change of volume (V) with respect to time (t) is \\ obtained by differentiating both sides with respect to t, so \\ \frac{dV}{dt} = \frac{d}{dt} (\frac{4}{3} {πr}^{3}) \\ \frac{dV}{dt} = \frac{4}{3} \times 3 {πr}^{2} \frac{dr}{dt} \\ = 4 π {(1)}^{2} (\frac{1}{2}) [\begin{array}{l} ∵ \frac{dr}{dt} = \frac{1}{2} cm / s \\ and r = 1 cm \end{array}] \\ = 2 π {cm}^{3} / s \\ Hence, the rate at which the volume of the bubble increases \\ is 2 π {cm}^{3} / 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} The volume (V) of a sphere with radius (r) is given by, \\ V = \frac{4}{3} {πr}^{3} \\ Diameter of balloon = \frac{3}{2} (2 x +1) \\ So, the radiusof balloon = \frac{3}{4} (2 x +1) \\ ∴ V = \frac{4}{3} π {\{\frac{3}{4} (2 x +1)\}}^{3} \\ = \frac{4}{3} π \times \frac{27}{64} (2 x {+1)}^{3} \\ V = \frac{9}{16} π (2 x {+1)}^{3} \\ Hence, the rate of change of volume with respect to x is \\ given by \\ \frac{dV}{dx} = \frac{9}{16} π \frac{d}{dx} (2 x {+1)}^{3} \\ = \frac{9}{16} π \times3 (2 x {+1)}^{2} \frac{d}{dx} (2 x +1) \\ [By chain rule] \\ \frac{dV}{dx} = \frac{27}{16} π (2 x {+1)}^{2} \times 2 \\ \Rightarrow \frac{dV}{dx} = \frac{27}{8} π (2 x {+1)}^{2} \\ Thus, the rate of change of volume with respect to x is \\ \frac{27}{8} π (2 x {+1)}^{2} . \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 one–sixth 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} The volume of a cone with radius (r) and height (h) is given by, \\ V = \frac{1}{3} {πr}^{2} h \\ Given, height of cone = \frac{1}{6} r \Rightarrow h = \frac{1}{6} r \Rightarrow r = 6 h \\ ∴ V = \frac{1}{3} π {(6 h)}^{2} h \\ = 12 {πh}^{3} \\ Differentiating both sides with respect to time (t), we get \\ \frac{dV}{dt} = \frac{d}{dt} (12 {πh}^{3}) \\ = 12 π \frac{d}{dt} h^{3} \\ = 36 {πh}^{2} \frac{dh}{dt} [By chain rule] \\ 12 = 36 π {(4)}^{2} \frac{dh}{dt} [∵ \frac{dV}{dt} = 12 and h = 4 cm] \\ \frac{dh}{dt} = \frac{12}{36 π {(4)}^{2}} = \frac{1}{48 π} \\ Hence, when the height of the sand cone is 4 cm, its height is \\ increasing at the rate of \frac{1}{48 π} cm / s . \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.

$\begin{array}{l} The total cost C (x) = 0.007 x^{3} - 0.003 x^{2} + 15 x + 4000 \\ Marginal cost is the rate of change of total cost with respect \\ to output . \\ ∴ Marginal cost (MC) = \frac{dC}{dx} \\ = 0.007 (3 x^{2}) - 0.003 (2 x) + 15 \\ When x = 17, MC = 0.021 {(17)}^{2} - 0.006 (17) + 15 \\ = 6 .069 - 0 .102 + 15 \\ = 20 .967 \\ Hence, when 17 units are produced, the marginal cost \\ is Rs . 20.967 . \end{array}$

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.

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

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.

$\begin{array}{l} T h e a r e a (A) o f a c i r c l e w i t h r a d i u s (r) i s g i v e n b y, \\ A = π r^{2} \\ D i f f e r e n t i a t i n g b o t h s i d e s w i t h r e s p e c t t o r, w e g e t \\ \frac{d A}{d r} = \frac{d}{d r} π r^{2} \\ = 2 π r \\ {(\frac{d A}{d r})}_{r = 6} = 2 π (6) \\ = 12 π \\ H e n c e, t h e r e q u i r e d r a t e o f c h a n g e o f t h e a r e a o f a c i r c l e \\ i s 12 π . \\ T h e c o r r e c t a n s w e r i s B . \end{array}$

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.

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

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

Ans.

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

Q.20 Show that the function given by f(x) = e^2x is strictly increasing on R

Ans.

\begin{array}{l} Let x_{1} and x_{2} be any two numbers in R . \\ Then, we have : \\ x_{1} < x_{2} \Rightarrow 2 x_{1} < 2 x_{2} \\ \Rightarrow e^{2 x_{1}} < e^{2 x_{2}} \\ \Rightarrow f (x_{1}) < f (x_{2}) \\ Hence, f is strictly increasing on R . \end{array}

Q.21

\begin{array}{l} Show that the function given by f(x) = sin x is \\ (a) strictly increasing in (0, \frac{π}{2}), \\ (b) strictly decreasing in (\frac{π}{2}, π), \\ (c) neither increasing nor decreasing in (0, π) \end{array}

Ans.

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

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

$\begin{array}{l} In the interval (- \infty, \frac{3}{4}), \\ f’ (x) = 4 x - 3 < 0 \\ T h e r e f o r e, f is strictly deacreasing in this interval . Also, in \\ the interval (\frac{3}{4}, \infty), f’ (x) > 0 and so the function f is strictly \\ increasing in this interval . \end{array}$

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.

