# NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives (Ex 6.1) Exercise 6.1

Class 12 is a challenging year for numerous students. Class 12 also marks one of the major academic turning points for students, as it sets the basis for the future education foundation of students.

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On the website, the students can easily access:

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## NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives (Ex 6.1) Exercise 6.1

Class 12 students can find it very challenging to appear for the board examination.

Extramarks has tools like the NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.1 to help deal with such pressure situations.

### An Overview of Application of Derivatives Exercise 6.1

Rate of Change, Intervals of Increase or Decrease, Lagrange’s Mean Value Theorem, Rolle’s Theorem, Point where a Tangent is Parallel or Perpendicular, Slope of a Tangent or Normal,  Approximate Value of Numbers, etc. are all determined by Derivatives. Class 12 Maths Ch 6 Ex 6.1 and other exercises of this chapter teach all these topics deeply. The Extramarks experts explain these applications in straightforward and understandable language to aid students in developing a solid understanding of the key ideas. The NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.1 are prepared similarly. The Extramarks experts create solutions like the NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.1 to help students revise thoroughly for their exams.

The NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.1 solve all the questions listed under this chapter that are part of the Class 12 Maths CBSE Syllabus.

These NCERT Solutions allow students to put their subject knowledge to use by including a comprehensive set of questions and answers arranged at an advanced degree of difficulty. Students can sharpen their preparation skills by making use of the NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.1.

### Class 12 Maths NCERT Solutions Chapter 6 Exercise 6.1 – Free PDF Download

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### Class 12 Maths Chapter 6 Exercise 6.1

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### Important Concepts Covered in Exercise 6.1 of Class 12 Maths NCERT Solutions

Exercise 6.1 for Class 12 Mathematics can be incredibly easy to comprehend and solve.

The majority of children can complete all the exercises promptly. Any topic or problem might vary in complexity for each student. Therefore, the NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.1 can be easy or difficult depending on the calibre of the student and according to the level of their preparation. If students revise in a steady manner, they can easily accomplish their desired scores in the exam, but it is important to analyse what step of preparation they are in. Students who begin preparing early have a head start while revising during the end. This allows them to not only identify their errors and weaknesses but also, strengthen their command over the topics in which they lack confidence.

The primary topics covered in this chapter the NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.1 are: Finding the Derivative of the Equations, Rate of Change of Quantity,  Increasing and Decreasing Functions, Tangents and Normal, Approximations, Maxima, and Minima (first derivative test motivated geometrically and second derivative test given as a probable tool), and simple problems that illustrate the basic principle and understanding of the subject as real-life situations. The NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.1 are available for students to make use of and practice these important concepts.

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### Weightage Marks of Class 12 Maths

To pass each topic on the CBSE Class 12 board exam, a candidate must receive a minimum mark of 33%. In addition, the candidate must obtain a minimum of 33% overall to pass the board exam. Candidates who receive aggregate marks of 33% or higher, but less than 33% in any subject must take the supplemental exam.

The NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.1 can help students in preparing to achieve a good score in the examinations. A minimum of 30% of the questions might be competency-based, whether they were multiple-choice, case-based, integrated from sources, or any other type. There might be 20% of objective questions. Questions with both short and extensive answers might make up the remaining 50% (as per the existing pattern).

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### Benefits of Exercise 6.1 Class 12 Maths NCERT Solutions

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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

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

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

$\begin{array}{l}\mathrm{A}\mathrm{balloon},\mathrm{which}\mathrm{always}\mathrm{remains}\mathrm{spherical},\mathrm{has}\mathrm{a}\mathrm{variable}\\ \mathrm{diameter}\frac{\mathrm{3}}{\mathrm{2}}\mathrm{ }\left(2\mathrm{x}+1\right). \mathrm{Find}\mathrm{the}\mathrm{rate}\mathrm{of}\mathrm{change}\mathrm{of}\mathrm{its}\mathrm{volume}\mathrm{with}\\ \mathrm{respect}\mathrm{to}\mathrm{x}\mathrm{.}\end{array}$

