NCERT Exemplar for Class 12 Maths Chapter 6

NCERT Exemplar for Class 12 Maths focuses on preparing students for the upcoming examination by boosting their knowledge and understanding on the various concepts discussed in the mathematics chapters. Students will gain in-depth understanding through the provided step-wise solutions. It will ensure that students cover all the sections of the chapter. Concepts, formulas, theorems, principles, are explained in a detailed manner in this provided exemplar.

Answers of the NCERT exemplar for class 12 Maths Chapter 6 Application of Derivatives make sure to fulfil and touch all the doubt areas of students, so they are left with no confusions. This boosts their confidence and gets them ready for the upcoming 12th board examination. NCERT exemplar for Class 12 Maths Chapter 6 Application of Derivatives contains a total of 32 important questions of short and long type. Students can easily access them by scrolling below and practise them consistently to score maximum marks.

Along with this, there are several NCERT exemplar for class 12 maths of all the chapters provided on this platform.

Access Solutions of the NCERT Exemplar Class 12 Maths Chapter 6 Application of Derivatives

Short Answer (S.A.)

1. A spherical ball of salt is dissolving in water in such a manner that the rate of decrease of the volume at any instant is proportional to the surface. Prove that the radius is decreasing at a constant rate.

Solution:

Given, a spherical ball of salt

Then, the volume of ball V = 4/3 πr3 where r = radius of the ball

Now, according to the question we have

dV/dt ∝ S, where S = surface area of the ball

Therefore, the radius of the ball is decreasing at constant rate.

2. If the area of a circle increases at a uniform rate, then prove that the perimeter varies inversely as the radius. \section*

{Solution} Let the radius of the circle be \( r \). Then the area \( A \) of the circle is given by: \[ A = \pi r^2 \] Differentiating both sides with respect to \( t \), we get: \[ \frac{dA}{dt} = \frac{d}{dt} (\pi r^2) \] \[ \frac{dA}{dt} = 2 \pi r \frac{dr}{dt} \quad \text{(1)} \] Since the area of the circle increases at a uniform rate, we have: \[ \frac{dA}{dt} = k \quad \text{(2)} \] Substitute (2) into (1): \[ 2 \pi r \frac{dr}{dt} = k \] \[ \frac{dr}{dt} = \frac{k}{2 \pi r} \] The perimeter \( P \) of the circle is given by: \[ P = 2 \pi r \] Differentiating \( P \) with respect to \( t \): \[ \frac{dP}{dt} = 2 \pi \frac{dr}{dt} \] Substitute \(\frac{dr}{dt}\) from above: \[ \frac{dP}{dt} = 2 \pi \left(\frac{k}{2 \pi r}\right) \] \[ \frac{dP}{dt} = \frac{k}{r} \] Thus, \(\frac{dP}{dt} \propto \frac{1}{r}\). This shows that the perimeter varies inversely as the radius of the circle. Hence proved.

 

4. Two men A and B start with velocities v at the same time from the junction of two roads inclined at 45° to each other. If they travel by different roads, find the rate at which they are being separated.

Solution:

Let’s consider P to be any point at which the two roads are inclined at an angle of 45o.

Now, two men A and B are moving along the roads PA and PB respectively with same speed ‘V’.

And, let A and B be their final positions such that AB = y

∠APB = 45o and they move with the same speed.

So, ∆APB is an isosceles triangle.

Now, draw PQ ⊥ AB.

We have, AB = y

So, AQ = y/2 and PA = PA = x (assumption)

And, ∠APQ = ∠BPQ = 45o/2 = 22.5o

[As the altitude drawn from the vertex of an isosceles ∆, bisects the base]

Now, in right ∆APQ

sin 22.5o = AQ/AP

Therefore, the rate of their separation is

5. Find an angle \( \theta \), where \( 0 < \theta < \frac{\pi}{2} \), which increases twice as fast as its sine. \section*

{Solution} Let \( \theta \) increase twice as fast as its sine: \[ \theta = 2 \sin \theta \] Differentiate both sides with respect to \( t \): \[ \frac{d \theta}{d t} = 2 \cos \theta \cdot \frac{d \theta}{d t} \] Rearrange to solve for \( \frac{d \theta}{d t} \): \[ 1 = 2 \cos \theta \] \[ \frac{1}{2} = \cos \theta \] \[ \cos \theta = \cos \frac{\pi}{3} \] Therefore: \[ \theta = \frac{\pi}{3} \] So, the required angle is \( \frac{\pi}{3} \).