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

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

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} (a) We have, \\ f (x) = x^{2} + 2 x - 5 \\ Differentiating both sides with respect to x, we get \\ f ‘ (x) = 2 x + 2 \\ ∴ f ‘ (x) = 0 \Rightarrow 2 x + 2 = 0 \\ \Rightarrow x = - 1 \\ Point x = - 1 divides the real line into two disjoint \\ intervals i . e ., (- \infty, - 1) and (- 1, \infty) . \\ In interval (- \infty, - 1), f ‘ (x) = 2 x + 2 < 0 \\ ∴ f is strictly decreasing in interval (- \infty, - 1) . \\ Thus, f is strictly decreasing for x < - 1. \\ In interval (- 1, \infty), f ‘ (x) = 2 x + 2 > 0 \\ ∴ f is strictly increasing in interval (- 1, \infty) . \\ Thus, f is strictly increasing for x > - 1. \\ (b) We have, \\ f (x) = 10 - 6 x - 2 x^{2} \\ Differentiating both sides with respect to x, we get \\ f ‘ (x) = - 6 - 4 x \\ ∴ f ‘ (x) = 0 \Rightarrow - 6 - 4 x = 0 \\ \Rightarrow x = - \frac{3}{2} \\ Point x = - \frac{3}{2} divides the real line into two disjoint intervals \\ i . e ., (- \infty, - \frac{3}{2}) and (- \frac{3}{2}, \infty) . \\ In interval (- \infty, - \frac{3}{2}), f ‘ (x) = - 6 - 4 x > 0 \\ ∴ f is strictly increasing in interval (- \infty, - \frac{3}{2}) . \\ Thus, f is strictly increasing for x < - \frac{3}{2} . \\ In interval (- \frac{3}{2}, \infty), f ‘ (x) = - 6 - 4 x < 0 \\ ∴ f is strictly decreasing in interval (- \frac{3}{2}, \infty) . \\ Thus, f is strictly decreasing for x > - \frac{3}{2} . \\ (c) We have, \\ f (x) = - 2 x^{3} - 9 x^{2} - 12 x + 1 \\ Differentiating both sides with respect to x, we get \\ f ‘ (x) = - 6 x^{2} - 18 x - 12 = - 6 (x^{2} + 3 x + 2) \\ ∴ f ‘ (x) = 0 \Rightarrow - 6 (x^{2} + 3 x + 2) = 0 \\ \Rightarrow (x + 1) (x + 2) = 0 \\ \Rightarrow x = - 1, - 2 \\ The points x = - 1 and x = - 2 divide the real line into three \\ disjoint intervals i . e ., (- \infty, - 2), (- 2, - 1) and (- 1, \infty) . \\ In intervals (- \infty, - 2) and (- 1, \infty), f ‘ (x) is negative . \\ Hence, the given function (f) is strictly decreasing in intervals \\ (- \infty, - 2) and (- 1, \infty), \\ i . e ., function (f) is strictly decreasing when x < - 2 and x > - 1. \end{array}$ $\begin{array}{l} In intervals (- 2, - 1), f ‘ (x) is positive . \\ So, function (f) is strictly increasing in interval (- 2, - 1), \\ i . e ., function (f) is strictly increasing w h e n - 2 < x < - 1. \\ (d) We have, \\ f (x) = 6 - 9x - x^{2} \\ Differentiating both sides with respect to x, we get \\ f’ (x) = - 9 - 2 x \\ ∴ f’ (x) = 0 \Rightarrow - 9 - 2 x = 0 \\ \Rightarrow x = - \frac{9}{2} \\ Point x = - \frac{9}{2} divides the real line into two disjoint intervals \\ i .e ., (- \infty, - \frac{9}{2}) a n d (- \frac{9}{2}, \infty) . \\ In interval (- \infty, - \frac{9}{2}), f ‘ (x) = - 9 - 2 x > 0 \\ ∴ f is strictly increasing in interval (- \infty, - \frac{9}{2}) . \\ Thus, f is strictly increasing for x < - \frac{9}{2} . \\ In interval (- \frac{9}{2}, \infty), f ‘ (x) = - 9 - 2 x < 0 \\ ∴ f is strictly decreasing in interval (- \frac{9}{2}, \infty) . \\ Thus, f is strictly decreasing for x > - \frac{9}{2} . \end{array}$ $\begin{array}{l} (e) We have, \\ f(x) = {(x + 1)}^{3} (x - {3)}^{3} \\ D i f f e r e n t i a t i n g both sides with respect to x, we get \\ f’ (x) = {(x + 1)}^{3} \frac{d}{d x} {(x - 3)}^{3} + {(x - 3)}^{3} \frac{d}{d x} {(x + 1)}^{3} \\ [B y product rule] \\ \Rightarrow f’ (x) = {(x + 1)}^{3} 3 {(x - 3)}^{2} \frac{d}{d x} (x - 3) + {(x - 3)}^{3} 3 {(x + 1)}^{2} \frac{d}{d x} (x + 1) \\ [B y chain rule] \\ \Rightarrow f’ (x) = 3 {(x + 1)}^{3} {(x - 3)}^{2} (1 - 0) + 3 {(x - 3)}^{3} {(x + 1)}^{2} (1 + 0) \\ \Rightarrow f’ (x) = 3 {(x + 1)}^{2} {(x - 3)}^{2} (x + 1 + x - 3) \\ \Rightarrow f’ (x) = 3 {(x + 1)}^{2} {(x - 3)}^{2} (2 x - 2) \\ = 6 {(x + 1)}^{2} {(x - 3)}^{2} (x - 1) \\ N o w, f’ (x) = 0 \Rightarrow x = - 1, 1, 3 \\ The points x = - 1, x = 1, a n d x = 3 divide the real line into \\ four disjoint intervals i .e ., (- \infty, - 1), (- 1, 1), (1, 3) and (3, \infty) . \\ In intervals (- \infty, - 1) and (- 1, 1), \end{array}$ $\begin{array}{l} f’ (x) = 6 {(x + 1)}^{2} {(x - 3)}^{2} (x - 1) < 0 \\ ∴ f is strictly decreasing in intervals (- \infty, - 1) and (- 1, 1) . \\ In intervals (1, 3) and (3, \infty), f’ (x) = 6 {(x + 1)}^{2} {(x - 3)}^{2} (x - 1) > 0 \\ ∴ f is strictly increasing in intervals (1, 3) and (3, \infty) . \end{array}$

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

Ans.

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

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

Ans.

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

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

Ans.

