Class 12 - Maths - Application of Derivatives

                                                                                   Exercise 6.1

Question 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

Answer:

The area of a circle (A) with radius (r) is given by

A = πr2

Now, the rate of change of the area with respect to its radius is given by

dA/dr = d(πr2 )/dr = 2πr

  1. When r = 3 cm,

dA/dr = 2π * 3 = 6π

Hence, the area of the circle is changing at the rate of 6π cm2/s when its radius is 3 cm.

  1. When r = 4 cm,

dA/dr = 2π * 4 = 8π

Hence, the area of the circle is changing at the rate of 8π cm2/s when its radius is 4 cm.

Question 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?

Answer:

Let x be the length of a side, V be the volume, and s be the surface area of the cube.

Then, V = x3 and S = 6x2 where x is a function of time t.

It is given that dV/dt = 8 cm3/s

Then, by using the chain rule, we have:

      8 = dV/dt

=> 8 = d(x3)/dt

=> 8 = d(x3)/dx * dx/dt                    [By chain rule]

=> 8 = 3x2 * dx/dt

=> dx/dt = 8/3x2

Now, dS/dt = d(6x2)/dt

=> dS/dt = d(6x2)/dx * dx/dt            [By chain rule]

=> dS/dt = 12x * dx/dt

=> dS/dt = 12x * 8/3x2

=> dS/dt = 32/x

Thus, when x = 12 cm,

=> dS/dt = 32/12

=> dS/dt = 8/3 cm2/s

Hence, if the length of the edge of the cube is 12 cm, then the surface area is increasing at the

rate of 8/3 cm2/s.

Question 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.

Answer:

The area of a circle (A) with radius (r) is given by

A = πr2

Now, the rate of change of area (A) with respect to time (t) is given by,

      dA/dt = d(πr2)/dt

=> dA/dt = d(πr2)/dr * dr/dt              [By chain rule]

=> dA/dt = 2πr * dr/dt

It is given that,

dr/dt = 3 cm/s

Now, dA/dt = 2πr * 3 = 6πr

Thus, when r = 10 cm,

=> dA/dt = 6π * 10 = 60π cm2/s

Hence, the rate at which the area of the circle is increasing when the radius is 10 cm is

60π cm2/s.

Question 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?

Answer:

Let x be the length of a side and V be the volume of the cube. Then,

V = x3

Now, dV/dt = d(x3)/dt

=> dV/dt = d(x3)/dx * dx/dt                   [By chain rule]

=> dV/dt = 3x2 * dx/dt

It is given that,

dx/dt = 3 cm/s

Now, dV/dt = 3x2 * 3 = 9x2

Thus, when x = 10 cm,

=> dV/dt = 9 * 102 = 9 * 100 = 900 cm3/s

Hence, the volume of the cube is increasing at the rate of 900 cm3/s when the edge is 10 cm

long.

Question 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?

Answer:

The area of a circle (A) with radius (r) is given by

A = πr2

Therefore, the rate of change of area (A) with respect to time (t) is given by

      dA/dt = d(πr2)/dt

=> dA/dt = d(πr2)/dr * dr/dt              [By chain rule]

=> dA/dt = 2πr * dr/dt

It is given that dr/dt = 5 cm/s

Thus, when r = 8 cm,

=> dA/dt = 2π * 8 * 5 = 80π

Hence, when the radius of the circular wave is 8 cm, the enclosed area is increasing at the rate

of 80π cm2/s.

Question 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?

Answer:

The circumference of a circle (C) with radius (r) is given by

C = 2πr

Therefore, the rate of change of circumference (C) with respect to time (t) is given by

      dC/dt = d(2πr)/dt

=> dC/dt = d(2πr)/dr * dr/dt             [By chain rule]

=> dC/dt = 2π * dr/dt

It is given that dr/dt = 0.7 cm/s

Hence, the rate of increase of the circumference = 2π * 0.7 = 1.4π cm/s

Question 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.

Answer:

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:

dx/dt = -5 cm/min and dy/dt = 4 cm/min

(a) The perimeter (P) of a rectangle is given by,

P = 2(x + y)

Now, dP/dt = 2(dx/dt + dy/dt)

=> dP/dt = 2(-5 + 4)

=> dP/dt = -2 cm/min

Hence, the perimeter is decreasing at the rate of 2 cm/min.