6. Find the approximate value of (1.999)5.

Solution:

(1.999)5 = (2 – 0.001)5

Let x = 2 and ∆x = -0.001

Also, let y = x5

Differentiating both sides w.r.t, x, we get

dy/dx = 5×4 = 5(2)4 = 80

Now, ∆y = (dy/dx). ∆x = 80. (-0.001) = -0.080

And, (1.999)5 = y + ∆y

= x5 – 0.080 = (2)5 – 0.080 = 32 – 0.080 = 31.92

Therefore, approximate value of (1.999)5 is 31.92

7. Find the approximate volume of metal in a hollow spherical shell whose internal and external radii are 3 cm and 3.0005 cm, respectively. \section*

{Solution} Let the internal radius be \( r = 3 \) cm and the external radius be \( R = 3.0005 \) cm. The volume of the hollow spherical shell is given by: \[ V = \frac{4}{3} \pi \left( R^3 – r^3 \right) \] Substitute \( R = 3.0005 \) and \( r = 3 \): \[ V = \frac{4}{3} \pi \left[ (3.0005)^3 – 3^3 \right] \] To find the approximate value of \( (3.0005)^3 \), use differentiation: Let \( (3.0005)^3 = y + \Delta y \) and \( x = 3 \), \( \Delta x = 0.0005 \). Let \( y = x^3 \). Differentiating both sides with respect to \( x \): \[ \frac{d y}{d x} = 3 x^2 \] Thus: \[ \Delta y = \frac{d y}{d x} \cdot \Delta x = 3 x^2 \cdot 0.0005 = 3 \cdot 3^2 \cdot 0.0005 = 27 \cdot 0.0005 = 0.0135 \] Hence: \[ (3.0005)^3 = y + \Delta y = 3^3 + 0.0135 = 27 + 0.0135 = 27.0135 \] So, the volume \( V \) is: \[ V = \frac{4}{3} \pi \left[ 27.0135 – 27.000 \right] = \frac{4}{3} \pi \left[ 0.0135 \right] = \frac{4 \pi \times 0.0135}{3} = 0.0180 \pi \text{ cm}^3 \]

 

9. A swimming pool is to be drained for cleaning. If L represents the number of litres of water in the pool t seconds after the pool has been plugged off to drain and L = 200 (10 – t)2. How fast is the water running out at the end of 5 seconds? What is the average rate at which the water flows out during the first 5 seconds?

Solution:

Given, L = 200(10 – t)2 where L represents the number of liters of water in the pool.

On differentiating both the sides w.r.t, t, we get

dL/dt = 200 x 2(10 – t) (-1) = -400(10 – t)

But, the rate at which the water is running out

= -dL/dt = 400(10 – t)

Now, rate at which the water is running after 5 seconds will be

= 400 x (10 – 5) = 2000 L/s (final rate)

T = 0 for initial rate

= 400 (10 – 0) = 4000 L/s

So, the average rate at which the water is running out is given by

= (Initial rate + Final rate)/ 2 = (4000 + 2000)/ 2 = 6000/2 = 3000 L/s

Therefore, the required rate = 3000 L/s

10. The volume of a cube increases at a constant rate. Prove that the increase in its surface area varies inversely as the length of the side. \section*