$\begin{array}{l} We have, y = \frac{4 sinθ}{(2 + cosθ)} - θ \\ Differentiating w . r . t . θ, we get \\ \frac{dy}{dθ} = \frac{d}{dθ} {\frac{4 sinθ}{(2 + cosθ)} - θ} \\ = \frac{(2 + cosθ) \frac{d}{dθ} (4 sinθ) - 4 sinθ \frac{d}{dθ} (2 + cosθ)}{{(2 + cosθ)}^{2}} - \frac{d}{dθ} θ \\ = \frac{4 (2 + cosθ) cosθ - 4 sinθ (0 - sinθ)}{{(2 + cosθ)}^{2}} - 1 \\ = \frac{4 (2 + cosθ) cosθ + 4 \sin^{2} θ}{{(2 + cosθ)}^{2}} - 1 \\ = \frac{8 cosθ + 4 (\cos^{2} θ + \sin^{2} θ)}{{(2 + cosθ)}^{2}} - 1 \\ = \frac{8 cosθ + 4}{{(2 + cosθ)}^{2}} - 1 \\ Now, \frac{dy}{dθ} = 0 \\ \Rightarrow \frac{8 cosθ + 4}{{(2 + cosθ)}^{2}} - 1 = 0 \Rightarrow \frac{8 cosθ + 4}{{(2 + cosθ)}^{2}} = 1 \\ \Rightarrow 8 cosθ + 4 = {(2 + cosθ)}^{2} \\ \Rightarrow 8 cosθ + 4 = 4 + 4 cosθ + \cos^{2} θ \\ \Rightarrow \cos^{2} θ - 4 cosθ = 0 \\ \Rightarrow cosθ (cosθ - 4) = 0 \\ \Rightarrow cosθ = 0 or cosθ = 4 \\ Since \cos θ \neq 4 \\ ∴ cosθ = 0 \Rightarrow θ = \frac{π}{2} \\ Now, \frac{dy}{dθ} = \frac{8 cosθ + 4}{{(2 + cosθ)}^{2}} - 1 \\ = \frac{8 cosθ + 4 - {(2 + cosθ)}^{2}}{{(2 + cosθ)}^{2}} \\ = \frac{8 cosθ + 4 - 4 - 4 cosθ - \cos^{2} θ}{{(2 + cosθ)}^{2}} \\ = \frac{cosθ (4 - cosθ)}{{(2 + cosθ)}^{2}} \\ In interval (0, \frac{π}{2}), we have cosθ > 0 . Also, 4 > cosθ \\ \Rightarrow 4 - cosθ > 0 . \\ ∴ cosθ (4 - cosθ) > 0 and also {(2 + cosθ)}^{2} > 0 \\ \Rightarrow \frac{cosθ (4 - cosθ)}{{(2 + cosθ)}^{2}} > 0 \Rightarrow \frac{dy}{dθ} > 0 \\ Therefore, y is strictly increasing in interval (0, \frac{π}{2}) . \\ Also, the given function is continuous at x = 0 and x = \frac{π}{2} . \\ Hence, y is increasing in interval (0, \frac{π}{2}) . \end{array}$

Q.28 Provethatthelogarithmicfunctionisstrictly increasingon (0, ∞).

Ans.

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

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} {The given function is f(x) = x}^{2} - x + 1 . \\ D i f f e r e n t i a t i n g with respect to x, we get \\ f’ (x) = 2 x - 1 \\ N o w, f ‘ (x) = 0 \Rightarrow 2 x - 1 = 0 \Rightarrow x = \frac{1}{2} \\ The point \frac{1}{2} divides the interval (- 1, 1) into two disjoint \\ intervals i .e ., (- 1, \frac{1}{2}) a n d (\frac{1}{2}, 1) . \\ Now, for interval (- 1, \frac{1}{2}), f ‘ (x) = 2 x - 1 < 0 \\ Therefore, f is strictly decreasing in interval (- 1, \frac{1}{2}) . \\ Now, for interval (\frac{1}{2}, 1), f ‘ (x) = 2 x - 1 > 0 \\ So, f is strictly increasing in interval (\frac{1}{2}, 1) . \\ Hence, f is neither strictly increasing nor decreasing in \\ interval (- 1, 1) . \end{array}

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

Ans.

$\begin{array}{l} (A) L e t f_{1} (x) = \cos x \\ Differentiating w .r .t . x, we get \\ {f’}_{1} (x) = \frac{d}{d x} \cos x \\ = - \sin x \\ I n interval (0, \frac{π}{2}), sinx is positive in first quadrant . \\ So, {f’}_{1} (x) < 0 \\ T h e r e f o r e, f_{1} (x) = \cos x is strictly decreasing in interval (0, \frac{π}{2}) . \\ (B) L e t f_{2} (x) = \cos 2 x \\ Differentiating w .r .t . x, we get \\ {f’}_{2} (x) = \frac{d}{d x} \cos 2 x \\ = - 2 \sin 2 x \\ I n interval (0, \frac{π}{2}), sin2x is positive in first quadrant . \\ So, {f’}_{2} (x) < 0 \\ T h e r e f o r e, f_{2} (x) = \cos 2 x is strictly decreasing in interval (0, \frac{π}{2}) . \\ (C) L e t f_{3} (x) = \cos 3 x \\ Differentiating w .r .t . x, we get \\ {f’}_{3} (x) = \frac{d}{d x} \cos 3 x \\ = - 3 \sin 3 x \\ N o w, {f’}_{3} (x) = 0 \Rightarrow - 3 \sin 3 x = 0 \\ \Rightarrow 3 x = π \Rightarrow x = \frac{π}{3} \\ The point x = \frac{π}{3} divides the interval (0, \frac{π}{2}) into two disjoint \\ intervals i .e ., (0, \frac{π}{3}) and (\frac{π}{3}, \frac{π}{2}) . \\ N o w, in interval (0, \frac{π}{3}), \\ {f’}_{3} (x) = - 3 \sin 3 x < 0 [∵ 0 < x < \frac{π}{3} \Rightarrow 0 < 3 x < π] \\ ∴ f_{3} is strictly decreasing in interval (0, \frac{π}{3}) . \\ N o w, i n interval (\frac{π}{3}, \frac{π}{2}), \\ \frac{π}{3} < x < \frac{π}{2} \Rightarrow π < 3 x < \frac{3 π}{2} \\ S o, sin3x is in third quadrant, where sin3x is negative . \\ ∴ f_{3} (x) = - 3 \sin 3 x > 0 \\ T h e r e f o r e, f_{3} is strictly increasing in interval (\frac{π}{3}, \frac{π}{2}) . \\ {Hence, f}_{3} is neither increasing nor decreasing in interval (0, \frac{π}{2}) . \\ (D) L e t f_{4} (x) = \tan x \\ D i f f e r e n t i a t i n g w .r .t . x, we get \\ {f’}_{4} (x) = \frac{d}{d x} \tan x \\ = \sec^{2} x \\ I n int e r v a l (0, \frac{π}{2}), \sec x > 0 \Rightarrow \sec^{2} x > 0 \\ ∴ f_{4} (x) > 0 \\ ∴ f4 is strictly increasing in interval (0, \frac{π}{2}) . \\ Therefore, functions cos x and cos 2x are strictly decreasing \\ in (0, \frac{π}{2}) . \\ Hence, the correct answers are A and B . \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.

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

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.

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

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.

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

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

Ans.

