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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.

Class 12 students are constantly reminded to practice for the Board examinations that they are expected to excel in. The students, even before entering Class 12, are told how important the Class 12 examinations are. The NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.1 can be very helpful in the preparation for the Mathematics board examinations.

The Central Board of Secondary Education or the CBSE is responsible for conducting these board examinations. Students are advised to make use of the tools that are available on the Extramarks’ website, such as the NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.1, for their board examination preparation. Class 12 Maths Ch. 6 Ex 6.1 is one exercise for which the solutions are available on the website. Students are advised to practice the Ex 6.1 Class 12 Maths NCERT Solutions along with the solutions of other chapters to have a good revision of the topics before the examinations.

The Extramarks also has NCERT Solutions for most classes. NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.1 can be of use here as well.

On the website, the students can easily access:

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The experts at Extramarks understand the importance of performing well in all the classes.

And the importance of having a clear understanding of the study material of these classes if the student wishes to excel in Class 12. The experts come up with these NCERT Solutions for all the classes from Class 1 to Class 12. The NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.1 is a part of those solutions.

There is no denying that Class 12 examinations hold a lot of importance in a student’s life. It is based on these scores that the students get to decide which college they want to get admission into. Making use of the NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.1 can be very useful in preparing for the final exam that could allow students to pursue their desired stream in their preferred college.

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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.

Access NCERT solutions for Class 12 Maths Chapter 6 – Application of Derivatives

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

The NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.1 are created by the experts keeping in mind the differences in the learning abilities of students. Every student excels in different departments and different students have different topics that they might find interesting. Using the right tool for preparation and revision can ensure good marks in the board examinations.

The NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.1 is just one of many such tools on the Extramarks website. Exercise 6.1 Class 12 Maths deals with the topic ‘Application of Derivatives’. The NCERT books of Mathematics follow a pattern where the chapters are further divided into exercises. Different exercises contain different types of questions for the students to practice, related to the topics of the chapter.

Practising the exercises with the help of tools like the NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.1 can help the students understand these exercises efficiently.

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Class 12 Maths Chapter 6 Exercise 6.1

The NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.1 can therefore be very useful for students. The students can verify and try to understand the part of the solutions where they get stuck with the help of the solutions that are prepared by the experts at the Extramarks, just like the NCERT Solutions for 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).

To practice efficiently for the examinations the students are advised to make sure to use helpful tools like the NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.1.

Benefits of Exercise 6.1 Class 12 Maths NCERT Solutions

The students of Class 12 take help from a lot of other books apart from their NCERT books when preparing for their examinations. What other books might sometimes lack are authentic tools, like the NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.1, based on the NCERT books.

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The NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.1 are prepared by the experts of Mathematics which ensures that they are the correct solutions.

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The NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.1 and other solutions, that are prepared by the experts, can be used by the students for revision as well. For subjects like Mathematics, especially with topics that are mathematical and not theoretical, it can sometimes be difficult for students to revise those chapters. The NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.1 can help solve this problem as well.

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When students make use of other books apart from NCERT, they can find different ways to solve the same questions. This can be very confusing for the students. Students, sometimes, may find these other methods of solving to be easier or quicker, but they need to remember to follow the format that is given by the NCERT.

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Before the board examination of Mathematics, the students usually get a long stretch to prepare for the final exam. Being able to manage this time with efficiency is one of the most important skills that the students require. One of the most reliable sources for revision is the NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.1 and other such tools made available by the Extramarks website. Extramarks allows students to prepare for their examinations holistically.

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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

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

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

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

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

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

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

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

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

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

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

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

Ans

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

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FAQs (Frequently Asked Questions)

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.

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.

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.

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

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

Access NCERT solutions for Class 12 Maths Chapter 6 – Application of Derivatives

An Overview of Application of Derivatives Exercise 6.1

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

Class 12 Maths Chapter 6 Exercise 6.1

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

Weightage Marks of Class 12 Maths

Benefits of Exercise 6.1 Class 12 Maths NCERT Solutions

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

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

Access NCERT solutions for Class 12 Maths Chapter 6 – Application of Derivatives

An Overview of Application of Derivatives Exercise 6.1

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

Class 12 Maths Chapter 6 Exercise 6.1

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

Weightage Marks of Class 12 Maths

Benefits of Exercise 6.1 Class 12 Maths NCERT Solutions

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FAQs (Frequently Asked Questions)

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

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

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

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