{Solution} Let the side of the cube be \( x \) units. The volume of the cube is given by: \[ V = x^3 \] Differentiating both sides with respect to \( t \), we get: \[ \frac{dV}{dt} = 3x^2 \frac{dx}{dt} \] Let \(\frac{dV}{dt} = k\), where \( k \) is a constant. Therefore: \[ 3x^2 \frac{dx}{dt} = k \] \[ \frac{dx}{dt} = \frac{k}{3x^2} \quad \text{(1)} \] Now, the surface area of the cube is: \[ S = 6x^2 \] Differentiating with respect to \( t \), we get: \[ \frac{dS}{dt} = 12x \frac{dx}{dt} \] Substituting \(\frac{dx}{dt}\) from equation (1): \[ \frac{dS}{dt} = 12x \cdot \frac{k}{3x^2} \] \[ \frac{dS}{dt} = \frac{12k}{3x} \] \[ \frac{dS}{dt} = \frac{4k}{x} \] Thus, we can see that: \[ \frac{dS}{dt} \propto \frac{1}{x} \] Hence, the increase in the surface area varies inversely as the length of the side.

 

14. Find the coordinates of the point on the curve \(\sqrt{x} + \sqrt{y} = 4\) at which the tangent is equally inclined to the axes. \section*

{Solution} Let the required point be \((x_1, y_1)\). The given equation of the curve is: \[ \sqrt{x} + \sqrt{y} = 4 \] Since \((x_1, y_1)\) lies on the curve, we have: \[ \sqrt{x_1} + \sqrt{y_1} = 4 \quad \text{(1)} \] Differentiate the curve implicitly with respect to \(x\): \[ \frac{d}{dx}\left(\sqrt{x} + \sqrt{y}\right) = \frac{d}{dx}(4) \] \[ \frac{1}{2\sqrt{x}} + \frac{1}{2\sqrt{y}} \frac{dy}{dx} = 0 \] \[ \frac{dy}{dx} = -\frac{\sqrt{y}}{\sqrt{x}} \] Given that the tangent is equally inclined to the axes, the angle \(\theta\) of the tangent with the x-axis is \(45^\circ\). Therefore: \[ \tan \theta = 1 \] \[ \frac{dy}{dx} = 1 \] \[ -\frac{\sqrt{y_1}}{\sqrt{x_1}} = 1 \] Squaring both sides: \[ \frac{y_1}{x_1} = 1 \] \[ x_1 = y_1 \] Substitute \(x_1 = y_1\) into equation (1): \[ \sqrt{x_1} + \sqrt{x_1} = 4 \] \[ 2\sqrt{x_1} = 4 \] \[ \sqrt{x_1} = 2 \] \[ x_1 = 4 \] \[ y_1 = x_1 = 4 \] Thus, the required point is \((4, 4)\).

 

16. Prove that the curves y2 = 4x and x2 + y2 – 6x + 1 = 0 touch each other at the point (1, 2).

Solution:

Given curve equations are: y2 = 4x …. (1) and x2 + y2 – 6x + 1 = 0 ….. (2)

Now, differentiating (i) w.r.t. x, we get

2y.(dy/dx) = 4 ⇒ dy/dx = 2/y

Slope of tangent at (1, 2), m1 = 2/2 = 1

Differentiating (ii) w.r.t. x, we get

2x + 2y.(dy/dx) – 6 = 0

2y. dy/dx = 6 – 2x ⇒ dy/dx = (6 – 2x)/ 2y

Hence, the slope of the tangent at the same point (1, 2)

⇒ m2 = (6 – 2 x 1)/ (2 x 2) = 4/4 = 1

It’s seen that m1 = m2 = 1 at the point (1, 2).

Therefore, the given circles touch each other at the same point (1, 2).

17. Find the equation of the normal lines to the curve 3x2 – y2 = 8 which are parallel to the line x + 3y = 4.

Solution:

Given curve, 3x2 – y2 = 8

Differentiating both sides w.r.t. x, we get

6x – 2y. dy/dx = 0 ⇒ -2y(dy/dx) = -6x ⇒ dy/dx = 3x/y

So, slope of the tangent to the given curve = 3x/y

Thus, the normal to the curve = -1/(3x/y) = -y/3x

Now, differentiating both sides of the given line x + 3y = 4, we have

1 + 3.(dy/dx) = 0

dy/dx = -1/3

As the normal to the curve is parallel to the given line x + 3y = 4

We have, -y/3x = -1/3 ⇒ y = x

On putting the value of y in 3×2 – y2 = 8, we get

3×2 – x2 = 8

2×2 = 8 ⇒ x2 = 4 ⇒ x = ±2

So, y = ±2

Thus, the points on the curve are (2, 2) and (-2, -2).