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

Q.35

\begin{array}{l} Prove that the function f given by f(x) = logcos x is strictly \\ decreasing on (0, \frac{π}{2}) and strictly increasing on (\frac{π}{2}, π) . \end{array}

Ans.

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

Q.36 Prove that the function f given byf(x) = x2−3x2+3x−100is increasinginR.

Ans.

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

Q.37 The interval in which y=x2e−xis increasing is(A) (−∞, ∞) (B )(−2, 0) (C) (2, ∞) (D) (0, 2)

Ans.

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

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

Ans.

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

Q.39

\begin{array}{l} Find the slope of the tangent to the curve y = \frac{x - 1}{x - 2}, x \neq 2 \\ a t x = 10 . \end{array}

Ans.

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

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

Ans.

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

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

Ans.

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

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

Ans.

\begin{array}{l} Given that x = {acos}^{3} θ and y = {asin}^{3} θ \\ Differentiating w . r . t . θ, we get \\ \frac{dx}{dθ} = a \frac{d}{dθ} \cos^{3} θ \\ = 3 {acos}^{2} θ \frac{d}{dθ} cosθ = - 3 {acos}^{2} θsinθ \\ \frac{dy}{dθ} = a \frac{d}{dθ} \sin^{3} θ \\ = 3 {asin}^{2} θ \frac{d}{dθ} sinθ \\ = 3 {asin}^{2} θcosθ \\ ∴ \frac{dy}{dx} = \frac{(\frac{dy}{dθ})}{(\frac{dx}{dθ})} \\ = \frac{3 {asin}^{2} θcosθ}{- 3 {acos}^{2} θsinθ} = - tanθ \\ Therefore, the slope of the tangent at θ = \frac{π}{4}, is \\ {(\frac{dy}{dx})}_{(θ = \frac{π}{4})} = {(- tanθ)}_{(θ = \frac{π}{4})} \\ = - \tan \frac{π}{4} \\ = - 1 \\ Hence, the slope of the normal at θ = \frac{π}{4}, is \\ slope of normal (M) = - \frac{1}{slope of tangent (m)} \\ = - \frac{1}{- 1} \\ = 1 \end{array}

Q.43

\begin{array}{l} Find the slope of the normal to the curve x = 1 - a s i n θ, \\ y = b c o s^{2} θ a t θ = \frac{π}{2} \end{array}

Ans.

$\begin{array}{l} Given that : \\ x = 1 - asinθ and y = {bcos}^{2} θ \\ Differentiating w . r . t . θ, we get \\ \frac{dx}{dθ} = \frac{d}{dθ} (1 - asinθ) \\ = 0 - acosθ \\ = - acosθ \\ \frac{dy}{dθ} = \frac{d}{dθ} ({bcos}^{2} θ) \\ = 2 bcosθ \frac{d}{dθ} cosθ \\ = - 2 bcosθsinθ \\ ∴ \frac{dy}{dx} = \frac{(\frac{dy}{dθ})}{(\frac{dx}{dθ})} \\ = \frac{- 2 bcosθsinθ}{- acosθ} \\ = \frac{2 b}{a} sinθ \\ Slope of tangent at θ = \frac{π}{2}, is given by \\ m = {(\frac{dy}{dx})}_{θ = \frac{π}{2}} \\ = \frac{2 b}{a} \sin \frac{π}{2} = \frac{2 b}{a} \\ slope of normal \\ (M) = - \frac{1}{m} \\ = - \frac{1}{(\frac{2 b}{a})} \\ = - \frac{a}{2 b} \\ Thus, slope of normal is - \frac{a}{2 b} . \end{array}$

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

Ans.

$\begin{array}{l} The equation of the given curve is \\ y = x^{3} - {3x}^{2} - 9x + 7 \\ Differentiating w .r .t . x, we get \\ \frac{d y}{d x} = \frac{d}{d x} (x^{3} - {3x}^{2} - 9x + 7) \\ = 3 x^{2} - 6 x - 9 \\ Now, the tangent is parallel to the x-axis if the slope of the \\ tangent is zero . \\ 3 x^{2} - 6 x - 9 = 0 \Rightarrow 3 (x^{2} - 2 x - 3) = 0 \\ \Rightarrow 3 (x - 3) (x + 1) = 0 \\ \Rightarrow x = 3, - 1 \\ W h e n x = 3, then \\ y = {(3)}^{3} - 3 {(3)}^{2} - 9 (3) + 7 \\ = 27 - 27 - 27 + 7 \\ = - 20 \\ W h e n x = - 1, then \\ y = {(- 1)}^{3} - 3 {(- 1)}^{2} - 9 (- 1) + 7 \\ = - 1 - 3 + 9 + 7 \\ = 12 \\ Hence, the points at which the tangent is parallel to the x-axis \\ are (3, - 20) and (- 1, 12) . \end{array}$

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.

$\begin{array}{l} G i v e n curve is \\ y = {(x - 2)}^{2} \\ D i f f e r e n t i a t i n g both sides, w .r .t . x, we get \\ \frac{d y}{d x} = \frac{d}{d x} {(x - 2)}^{2} \\ = 2 (x - 2) \frac{d}{d x} (x - 2) \\ = 2 (x - 2) \\ The slope of the chord \\ = \frac{4 - 0}{4 - 2} [∵ m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}] \\ = 2 \\ Since, \\ the slope of the tangent = slope of the chord, we have: \\ 2 (x - 2) = 2 \\ \Rightarrow x = 3 \\ W h e n x = 3, y = {(3 - 2)}^{2} = 1 \\ Hence, the required point is (3, 1) . \end{array}$

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

Ans.

$\begin{array}{l} The equation of the given curve is \\ y = x^{3} - 11 x + 5 \\ The equation of the tangent to the given curve is \\ y = x - 11 \\ comparing it with y = mx + c, we get \\ m = 1 \\ ∴ slope of the tangent (m) = 1 \\ Now, the slope of the tangent to the given curve at the \\ point (x, y) is, \\ \frac{d y}{d x} = \frac{d}{d x} (x^{3} - 11 x + 5) \\ = 3 x^{2} - 11 \\ ∵ 3 x^{2} - 11 = 1 [G i v e n] \\ \Rightarrow x^{2} = \frac{12}{3} = 4 \\ \Rightarrow x = \pm 2 \\ When x = 2, y = {(2)}^{3} - 11(2) + 5 \\ = 8 - 22 + 5 \\ = - 9 . \\ When x = - 2, y = (- 2)3 - 11 (- 2) + 5 \\ = - 8 + 22 + 5 \\ = 19 . \\ Hence, the required points are (2, - 9) and (- 2, 19) . \end{array}$

Q.47

\begin{array}{l} Find the equation of all lines having slope - 1 that are \\ tangents to the curve y = \frac{1}{x - 1}, x \neq 1 . \end{array}

Ans.