(b) The area (A) of a rectangle is given by,

A = x * y

Now, dA/dt = dx/dt * y + x * dy/dt

When x = 8 cm and y = 6 cm,

=> dA/dt = -5 * 6 + 4 * 8 = -30 + 32 = 2 cm2/min

Hence, the area of the rectangle is increasing at the rate of 2 cm2/min.

 

Question 8:

A balloon, which always remains spherical on inflation, is being inflated by pumping in 900 cubic centimeters of gas per second.

Find the rate at which the radius of the balloon increases when the radius is 15 cm.

Answer:

The volume of a sphere (V) with radius (r) is given by

V = 4πr3/3

Rate of change of volume (V) with respect to time (t) is given by

      dV/dt = dV/dr * dr/dt       [By chain rule]

=> dV/dt = d(4πr3/3)/dr * dr/dt

=> dV/dt = (3 * 4πr2)/3 * dr/dt

=> dV/dt = 4πr2 * dr/dt

It is given that dV/dt = 900 cm3/s

Therefore, when radius = 15 cm,

=> dr/dt = 225/π(15)2 = 225/225 π = 1/ π

Hence, the rate at which the radius of the balloon increases when the radius is 15 cm is 1/π

cm/s.

Question 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.

Answer:

The volume of a sphere (V) with radius (r) is given by .

Rate of change of volume (V) with respect to its radius (r) is given by,

    dV/dr = d(4πr3/3)/dr

=> dV/dr = (3 * 4πr2)/3

=> dV/dr = 4πr2

Therefore, when radius = 10 cm,

=> dV/dr = 4π * (10)2 = 4π * 100 = 400π

Hence, the volume of the balloon is increasing at the rate of 400π cm3/s.

Question 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?

Answer:

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, by Pythagoras theorem, we have:

     x2 + y2 = 25               [Length of the ladder = 5 m]

=> y2 = 25 - x2

=> y = √(25 - x2)

Then, the rate of change of height (y) with respect to time (t) is given by

=> dy/dt = -x/√(25 - x2) * dx/dt

It is given that dx/dt = 2 cm/s

=> dy/dt = -x/√(25 - x2) * 2

=> dy/dt = -2x/√(25 - x2)

Now, when x = 4 m, we have:

=> dy/dt = (-2 * 4)/√(25 - 42)

=> dy/dt = -8/√(25 - 16)

=> dy/dt = -8/√9

=> dy/dt = -8/3

Hence, the height of the ladder on the wall is decreasing at the rate of 8/3 cm/s .

Question 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.

Answer:

The equation of the curve is given as:

6y = x3 + 2

The rate of change of the position of the particle with respect to time (t) is given by,

      6 * dy/dt = 3x2 * dx/dt + 0

=> 2 * dy/dt = x2 * dx/dt

When the y-coordinate of the particle changes 8 times as fast as the (dy/dt = 8 * dx/dt) x-

Coordinate i.e.

=> 2 * (8 * dx/dt) = x2 * dx/dt

=> 16 * dx/dt = x2 * dx/dt

=> (x2 – 16) * dx/dt = 0

=> x2 – 16 = 0

=> x2 = 16

=> x = ±4

When x = 4, we have

y = (43 + 2)/6 = (64 + 2)/6 = 66/6 = 11

When x = -4, we have

y = {(-4)3 + 2}/6 = (-64 + 2)/6 = -62/6 = -31/3

Hence, the points required on the curve are (4, 11) and (-4, -31/3).

Question 12:

The radius of an air bubble is increasing at the rate of 1/2 cm/s. At what rate is the volume of the bubble increasing when the radius is 1 cm?

Answer:

The air bubble is in the shape of a sphere.

Now, the volume of an air bubble (V) with radius (r) is given by,

V = 4πr3/3

The rate of change of volume (V) with respect to time (t) is given by,

      dV/dt = dV/dr * dr/dt                     [By chain rule]

=> dV/dt = d(4πr3/3)/dr * dr/dt

=> dV/dt = (3 * 4πr2)/3 * dr/dt

=> dV/dt = 4πr2 * dr/dt

It is given that dr/dt = 1/2 cm/s

Therefore, when r = 1 cm,

=> dV/dt = 4π * 12 * 1/2 = 2π cm3/s

Hence, the rate at which the volume of the bubble increases is 2π cm3/s.