Now, the equation of the normal to the curve at (2, 2) is given by

Therefore, the required equations are x + 3y = 8 and x + 3y = -8.

18. At what points on the curve \[ x^2 + y^2 – 2x – 4y + 1 = 0 \] are the tangents parallel to the \( y \)-axis? \section*

{Solution} Given the curve: \[ x^2 + y^2 – 2x – 4y + 1 = 0 \] Differentiate both sides with respect to \( x \): \[ 2x + 2y \frac{dy}{dx} – 2 – 4 \frac{dy}{dx} = 0 \] Rearrange the terms: \[ 2x – 2 – 2y \frac{dy}{dx} – 4 \frac{dy}{dx} = 0 \] \[ 2x – 2 = (2y + 4) \frac{dy}{dx} \] \[ \frac{dy}{dx} = \frac{2x – 2}{2y + 4} \] For the tangent to be parallel to the \( y \)-axis, \(\frac{dy}{dx}\) must be undefined, which occurs when the denominator is zero: \[ 2y + 4 = 0 \] \[ y = -2 \] Substitute \( y = -2 \) into the original curve equation: \[ x^2 + (-2)^2 – 2x – 4(-2) + 1 = 0 \] \[ x^2 + 4 – 2x + 8 + 1 = 0 \] \[ x^2 – 2x + 13 = 0 \] Solve for \( x \): \[ x^2 – 2x + 13 = 0 \] The discriminant of this quadratic equation is: \[ (-2)^2 – 4 \cdot 1 \cdot 13 = 4 – 52 = -48 \] Since the discriminant is negative, there are no real solutions for \( x \). Hence, the curve does not have any points where the tangents are parallel to the \( y \)-axis. Thus, there are no points on the given curve where the tangents are parallel to the \( y \)-axis.

19. Show that the line \[ \frac{x}{a} + \frac{y}{b} = 1 \] touches the curve \[ y = b e^{-\frac{x}{a}} \] at the point where the curve intersects the \( y \)-axis. \section*

{Solution} We are given the line: \[ \frac{x}{a} + \frac{y}{b} = 1 \] and the curve: \[ y = b e^{-\frac{x}{a}} \] First, find the point where the curve intersects the \( y \)-axis. This occurs when \( x = 0 \): \[ y = b e^{-\frac{0}{a}} = b \] Thus, the point of intersection on the \( y \)-axis is \( (0, b) \). Now, differentiate the equation of the curve with respect to \( x \): \[ y = b e^{-\frac{x}{a}} \] \[ \frac{dy}{dx} = b \left(-\frac{1}{a}\right) e^{-\frac{x}{a}} = -\frac{b}{a} e^{-\frac{x}{a}} \] At the point \( (0, b) \): \[ \frac{dy}{dx} \bigg|_{(0, b)} = -\frac{b}{a} e^{-\frac{0}{a}} = -\frac{b}{a} \] The slope of the line \(\frac{x}{a} + \frac{y}{b} = 1\) is: \[ \frac{dy}{dx} = -\frac{b}{a} \] Rewrite the line in the slope-intercept form: \[ \frac{y}{b} = 1 – \frac{x}{a} \] \[ y = -\frac{b}{a} x + b \] The slope of this line is \( -\frac{b}{a} \), which matches the slope of the tangent to the curve at the point \( (0, b) \). Since the slopes match, the line touches the curve at the point where the curve intersects the \( y \)-axis.

20. Show that f (x) = 2x + cot-1 x + log [√(1 + x2) – x] is increasing in R.

Solution:

Given,

f (x) = 2x + cot-1 x + log [√(1 + x2) – x]

Differentiating both sides w.r.t. x, we get

On squaring both the sides, we get

4×4 + 1 + 4×2 ≥ 1 + x2

4×4 + 4×2 – x2 ≥ 0

4×4 + 3×2 ≥ 0

x2(4×2 + 3) ≥ 0

The above is true for any value of x ∈ R.