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

Q.48

\begin{array}{l} Find the equation of all lines having slope 2 which are \\ tangents to the curve y = \frac{1}{x - 3}, x \neq 3 . \end{array}

Ans.

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

Q.49

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

Ans.

$\begin{array}{l} The equation of the given curve is y = \frac{1}{x^{2} - 2 x + 3} \\ The slope of the tangent to the given curve at any point (x, y) \\ {(\frac{d y}{d x})}_{(x, y)} = \frac{d}{d x} \frac{1}{x^{2} - 2 x + 3} \\ = - \frac{1}{{(x^{2} - 2 x + 3)}^{2}} \frac{d}{d x} (x^{2} - 2 x + 3) \\ = - \frac{(2 x - 2)}{{(x^{2} - 2 x + 3)}^{2}} \\ Since, the slope of the tangent is 0, then we have: \\ - \frac{(2 x - 2)}{{(x^{2} - 2 x + 3)}^{2}} = 0 \Rightarrow (2 x - 2) = 0 \\ \Rightarrow x = 1 \\ W h e n x = 1, y = \frac{1}{1 - 2 + 3} = \frac{1}{2} \\ T h e point on the curve is (1, \frac{1}{2}) . \\ ∴ The equation of the tangent through (1, \frac{1}{2}) i s \\ y - \frac{1}{2} = 0 (x - 1) \\ y = \frac{1}{2} \\ Hence, the equation of the required line is y = \frac{1}{2} . \end{array}$

Q.50

\begin{array}{l} Findpointsonthe curve \frac{x^{2}}{9} + \frac{y^{2}}{16} = 1 atwhichthe tangents \\ are (i) paralleltox - axis (ii) paralleltoy - axis . \end{array}

Ans.

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

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)

$(v) x = c o s t, y = s i n t a t t = \frac{π}{4} .$

Ans.

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

$\begin{array}{l} y - 3 = 2 (x - 1) \\ \Rightarrow y - 3 = 2 x - 2 \\ \Rightarrow y = 2 x + 1 \\ T h e s l o p e o f t h e n o r m a l (M) a t (1, 3) = - \frac{1}{2} \\ T h e r e f o r e, t h e e q u a t i o n o f t h e n o r m a l a t (1, 3) i s g i v e n a s : \\ y - 3 = - \frac{1}{2} (x - 1) [∵ y - y_{1} = M (x - x_{1})] \\ \Rightarrow x + 2 y - 7 = 0 \\ (i i) T h e e q u a t i o n o f t h e c u r v e i s y = x^{3} \\ O n d i f f e r e n t i a t i n g w i t h r e s p e c t t o x, w e g e t : \\ \frac{d y}{d x} = \frac{d}{d x} (x^{3}) = 3 x^{2} \\ {(\frac{d y}{d x})}_{(1, 1)} = 3 {(1)}^{2} = 3 \\ T h u s, t h e s l o p e o f t h e t a n g e n t a t (1, 1) i s 3 . \\ E q u a t i o n o f t a n g e n t a t (1, 1) \\ o f t h e t a n g e n t i s : y - 1 = 3 (x - 1) \Rightarrow y = 3 x - 2 \\ T h e s l o p e o f t h e n o r m a l a t (1, 1) = - \frac{1}{3} \\ T h e r e f o r e, t h e e q u a t i o n o f t h e n o r m a l a t (1, 1) i s : \\ y - 1 = - \frac{1}{3} (x - 1) \\ \Rightarrow x + 3 y - 4 = 0 \\ (i v) T h e e q u a t i o n o f t h e c u r v e i s y = x^{2} \\ O n d i f f e r e n t i a t i n g w i t h r e s p e c t t o x, w e g e t : \end{array}$ $\begin{array}{l} \frac{d y}{d x} = \frac{d}{d x} (x^{2}) = 2 x \\ {(\frac{d y}{d x})}_{(0, 0)} = 2 (0) = 0 \\ E q u a t i o n o f t a n g e n t a t (0, 0) : \\ y - 0 = 0 (x - 0) \\ \Rightarrow y = 0 \\ T h e s l o p e o f t h e n o r m a l a t (0, 0) = - \frac{1}{0} = \frac{1}{0} \\ T h e r e f o r e, t h e e q u a t i o n o f t h e n o r m a l a t (0, 0) i s : \\ y - 0 = \frac{1}{0} (x - 0) \Rightarrow x = 0 \\ (v) T h e e q u a t i o n o f t h e c u r v e i s x = c o s t, y = s i n t . \\ D i f f e r e n t i a t i n g w . r . t . t, w e g e t \\ \frac{d x}{d t} = - \sin t a n d \frac{d y}{d t} = \cos t \\ ∴ \frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}} \\ = \frac{\cos t}{- \sin t} = - \cot t \\ {(\frac{d y}{d x})}_{x = \frac{π}{4}} = - \cot (\frac{π}{4}) \\ = - 1 \\ ∴ E q u a t i o n o f t a n g e n t a t (\cos \frac{π}{4}, \sin \frac{π}{4}) i . e ., (\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}) : \end{array}$ $\begin{array}{l} y - \frac{1}{\sqrt{2}} = - 1 (x - \frac{1}{\sqrt{2}}) \\ \Rightarrow x + y = \sqrt{2} \\ T h e s l o p e o f t h e n o r m a l a t t = \frac{π}{4} : \\ M = - \frac{1}{{(\frac{d y}{d x})}_{x = \frac{π}{4}}} \\ = - \frac{1}{- 1} = 1 \\ T h e r e f o r e, t h e e q u a t i o n o f t h e n o r m a l a t (\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}) : \\ y - \frac{1}{\sqrt{2}} = 1 (x - \frac{1}{\sqrt{2}}) \Rightarrow y = x \end{array}$

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.

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

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

Ans.

$\begin{array}{l} W e have, \\ y = {7x}^{3} + 11 \\ Differentiating w .r .t . x, we get \\ \frac{d y}{d x} = 21 x^{2} \\ S l o p e of tangent at x=2, \\ m_{1} = {(\frac{d y}{d x})}_{x = 2} \\ = 21 {(2)}^{2} \\ = 84 \\ S l o p e of tangent at x= - 2, \\ m_{2} = {(\frac{d y}{d x})}_{x = - 2} \\ = 21 {(- 2)}^{2} \\ = 84 \\ S i n c e, m_{1} = m_{2} \\ Hence, the two tangents are parallel . \end{array}$

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.