Question 13:

A balloon, which always remains spherical, has a variable diameter 3(2x + 1)/2. Find the rate of change of its volume with respect to x.

Answer:

The volume of a sphere (V) with radius (r) is given by

V = 4πr3/3

It is given that:

      Diameter = 3(2x + 1)/2

=> 2r = 3(2x + 1)/2

=> r = 3(2x + 1)/4

Now, V = 4π{3(2x + 1)/4}3/3 = 9π(2x + 1)3/16

Hence, the rate of change of volume with respect to x is as

      dV/dx = 9π/16 * d(2x + 1)3/dx

=> dV/dx = 9π/16 * 3 * (2x + 1)2 * 2

=> dV/dx = 27π(2x + 1)2/8

Question 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?

Answer:

The volume of a cone (V) with radius (r) and height (h) is given by,

V = πr2h/3

It is given that

      h = r/6

=> r = 6h

Now, V = π(6h)2h/3

=> V = 12πh3

The rate of change of volume with respect to time (t) is given by

      dV/dt = 12π * d(h3)/dh * dh/dt

=> dV/dt = 12π * 3h2 * dh/dt

=> dV/dt = 36πh2 * dh/dt

It is also given that dV/dt = 12 cm3/s

Therefore, when h = 4 cm, we have:

     12 = 36π * 42 * dh/dt

=> dh/dt = 12/(36π * 16)

=> dh/dt = 1/48π

Hence, when the height of the sand cone is 4 cm, its height is increasing at the rate of 1/48π

cm/s.

Question 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 – 0.003x2 + 15x + 4000

Find the marginal cost when 17 units are produced.

Answer:

Marginal cost is the rate of change of total cost with respect to output.

Marginal cost (MC) = dC/dt

                                   = 0.007 * 3x2 – 0.003 * 2x + 15

                                   = 0.021x2 – 0.006x + 15   

When x = 17,

MC = 0.021 * (17)2 − 0.006 * 17 + 15

       = 0.021 * 289 − 0.006 * 17 + 15

       = 6.069 − 0.102 + 15

       = 20.967

Hence, when 17 units are produced, the marginal cost is Rs. 20.967.

 

Question 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.

Answer:

Marginal revenue is the rate of change of total revenue with respect to the number of units

sold.

So, Marginal Revenue (MR) = dR/dx

                                                   = 13 * 2x + 26

                                                   = 26x + 26

When x = 7

MR = 26 * 7 + 26 = 182 + 26 = 208

Hence, the required marginal revenue is Rs. 208.

Question 17:

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

Answer:

The area of a circle (A) with radius (r) is given by

A = πr2

Therefore, the rate of change of the area with respect to its radius r is

dA/dr = d(πr2)/dr = 2πr

When r = 6 cm

dA/dr = 2π * 6 = 12π

So, the required rate of change of the area of a circle is 12π cm2/s.

Hence, the correct answer is option B.

Question 18:

The total revenue in Rupees received from the sale of x units of a product is given by

R(x) = 3x2 + 36x + 5

The marginal revenue, when x =15 is   (A) 116   (B) 96   (C) 90 (D) 126

Answer:

Marginal revenue is the rate of change of total revenue with respect to the number of units

sold.

Marginal Revenue (MR) = dR/dx

                                            = 3 * 2x + 36

                                            = 6x + 36

When x = 15,

MR = 6 * 15 + 36 = 90 + 36 = 126

Hence, the required marginal revenue is Rs. 126.

So, the correct answer is option D.

 

                                                                      Exercise 6.2

Question 1:

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

Answer:

Let x1 and x2 be any two numbers in R.

Then, we have:

x1 < x2 => 3x1 < 3x2 => 3x1 + 17 < 3x2 + 17 => f(x1) < f(3x2

Hence, f is strictly increasing on R.

Question 2:

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

Answer:

Let x1 and x2 be any two numbers in R.

Then, we have:

x1 < x2 => 2x1 < 2x2 => e2x1 < e2x2 => f(x1) < f(3x2)

Hence, f is strictly increasing on R.