Therefore, the given function is an increasing function over R.

21. Show that for a ³ 1, f (x) = √3 sin x – cos x – 2ax + b is decreasing in R.

Solution:

Given,

f (x) = √3 sin x – cos x – 2ax + b, a ³ 1

On differentiating both sides w.r.t. x, we get

f’ (x) = √3 cos x + sin x – 2a

For increasing function, f’ (x) < 0

Therefore, the given function is decreasing in R.

22. Show that \[ f(x) = \tan^{-1}(\sin x + \cos x) \] is an increasing function in \( (0, \frac{\pi}{4}) \). \section*

{Solution} Let \[ f(x) = \tan^{-1}(\sin x + \cos x) \] To determine if \( f(x) \) is increasing, we need to find the derivative \( \frac{df}{dx} \) and check if it is positive in the interval \( (0, \frac{\pi}{4}) \). First, differentiate \( f(x) \): \[ \frac{df}{dx} = \frac{1}{1 + (\sin x + \cos x)^2} \cdot \frac{d}{dx}(\sin x + \cos x) \] Compute the derivative of \( \sin x + \cos x \): \[ \frac{d}{dx}(\sin x + \cos x) = \cos x – \sin x \] Thus, \[ \frac{df}{dx} = \frac{\cos x – \sin x}{1 + (\sin x + \cos x)^2} \] In the interval \( (0, \frac{\pi}{4}) \), \( \cos x > \sin x \). Therefore, the numerator \( \cos x – \sin x \) is positive. The denominator \( 1 + (\sin x + \cos x)^2 \) is always positive because it is a sum of 1 and a square term. Since the numerator is positive and the denominator is positive, we have: \[ \frac{df}{dx} > 0 \] Therefore, \( f(x) \) is an increasing function in \( (0, \frac{\pi}{4}) \).

23. At what point, the slope of the curve y = – x3 + 3x2 + 9x – 27 is maximum? Also find the maximum slope.

Solution:

Given, curve y = – x3 + 3x2 + 9x – 27

Differentiating both sides w.r.t. x, we get

dy/dx = -3×2 + 6x + 9

Let slope of the curve dy/dx = z

So, z = -3×2 + 6x + 9

Differentiating both sides w.r.t. x, we get

dz/dx = -6x + 6

For local maxima and local minima,

dz/dx = 0

-6x + 6 = 0 ⇒ x = 1

d2z/dx2 = -6 < 0 Maxima

Putting x = 1 in equation of the curve y = (-1)3 + 3(1)2 + 9(1) – 27

= -1 + 3 + 9 – 27 = -16

Maximum slope = -3(1)2 + 6(1) + 9 = 12

Therefore, (1, -16) is the point at which the slope of the given curve is maximum and maximum slope = 12.

24. Prove that \[ f(x) = \sin x + \sqrt{3} \cos x \] has a maximum value at \( x = \frac{\pi}{6} \). \section*

{Solution} We are given the function \[ f(x) = \sin x + \sqrt{3} \cos x \] First, find the first derivative \( f'(x) \): \[ f'(x) = \cos x – \sqrt{3} \sin x \] To find the critical points, set \( f'(x) = 0 \): \[ \cos x – \sqrt{3} \sin x = 0 \] Rearrange to solve for \( \tan x \): \[ \cos x = \sqrt{3} \sin x \] \[ \tan x = \frac{1}{\sqrt{3}} = \tan \frac{\pi}{6} \] \[ x = \frac{\pi}{6} \] Now, find the second derivative \( f”(x) \): \[ f”(x) = -\sin x – \sqrt{3} \cos x \] Evaluate \( f”(x) \) at \( x = \frac{\pi}{6} \): \[ f”\left(\frac{\pi}{6}\right) = -\sin \frac{\pi}{6} – \sqrt{3} \cos \frac{\pi}{6} \] \[ = -\frac{1}{2} – \sqrt{3} \cdot \frac{\sqrt{3}}{2} \] \[ = -\frac{1}{2} – \frac{3}{2} \] \[ = -2 < 0 \] Since \( f”\left(\frac{\pi}{6}\right) < 0 \), the function \( f(x) \) has a local maximum at \( x = \frac{\pi}{6} \). Hence, \( x = \frac{\pi}{6} \) is the point of local maxima.