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

$\begin{array}{l}\mathrm{Sand}\mathrm{is}\mathrm{pouring}\mathrm{from}\mathrm{a}\mathrm{pipe}\mathrm{at}\mathrm{the}\mathrm{rate}\mathrm{of}12\mathrm{cm}3/\mathrm{s}.\mathrm{The}\\ \mathrm{falling}\mathrm{sand}\mathrm{forms}\mathrm{a}\mathrm{cone}\mathrm{on}\mathrm{the}\mathrm{ground}\mathrm{in}\mathrm{such}\mathrm{a}\mathrm{way}\mathrm{that}\\ \mathrm{the}\mathrm{height}\mathrm{of}\mathrm{the}\mathrm{cone}\mathrm{is}\mathrm{always}\mathrm{one}–\mathrm{sixth}\mathrm{of}\mathrm{the}\mathrm{radius}\mathrm{of}\\ \mathrm{the}\mathrm{base}.\mathrm{How}\mathrm{fast}\mathrm{is}\mathrm{the}\mathrm{height}\mathrm{of}\mathrm{the}\mathrm{sand}\mathrm{cone}\mathrm{increasing}\\ \mathrm{when}\mathrm{the}\mathrm{height}\mathrm{is}4\mathrm{cm}?\end{array}$

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

$\begin{array}{l}\mathrm{The}\mathrm{rate}\mathrm{of}\mathrm{change}\mathrm{of}\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{a}\mathrm{circle}\mathrm{with}\mathrm{respect}\mathrm{to}\mathrm{its}\\ \mathrm{radius}\mathrm{r}\mathrm{at}\mathrm{r}=6\mathrm{cm}is\\ \left(\mathrm{A}\right)\phantom{\rule{0ex}{0ex}}10\mathrm{\pi }\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\left(\mathrm{B}\right)\phantom{\rule{0ex}{0ex}}12\mathrm{\pi }\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\left(\mathrm{C}\right)\phantom{\rule{0ex}{0ex}}8\mathrm{\pi }\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\left(\mathrm{D}\right)11\mathrm{\pi }\end{array}$

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

$\begin{array}{l}\mathrm{The}\mathrm{total}\mathrm{revenue}\mathrm{in}\mathrm{Rupees}\mathrm{received}\mathrm{from}\mathrm{the}\mathrm{sale}\mathrm{of}\mathrm{x}\\ \mathrm{units}\mathrm{of}\mathrm{a}\mathrm{product}\mathrm{is}\mathrm{given}\mathrm{by} \mathrm{R}\left(\mathrm{x}\right) =\mathrm{ }3{\mathrm{x}}^{\mathrm{2}}+36\mathrm{x}+5\mathrm{.}\\ \mathrm{The}\mathrm{marginal}\mathrm{revenue},\mathrm{when}\mathrm{x}= 15\mathrm{is}:\\ \left(\mathrm{A}\right) 116 \left(\mathrm{B}\right) 96 \left(\mathrm{C}\right) 90 \mathrm{ }\left(\mathrm{D}\right) 126\end{array}$

Ans

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

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### 1. What is the pattern of the Mathematics board exam paper?

There are three sections in the CBSE Class 12 Maths Question Paper: A, B, and C. Students must attempt each of the three sections. Mathematics, just like all the other subjects, is given 100 marks weightage. 20 marks are for internal marking and 80 marks are for the written examination. There are a total of 36 questions in the question paper. 20 questions carry one mark each and are very short answer type questions. Then there is a short answer type of questions which carry to marks each. There are 6 short answer type questions. Next are long answer type questions Part 1. There are 6 questions in this section each carrying 4 marks. The last is the long answer type questions Part 2. There are four questions in this section each question carries 6 marks.

### 2. How much importance should be given by Class 12 students to Chapter 6 of NCERT Mathematics?

The students should go through the scheme of mark distribution to see how much weightage is given to individual chapters. When preparing for their examinations, students can take the help of the marks distribution scheme to understand how much time should be spent on each chapter. There can be some theoretical chapters from which only one mark question might appear. Then, there can be questions from which five marks can be formed. The weightage is for the students to understand how much in-depth knowledge of each chapter they are supposed to have. Students are advised to remember that to score well they need to focus on every chapter. Students should not limit their preparations to just those chapters which carry maximum marks. The board examination questions can sometimes be unpredictable, so it is always advised to go through the entire syllabus thoroughly. Making use of the NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.1 at such times can be helpful.

### 3. What can the Class 12 students do to efficiently prepare for the board examinations?

There is no denying that there is a lot of pressure that comes along with the preparation for the Class 12 examination. These also help to assess how much of the study material provided to students in Class 12, they could efficiently understand and rehearse. The NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.1 allow students to revise thoroughly and help them frame better answers in the final examination.