Question 3:

Show that the function given by f(x) = sin x is

(a) strictly increasing in (0, π/2)  (b) strictly decreasing in (π/2, π)

(c) neither increasing nor decreasing in (0, π)

Answer:

The given function is f(x) = sin x

Now f’(x) = cos x

(a) Since for each x є (0, π/2), cos x > 0, we have f’(x) > 0

Hence, f is strictly increasing in (0, π/2).

(b) Since for each x є (π/2, π), cos x < 0, we have f’(x) < 0

Hence, f is strictly decreasing in (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).

Question 4:

Find the intervals in which the function f given by f(x) = 2x2 − 3x is (a) strictly increasing  (b) strictly decreasing

Answer:

The given function is f(x) = 2x2 − 3x

Now, f’(x) = 4x – 3

      f’(x) = 0

=> 4x – 3 = 0

=> x = 3/4

Now, the point 3/4 divides the real line into two disjoint intervals i.e., (-∞, 3/4) and (3/4, ∞)

In interval (-∞, 3/4), f’(x) = 4x – 3 < 0

Hence, the given function (f) is strictly decreasing in interval (-∞, 3/4).

In interval (3/4, ∞), f’(x) = 4x – 3 > 0

Hence, the given function (f) is strictly increasing in interval (3/4, ∞).

Question 5:

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

Answer:

The given function is f(x) = 2x3 − 3x2 − 36x + 7

Now, f’(x) = 6x2 − 6x – 36

                   = 6(x2 − x – 6)

                   = 6(x + 2)(x - 3)

Now, f’(x) = 0

=> 6(x + 2)(x - 3) = 0         

=> x = − 2, 3

The points x = −2 and x = 3 divide the real line into three disjoint intervals i.e. (-∞, 2), (-2, 3)

and (3, ∞).

In intervals (-∞, 2) and (3, ∞), f’(x) is positive while in interval (−2, 3), f’(x) is negative.

Hence, the given function (f) is strictly increasing in intervals (-∞, 2) and (3, ∞), while function

(f) is strictly decreasing in interval (−2, 3).

Question 6:

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

Answer:

(a) We have,

f(x) = x2 + 2x – 5

f’(x) = 2x + 2

Now, f’(x) = 0

=> 2x + 2 = 0

=> x = -1

Point x = −1 divides the real line into two disjoint intervals i.e. (-∞, -1) and (-1, ∞)

In interval (-∞, -1), f’(x) = 2x + 2 < 0

So, f is strictly decreasing in interval (-∞, -1).

Thus, f is strictly decreasing for x < −1

In interval (-1, ∞), f’(x) = 2x + 2 > 0

So, f is strictly increasing in interval (-1, ∞).

Thus, f is strictly increasing for x > −1

(b) We have,

f(x) = 10 − 6x − 2x2

f’(x) = -6 – 4x

Now, f’(x) > 0

=> -6 – 4x = 0

=> x = -3/2

The point divides the real line into two disjoint intervals (-∞, -3/2) and (-3/2, ∞)

In interval (-∞, -3/2), f’(x) = -6 – 4x > 0

So, f is strictly increasing in interval (-∞, -3/2).

Thus, f is strictly increasing for x < −3/2

In interval (-3/2, ∞), f’(x) = -6 – 4x < 0

So, f is strictly decreasing in interval (-3/2, ∞).

Thus, f is strictly increasing for x > −3/2

(c) We have,

f(x) = −2x3 − 9x2 − 12x + 1

f’(x) = −6x2 − 18x – 12

Now, f’(x) = 0

=> −6x2 − 18x – 12 = 0

=> x2 + 3x + 2 = 0

=> (x + 1)(x + 2) = 0

=> x = -1, -2

Points x = −1 and x = −2 divide the real line into three disjoint intervals

i.e. (-∞, -2), (-2, -1) and (-1, ∞)

In intervals (-∞, -2) and (-1, ∞) i.e., when x < −2 and x > −1

f’(x) = −6x2 − 18x – 12 < 0

So, f is strictly decreasing for x < −2 and x > −1

f’(x) = −6x2 − 18x – 12 < 0

Now, in interval (−2, −1) i.e., when −2 < x < −1,

f’(x) = −6x2 − 18x – 12 > 0

So, f is strictly increasing for −2 < x < −1

(d) We have,

f(x) = 6 − 9x − x2   

f’(x) = -9 – 2x

Now, f’(x) = 0

=> -9 – 2x = 0

=> x = -9/2

The point divides the real line into two disjoint intervals i.e. (-∞, -9/2) and (-9/2, ∞).