Long Answer (L.A.)

27. A telephone company in a town has 500 subscribers on its list and collects fixed charges of Rs 300/- per subscriber per year. The company proposes to increase the annual subscription and it is believed that for every increase of Rs 1/- one subscriber will discontinue the service. Find what increase will bring maximum profit?

Solution:

Let’s consider that the company increases the annual subscription by Rs x.

So, x is the number of subscribers who discontinue the services.

Total revenue, R(x) = (500 – x) (300 + x)

= 150000 + 500x – 300x – x2

= -x2 + 200x + 150000

Differentiating both sides w.r.t. x, we get R’(x) = -2x + 200

For local maxima and local minima, R’(x) = 0

-2x + 200 = 0 ⇒ x = 100

R’’(x) = -2 < 0 Maxima

So, R(x) is maximum at x = 100

Therefore, in order to get maximum profit, the company should increase its annual subscription by Rs 100.

28. If the straight-line x cos α + y sin α = p touches the curve x2/a2 + y2/b2 = 1, then prove that a2 cos2 α + b2 sin2 α = p2.

Solution:

The given curve is x2/a2 + y2/b2 = 1 and the straight-line x cos α + y sin α = p

Differentiating equation (i) w.r.t. x, we get

Therefore, a2 cos2 α + b2 sin2 α = p2.

 

 

NCERT Exemplar For Class 12 Maths

The NCERT exemplars are an effective study material for scoring higher marks in the examination paper. Students must practise these additional questions for their own benefits, as these are curated by the best subject-matter experts to boost both knowledge and confidence. Students can easily access the ncert exemplar for class 12 maths by visiting our website SimplyAcad and solve all the questions listed to secure maximum marks.

Here are some other NCERT exemplar for class 12 maths:

NCERT exemplar for class 12 maths Chapter 1	NCERT exemplar for class 12 maths Chapter 7
NCERT exemplar for class 12 maths Chapter 2	NCERT exemplar for class 12 maths Chapter 8
NCERT exemplar for class 12 maths Chapter 3	NCERT exemplar for class 12 maths Chapter 9
NCERT exemplar for class 12 maths Chapter 4	NCERT exemplar for class 12 maths Chapter 10
NCERT exemplar for class 12 maths Chapter 5	NCERT exemplar for class 12 maths Chapter 11

FAQs

1. Why should I practice the NCERT Exemplar for Class 12 Maths Chapter 6?

   Practicing the NCERT Exemplar for Class 12 Maths Chapter 6 is essential because it provides an in-depth understanding of the Application of Derivatives. Moreover, it helps students cover all sections of the chapter comprehensively. By working through these problems, students can identify and overcome their doubts, ultimately boosting their confidence for the board examinations.

2. How are the questions in the NCERT Exemplar different from the textbook exercises?

   While the textbook exercises focus on fundamental problems, the NCERT Exemplar includes advanced and application-based questions. Additionally, the exemplar questions are designed to challenge students’ problem-solving skills, making them more prepared for competitive exams. As a result, practicing exemplar questions helps students think critically and understand the concepts on a deeper level.

3. Can I rely solely on the NCERT Exemplar for exam preparation?

   Although the NCERT Exemplar is a valuable resource, it should be used alongside your regular textbook and other reference materials. The exemplar provides additional problems that enhance conceptual understanding; however, it’s important to study the textbook first to grasp the basics. Furthermore, solving exemplar questions regularly will complement your preparation, helping you score higher in the exams.

4. What is the best approach to solving NCERT Exemplar problems effectively?

   To solve NCERT Exemplar problems effectively, start by thoroughly understanding the underlying concepts in the textbook. Next, attempt the exemplar problems step-by-step, ensuring you follow the solution methods provided. Moreover, reviewing your mistakes and clarifying doubts immediately is crucial for solidifying your knowledge. Finally, consistent practice and revision of these exemplar problems will significantly improve your problem-solving skills.