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

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

passes through the origin.

Ans.

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

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

Ans.

$\begin{array}{l} The equation of the given curve is \\ x^{2} {+ y}^{2} - 2x - 3 = 0 … (i) \\ Differentiating both sides, w .r .t . x, we get \\ 2x + 2y \frac{d y}{d x} - 2 = 0 \\ \Rightarrow \frac{d y}{d x} = \frac{2 - 2 x}{2 y} \\ = \frac{1 - x}{y} \\ S l o p e of tangent of the given curve at point (x_{1}, y_{1}) i s \\ \Rightarrow {(\frac{d y}{d x})}_{(x_{1}, y_{1})} = \frac{1 - x_{1}}{y_{1}} \\ I t is given that \\ {(\frac{d y}{d x})}_{(x_{1}, y_{1})} = 0 \\ \frac{1 - x_{1}}{y_{1}} = 0 \Rightarrow 1 - x_{1} = 0 \\ \Rightarrow x_{1} = 1 \\ S i n c e, point (x_{1}, y_{1}) lies on curve (i), we get \\ x_{1}^{2} + y_{1}^{2} - {2x}_{1} - 3 = 0 \\ {(1)}^{2} + y_{1}^{2} - 2 (1) - 3 = 0 \\ y_{1}^{2} = 4 \Rightarrow y_{1} = \pm 2 \\ Hence, the points at which the tangents are parallel to the \\ x-axis are (1, 2) and (1, - 2) . \end{array}$

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

Ans.

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

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.

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

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

Ans.

\begin{array}{l} The equation of parabola is y^{2} = 4 ax \\ Differentiating both sides w . r . t . x, we get \\ 2 y \frac{dy}{dx} = 4 a \\ \Rightarrow \frac{dy}{dx} = \frac{4 a}{2 y} = \frac{2 a}{y} \\ Slope of tangent to parabola at point ({at}^{2}, 2 at) \\ \Rightarrow {(\frac{dy}{dx})}_{({at}^{2}, 2 at)} = \frac{2 a}{2 at} = \frac{1}{t} \\ Slope of normal to parabola at point ({at}^{2}, 2 at) \\ = - \frac{1}{{(\frac{dy}{dx})}_{({at}^{2}, 2 at)}} \\ = - \frac{1}{(\frac{1}{t})} = - t \\ Then, the equation of the tangent at ({at}^{2}, 2 at) \\ y - 2 at = \frac{1}{t} (x - {at}^{2}) \\ \Rightarrow ty = x + {at}^{2} \\ And, the equation of the normal at ({at}^{2}, 2 at) \\ y - 2 at = - t (x - {at}^{2}) \\ \Rightarrow tx + y = 2 at + {at}^{3} \\ Thus, the required equations of tangent and normal are \\ ty = x + {at}^{2} and tx + y = 2 at + {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} Given equations of the curves are x = y^{2} and xy = k . \\ Putting x = y^{2} in xy = k, we get \\ y^{2} . y = k \Rightarrow y = k^{\frac{1}{3}} \\ Putting value of y in xy = k, we get \\ k^{\frac{1}{3}} x = k \Rightarrow x = k^{\frac{2}{3}} \\ Thus, the point of intersection of the given curves is (k^{\frac{2}{3}}, k^{\frac{1}{3}}) . \\ Differentiating ​ the curves x = y^{2} and xy = k w . r . t . x, we get \\ \frac{d}{dx} x = \frac{d}{dx} y^{2} \Rightarrow 1 = 2 y \frac{dy}{dx} \Rightarrow \frac{dy}{dx} = \frac{1}{2 y} \\ Slope of tangent to the curves x = y^{2} at point (k^{\frac{2}{3}}, k^{\frac{1}{3}}) \\ m_{1} = {(\frac{dy}{dx})}_{(k^{\frac{2}{3}}, k^{\frac{1}{3}})} \\ = \frac{1}{2 k^{\frac{1}{3}}} \\ and \frac{d}{dx} (xy) = \frac{d}{dx} k \Rightarrow x \frac{dy}{dx} + y \frac{d}{dx} x = 0 \Rightarrow \frac{dy}{dx} = - \frac{y}{x} \\ Slope of tangent to the curve xy = k at point (k^{\frac{2}{3}}, k^{\frac{1}{3}}) \\ m_{2} = {(\frac{dy}{dx})}_{(k^{\frac{2}{3}}, k^{\frac{1}{3}})} \\ = - \frac{k^{\frac{1}{3}}}{k^{\frac{2}{3}}} \\ = - \frac{1}{k^{\frac{1}{3}}} \\ Since, two curves intersect at right angles if \\ m_{1} . m_{2} = - 1 \\ \frac{1}{2 k^{\frac{1}{3}}} \times - \frac{1}{k^{\frac{1}{3}}} = - 1 \\ 1 = 2 k^{\frac{2}{3}} \\ Cubing both sides, we get \\ 1 = 8 k^{2} . \\ Hence, the given two curves cut at right angels if 8 k^{2} = 1. \end{array}

Q.61

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

Ans.

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

Q.62

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

Ans.

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

$\begin{array}{l} T h e equation of given curve is y = {2x}^{2} + 3sinx \\ Differentiating w .r .t . x, we get \\ \frac{d y}{d x} = \frac{d}{d x} ({2x}^{2} + 3sinx) \\ = 4 x + 3 \cos x \\ P u t t i n g x= 0, we get \\ m = {(\frac{d y}{d x})}_{x= 0} \\ = 4 (0) + 3 \cos 0 \\ = 3 \\ S l o p e of normal to the given curve is \\ M = - \frac{1}{m} \\ = - \frac{1}{3} \\ The correct answer is D . \end{array}$

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.

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

Q.65

\begin{array}{l} Using differentials, find the approximate value o f e a c h o f \\ the following up to 3 places of d e c i m a l : \\ (i) \sqrt{25 .3} (i i) \sqrt{49 .5} (i i i) \sqrt{0 .6} (i v) {(0 .009)}^{\frac{1}{3}} \\ (v) {(0 .999)}^{\frac{1}{10}} (v i) {(15)}^{\frac{1}{4}} (v i i) {(26)}^{\frac{1}{3}} (v i i i) {(255)}^{\frac{1}{4}} \\ (i x) {(82)}^{\frac{1}{4}} (x) {(401)}^{\frac{1}{2}} (x i) {(0 .0037)}^{\frac{1}{2}} (x i i) {(26 .57)}^{\frac{1}{3}} \\ (x i i i) {(81 .5)}^{\frac{1}{4}} (x i v) {(3 .968)}^{\frac{3}{2}} (x v) {(32 .15)}^{\frac{1}{5}} \end{array}

Ans.