In interval (-∞, -9/2)i.e. for x < -9/2

f’(x) = -9 – 2x > 0

So, f is strictly increasing for x < -9/2

In interval (-9/2, ∞) i.e., for x > -9/2

f’(x) = -9 – 2x < 0

So, f is strictly decreasing for x > -9/2

(e) We have,

f(x) = (x + 1)3(x − 3)3

f’(x) = 3(x + 1)2(x − 3)3 + 3 (x − 3)2(x + 1)3

         = 3(x + 1)2(x − 3)2 [x – 3 + x + 1]

         = 3(x + 1)2(x − 3)2(2x - 2)

         = 6(x + 1)2(x − 3)2(x - 1)    

Now, f’(x) = 0

=> 6(x + 1)2(x − 3)2(x - 1) = 0

=> x = -1, 3, 1

The points x = −1, x = 1, and x = 3 divide the real line into four disjoint intervals i.e. (-∞, -1),

(−1, 1), (1, 3) and (3, ∞).

In intervals (-∞, -1) and (−1, 1)

f’(x) = 6(x + 1)2(x − 3)2(x - 1) < 0

So, f is strictly decreasing in intervals (-∞, -1) and (−1, 1).

In intervals (1, 3) and (3, ∞)

f’(x) = 6(x + 1)2(x − 3)2(x - 1) > 0

So, f is strictly increasing in intervals (1, 3) and (3, ∞).

Question 7:

Show that y = log(1 + x) – 2x/(2 + x), x > -1, is an increasing function of x throughout its domain.

Answer:

Given, y = log(1 + x) – 2x/(2 + x)

Now, dy/dx = 1/(1 + x) – {(2 + x) * 2 – 2x * 1}/(2 + x)2

=> dy/dx = 1/(1 + x) – 4/(2 + x)2

=> dy/dx = x2/{(x + 1)(2 + x)2}

Now, dy/dx= 0

=> x2/{(x + 1)(2 + x)2} = 0

=> x2 = 0

=> x = 0

Since x > −1, point x = 0 divides the domain (−1, ∞) in two disjoint intervals i.e., −1 < x < 0

and x > 0.

When −1 < x < 0, we have:

x < 0 => x2 > 0

x > -1 => 2 + x > 0 => (2 + x)2 > 0

So, y’ = x2/{(1 + x)(2 + x)2} > 0

Also, when x > 0:

x > 0 => x2 > 0 ,(2 + x)2 > 0

So, y’ = x2/{(1 + x)(2 + x)2} > 0

Hence, function f is increasing throughout this domain.

Question 8:

Find the values of x for which y = [x(x - 2)]2 is an increasing function.

Answer:

We have, y = [x(x - 2)]2 = [x2 - 2x]2

Now, dy/dx = y’ = y = 2(x2 - 2x)(2x - 2) = 4x(x - 2)(x - 1)

dy/dx = 0

=> 4x(x - 2)(x - 1) = 0

=> x = 0, 1, 2

The points x = 0, x = 1, and x = 2 divide the real line into four disjoint intervals i.e. (-∞, 0), (0, 1)

(1, 2) and (2, ∞).

In intervals (-∞, 0) and (1, 2), dy/dx < 0

So, y is strictly decreasing in intervals (-∞, 0) and (1, 2).

However, in intervals (0, 1) and (2, ∞), dy/dx > 0

So, y is strictly increasing in intervals (0, 1) and (2, ∞).

Hence, y is strictly increasing for 0 < x < 1 and x > 2

Question 9:

Prove that y = 4 sin θ /(2 + cos θ) – θ is an increasing function of θ in [0, π/2].

Answer:

We have, y = 4 sin θ /(2 + cos θ) – θ

Now, dy/dx = {(2 + cos θ) * 4 cos θ – 4 sin θ (-sin θ)}/(2 + cos θ)2 - 1

                      = (8 cos θ + 4 cos2 θ + 4 sin2 θ)/(2 + cos θ)2 - 1

                      = (8 cos θ + 4)/(2 + cos θ)2 - 1

Now, dy/dx = 0

=> (8 cos θ + 4)/(2 + cos θ)2 – 1 = 0

=> (8 cos θ + 4)/(2 + cos θ)2 = 1

=> (8 cos θ + 4) = (2 + cos θ)2

=> 8 cos θ + 4 = 4 + cos2 θ + 4 cos θ

=> cos2 θ - 4 cos θ = 0

=> cos θ(cos θ – 4) = 0

=> cos θ = 0, 4

Since, cos θ ≠ 4

Hence, cos θ = 0

=> θ = π/2

Again, dy/dx = (8 cos θ + 4)/(2 + cos θ)2 – 1

=> dy/dx = {8 cos θ + 4 – (4 + cos2 θ + 4 cos θ)}/(2 + cos θ)2

=> dy/dx = (4 cos θ – cos2 θ)/(2 + cos θ)2

=> dy/dx = cos θ(4 – cos θ)/(2 + cos θ)2

In interval (0, π/2), we have cos θ > 0. Also, 4 > cos θ => 4 − cos θ > 0

So, cos θ(4 – cos θ) > 0 and (2 + cos θ)2 > 0

=> cos θ(4 – cos θ)/(2 + cos θ)2 > 0

=> dy/dx > 0

Therefore, y is strictly increasing in interval (0, π/2).

Also, the given function is continuous at x = 0 and x = π/2

Hence, y is increasing in interval [0, π/2].

Question 10:

Prove that the logarithmic function is strictly increasing on (0, ∞).

Answer:

The given function is f(x) = log x

So, f’(x) = 1/x

It is clear that for x > 0, f’(x) = 1/x > 0

Hence, f(x) = log x is strictly increasing in interval (0, ∞).

 

Question 11:

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

Answer:

The given function is f(x) = x2 − x + 1

So, f’(x) = 2x – 1

Now, f’(x) = 0

=> 2x – 1 = 0

=> x = 1/2

The point divides the interval (−1, 1) into two disjoint intervals i.e. (-1, 1/2) and (1/2, 1)

Now, in interval (-1, 1/2), f’(x) = 2x – 1 < 0

Therefore, f is strictly decreasing in interval (-1, 1/2).

However, in interval (1/2, 1), f’(x) = 2x – 1 > 0

Therefore, f is strictly increasing in interval (1/2, 1).

Hence, f is neither strictly increasing nor decreasing in interval (-1, 1).

Question 12:

Which of the following functions are strictly decreasing on (0, π/2)?

(A) cos x                         (B) cos 2x                           (C) cos 3x                              (D) tan x

Answer:

(A) Let f(x) = cos x

So, f’(x) = -sin x

In interval (0, π/2), f’(x) = -sin x < 0

So, f(x) = cos x is strictly decreasing in interval (0, π/2)

(B) Let f(x) = cos2 x

So, f’(x) = -2 sin2 x

Now, 0 < x < π/2

=> 0 < 2x < π

=> sin 2x > 0

=> -2 sin 2x < 0

So, f(x) = cos 2x is strictly decreasing in interval (0, π/2).

(C) Let f(x) = cos 3x

So, f’(x) = -3 sin 3x

Now, f’(x) = 0

=> -3 sin 3x = 0

=> sin 3x = 0

=> 3x = π as x є (0, π/2)

=> x = π/3

The point x = π/3 divides the interval (0, π/2) into two disjoint intervals i.e. (0, π/3) and

(π/3, π/2).

Now, in interval (0, π/3), f’(x) = -3 sin 3x < 0                [As 0 < x < π/3 => 0 < 3x < π]

So, f(x) = cos 3x is strictly decreasing in interval (0, π/3)

However in interval (π/3, π/2), f’(x) = -3 sin 3x > 0    [As π/3 < x < π/2 => π < 3x < 3π/2]

So, f(x) = cos 3x is strictly increasing in interval (π/3, π/2).

Hence, f(x) = cos 3x is neither increasing nor decreasing in the interval (0, π/2).

(A) Let f(x) = tan x

So, f’(x) = sec2 x

In interval (0, π/2), f’(x) = sec2 x > 0

So, f(x) = tan x is strictly increasing in interval (0, π/2).