$\begin{array}{l} (i) \sqrt{25 .3} \\ Let y = \sqrt{x}, x = 25 and Δx = 0.3 \\ \Rightarrow \frac{dy}{dx} = \frac{1}{2 \sqrt{x}} \\ Then, \\ Δy = \sqrt{x + Δx} - \sqrt{x} \\ = \sqrt{25 .3} - \sqrt{25} \\ = \sqrt{25 .3} - 5 \\ \sqrt{25 .3} = Δy + 5 \\ Now, dy is approximately equal to Δy and is given by, \\ dy = (\frac{dy}{dx}) Δx \\ = \frac{1}{2 \sqrt{x}} (0 .3) \\ = \frac{1}{2 \sqrt{25}} (0 .3) \\ = \frac{1}{2 \times 5} \times 0 .3 \\ = 0 .03 \\ Hence, the approximate value of \sqrt{25 .3} is 5 + 0.03 = 5.03 . \\ (ii) \sqrt{49 .5} \\ Let y = \sqrt{x}, x = 49 and Δx = 0.5 \\ \Rightarrow \frac{dy}{dx} = \frac{1}{2 \sqrt{x}} \\ Then, \\ Δy = \sqrt{x + Δx} - \sqrt{x} \\ = \sqrt{49 .5} - \sqrt{49} \\ = \sqrt{49 .5} - 7 \\ \sqrt{49 .5} = Δy + 7 \\ Now, dy is approximately equal to Δy and is given by, \\ dy = (\frac{dy}{dx}) Δx \\ = \frac{1}{2 \sqrt{x}} (0 .5) \\ = \frac{1}{2 \sqrt{49}} (0 .5) \\ = \frac{1}{2 \times 7} \times 0 .5 \\ = 0 .035 \\ Hence, the approximate value of \sqrt{49 .5} is 7 + 0.035 = 7.035 . \\ (iii) \sqrt{0 .6} \\ Let y = \sqrt{x}, x = 1 and Δx = - 0.4 \\ \Rightarrow \frac{dy}{dx} = \frac{1}{2 \sqrt{x}} \\ Then, \\ Δy = \sqrt{x + Δx} - \sqrt{x} \\ = \sqrt{0 .6} - 1 \\ \sqrt{0 .6} = Δy + 1 \\ Now, dy is approximately equal to Δy and is given by, \\ dy = (\frac{dy}{dx}) Δx \\ = \frac{1}{2 \sqrt{x}} (- 0.4) \\ = \frac{1}{2 \sqrt{1}} (- 0.4) \\ = - 0.2 \\ Hence, the approximate value of \sqrt{0 .6} is 1 - 0.2 = 0.8 \\ (iv) {(0 .009)}^{\frac{1}{3}} \\ Let y = x^{\frac{1}{3}}, x = 0 .008 and Δx = 0 .001 \\ \Rightarrow \frac{dy}{dx} = \frac{1}{3 x^{\frac{2}{3}}} \\ Then, \\ Δy = {(x + Δx)}^{\frac{1}{3}} - x^{\frac{1}{3}} \\ = {(0 .009)}^{\frac{1}{3}} - 0 .2 \\ {(0 .009)}^{\frac{1}{3}} = Δy + 0 .2 \\ Now, dy is approximately equal to Δy and is given by, \\ \Rightarrow \frac{dy}{dx} = \frac{1}{3 x^{\frac{2}{3}}} (Δx) \\ = \frac{1}{3 {(0 .008)}^{\frac{2}{3}}} (0 .001) \\ = \frac{1}{3 (0 .04)} (0 .001) = \frac{0 .001}{0 .12} \\ = 0 .008 \\ Hence, the approximate value of {(0 .009)}^{\frac{2}{3}} is 0.2 + 0 . 008 = 0.208 \end{array}$ $\begin{array}{l} (v) {(0.999)}^{\frac{1}{10}} \\ L e t y = x^{\frac{1}{10}}, x = 1 and Δ x = - 0.001 \\ \Rightarrow \frac{d y}{d x} = \frac{1}{10 x^{\frac{9}{10}}} \\ T h e n, \\ Δ y = {(x + Δ x)}^{\frac{1}{10}} - {(x)}^{\frac{1}{10}} \\ = {(0.999)}^{\frac{1}{10}} - {(1)}^{\frac{1}{10}} \\ {(0.999)}^{\frac{1}{10}} = Δ y + 1 \\ Now, dy is approximately equal to Δ y and is given by, \\ dy = (\frac{d y}{d x}) Δ x \\ = \frac{1}{10 x^{\frac{9}{10}}} \times - 0.001 \\ = \frac{1}{10 {(1)}^{\frac{9}{10}}} \times - 0.001 \\ = - \frac{0.001}{10} \\ = - 0.0001 \\ Hence, the approximate value of {(0.999)}^{\frac{1}{10}} i s \\ 1 + (- 0.001) = 0.9999 \\ (v i) {(15)}^{\frac{1}{4}} \\ L e t y = x^{\frac{1}{4}}, x = 16 and Δ x = - 1 \\ \Rightarrow \frac{d y}{d x} = \frac{1}{4 x^{\frac{3}{4}}} \\ Then, \\ Δ y = {(x + Δ x)}^{\frac{1}{4}} - x^{\frac{1}{4}} \\ = {(15)}^{\frac{1}{4}} - {(16)}^{\frac{1}{4}} \\ {(15)}^{\frac{1}{4}} = Δ y + 2 \\ Now, dy is approximately equal to Δ y and is given by, \\ dy = (\frac{d y}{d x}) Δ x \\ = \frac{1}{4 x^{\frac{3}{4}}} \times - 1 \\ = \frac{1}{4 {(16)}^{\frac{3}{4}}} \times - 1 \\ = \frac{1}{4 \times 8} \times - 1 = - 0.03125 \\ Hence, the approximate value of {(15)}^{\frac{1}{4}} i s \\ 2 + (- 0.03125) = 1.96875 \\ (v i i) {(26)}^{\frac{1}{3}} \\ L e t y = x^{\frac{1}{3}}, x = 27 and Δ x = - 1 \\ \Rightarrow \frac{d y}{d x} = \frac{1}{3 x^{\frac{2}{3}}} \\ Then, \\ Δ y = {(x + Δ x)}^{\frac{1}{3}} - x^{\frac{1}{3}} \\ = {(26)}^{\frac{1}{3}} - {(27)}^{\frac{1}{3}} \\ {(26)}^{\frac{1}{3}} = Δ y + 3 \\ Now, dy is approximately equal to Δ y and is given by, \\ dy = (\frac{d y}{d x}) Δ x \\ = \frac{1}{3 x^{\frac{2}{3}}} \times - 1 \\ = \frac{1}{3 {(27)}^{\frac{2}{3}}} \times - 1 \\ = \frac{1}{3 \times 9} \times - 1 \\ = - 0.0 \bar{370} \\ Hence, the approximate value of {(26)}^{\frac{1}{3}} i s \\ 3 + (- 0.0370) = 2 .9629 \\ (v i i i) {(255)}^{\frac{1}{4}} \\ L e t y = x^{\frac{1}{4}}, x = 256 and Δ x = - 1 \\ \Rightarrow \frac{d y}{d x} = \frac{1}{4 x^{\frac{3}{4}}} \\ Then, \\ Δ y = {(x + Δ x)}^{\frac{1}{4}} - x^{\frac{1}{4}} \\ = {(255)}^{\frac{1}{4}} - {(256)}^{\frac{1}{4}} \\ {(255)}^{\frac{1}{4}} = Δ y + 4 \\ Now, dy is approximately equal to Δ y and is given by, \\ dy = (\frac{d y}{d x}) Δ x \\ = \frac{1}{4 x^{\frac{3}{4}}} \times (- 1) \\ = \frac{1}{4 {(256)}^{\frac{3}{4}}} \times (- 1) \\ = \frac{- 1}{4 \times 4^{3}} = - 0.0039 \\ Hence, the approximate value of {(255)}^{\frac{1}{4}} i s \\ 4 + (- 0 .0039) = 3 .9961 \end{array}$ $\begin{array}{l} (i x) {(82)}^{\frac{1}{4}} \\ L e t y = x^{\frac{1}{4}}, x = 81 and Δ x = 1 \\ \Rightarrow \frac{d y}{d x} = \frac{1}{4 x^{\frac{3}{4}}} \\ Then, \\ Δ y = {(x + Δ x)}^{\frac{1}{4}} - x^{\frac{1}{4}} \\ = {(82)}^{\frac{1}{4}} - {(81)}^{\frac{1}{4}} \\ {(82)}^{\frac{1}{4}} = Δ y + 3 \\ Now, dy is approximately equal to Δ y and is given by, \\ dy = (\frac{d y}{d x}) Δ x \\ = \frac{1}{4 x^{\frac{3}{4}}} \times (1) \\ = \frac{1}{4 {(81)}^{\frac{3}{4}}} \times (1) \\ = \frac{1}{4 \times 3^{3}} = 0.009 \\ Hence, the approximate value of {(82)}^{\frac{1}{4}} i s \\ 3 + (0 .009) = 3 .009 \\ (x) {(401)}^{\frac{1}{2}} \\ L e t y = x^{\frac{1}{2}}, x = 400 and Δ x = 1 \\ \Rightarrow \frac{d y}{d x} = \frac{1}{2 \sqrt{x}} \\ Then, \\ Δ y = \sqrt{x + Δ x} - \sqrt{x} \\ = \sqrt{401} - \sqrt{400} \\ = \sqrt{401} - 20 \\ \sqrt{401} = Δ y + 20 \\ Now, dy is approximately equal to Δ y and is given by, \\ dy = (\frac{d y}{d x}) Δ x \\ = \frac{1}{2 \sqrt{x}} (1) \\ = \frac{1}{2 \sqrt{400}} (1) \\ = \frac{1}{2 \times 20} \times (1) \\ = 0.025 \\ Hence, the approximate value of \sqrt{401} is 20 + 0 .025 = 20 .025 . \\ (x i) {(0.0037)}^{\frac{1}{2}} \\ L e t y = x^{\frac{1}{2}}, x = 0.0036 and Δ x = 0.0001 \\ \Rightarrow \frac{d y}{d x} = \frac{1}{2 \sqrt{x}} \\ Then, \\ Δ y = \sqrt{x + Δ x} - \sqrt{x} \\ = \sqrt{0.0037} - \sqrt{0.0036} \\ = \sqrt{0.0037} - 0.06 \\ \sqrt{0.0037} = Δ y + 0.06 \\ Now, dy is approximately equal to Δ y and is given by, \\ dy = (\frac{d y}{d x}) Δ x \\ = \frac{1}{2 \sqrt{x}} (0.0001) \\ = \frac{1}{2 \sqrt{0.0036}} (0.0001) \\ = \frac{1}{2 \times 0.06} \times (0.0001) \\ = 0.00083 \\ Hence, the approximate value of \sqrt{0.0037} is \\ 0.06 + 0.00083 = 0 .06083 \\ (x i i) {(26.57)}^{\frac{1}{3}} \\ L e t y = x^{\frac{1}{3}}, x = 27 and Δ x = - 0.43 \\ \Rightarrow \frac{d y}{d x} = \frac{1}{3 x^{\frac{2}{3}}} \\ Then, \\ Δ y = {(x + Δ x)}^{\frac{1}{3}} - x^{\frac{1}{3}} \\ = {(26.57)}^{\frac{1}{3}} - {(27)}^{\frac{1}{3}} \\ {(26.57)}^{\frac{1}{3}} = Δ y + 3 \\ Now, dy is approximately equal to Δ y and is given by, \\ dy = (\frac{d y}{d x}) Δ x \\ = \frac{1}{3 x^{\frac{2}{3}}} \times - 0.43 \\ = \frac{1}{3 {(27)}^{\frac{2}{3}}} \times - 0.43 \\ = \frac{1}{3 (9)} \times - 0.43 = - 0.015 \\ Hence, the approximate value of {(26.57)}^{\frac{1}{3}} is \\ 3 - 0.015 = 2 .984 \\ (x i i i) {(81.5)}^{\frac{1}{4}} \\ L e t y = x^{\frac{1}{4}}, x = 81 and Δ x = 0.5 \\ \Rightarrow \frac{d y}{d x} = \frac{1}{4 x^{\frac{3}{4}}} \\ Then, \\ Δ y = {(x + Δ x)}^{\frac{1}{4}} - x^{\frac{1}{4}} \end{array}$

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