Therefore, functions cos x and cos 2x are strictly decreasing in (0, π/2).

Hence, the correct answers are A and B.

Question 13:

On which of the following intervals is the function f given by f(x) = x100 + sin x – 1 strictly decreasing?                                                                                                                                                 

(A) (0, 1)                           (B) (π/2, π)                         (C) (0, π/2)                        (D) None of these

Answer:

We have,

f(x) = x100 + sin x – 1

f’(x) = 100x99 + cos x

In interval (0, 1) cos x > 0 and 100x99 > 0

Thus, function f is strictly increasing in interval (0, 1).

In interval (π/2, π), cos x < 0 and 100x99 > 0

Also, 100x99 > cos x

So, f’(x) > 0 in (π/2, π)

Thus, function f is strictly increasing in interval (π/2, π).

In interval (0, π/2), cos x > 0 and 100x99 > 0

So, 100x99 + cos x > 0

=> f’(x) > 0 in (0, π/2)

Thus, function f is strictly increasing in interval (0, π/2).

Hence, function f is strictly decreasing in none of the intervals.

So, the correct answer is option D.

 

Question 14:

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

Answer:

We have, f(x) = x2 + ax + 1

So, f’(x) = 2x + a

Now, function f is increasing on [1, 2]

=> f’(x) ≥ 0 on [1, 2]

Now, we have 1 ≤ x ≤ 2

=> 2 ≤ 2x ≤ 4

=> 2 + a ≤ 2x + a ≤ 4 + a

  => 2 + a ≤ f’(x) ≤ 4 + a

Since f’(x) ≥ 0

=> 2 + a ≥ 0

=> a ≥ -2

So, the least value of a is −2

Question 15:

Let I be any interval disjoint from (−1, 1). Prove that the function f given by f(x) = x + 1/x is strictly increasing on I.

Answer:

We have, f(x) = x + 1/x

So, f’(x) = 1 - 1/x2

Now, f’(x) = 0

=> 1 - 1/x2 = 0

=> 1/x2 = 1

=> x2 = 1

=> x = ±1

The points x = 1 and x = −1 divide the real line in three disjoint intervals i.e. (-∞, -1), (-1, 1)

and (1, ∞).

In interval (−1, 1), it is observed that:

-1 < x < 1

=> x2 < 1

=> 1 < 1/x2, x ≠ 0

=> 1 - 1/x2 < 0, x ≠ 0

So, f’(x) = 1 - 1/x2 < 0 on (-1, 1) – {0}

Hence, f is strictly decreasing on (-1, 1) – {0}.

In intervals (-∞, -1) and (1, ∞), it is observed that:

      x < -1 or 1 < x

=> x2 > 1

=> 1 > 1/x2

=> 1 - 1/x2 > 0

So, f’(x) = 1 - 1/x2 > 0 on (-∞, -1) and (1, ∞)

Hence, f is strictly increasing on (-∞, -1) and (1, ∞).

Hence, function f is strictly increasing in interval I disjoint from (−1, 1).

Hence, the given result is proved.

Question 16:

Prove that the function f given by f(x) = log sin x is strictly increasing on (0, π/2) and strictly decreasing on (π/2, π).

Answer:

We have, f(x) = log sin x

So, f’(x) = (1/sin x) * cos x = cot x

In interval (0, π/2), f’(x) = cot x > 0

So, f is strictly increasing in (0, π/2).

In interval (π/2, π), f’(x) = cot x < 0

So, f is strictly decreasing in (π/2, π).

Question 17:

Prove that the function f given by f(x) = log cos x is strictly increasing on (0, π/2) and strictly decreasing on (π/2, π).

Answer:

We have, f(x) = log cos x

So, f’(x) = (1/cos x) * (- sin x) = -tan x

In interval (0, π/2),

      tan x > 0

=> -tan x < 0

=> f’(x) = -tan x < 0 on (0, π/2)

So, f is strictly decreasing in (0, π/2).

In interval (π/2, π),

      tan x < 0

=> -tan x > 0

=> f’(x) = -tan x > 0 on (π/2, π)

So, f is strictly increasing in (π/2, π).

 

Question 18:

Prove that the function given by f(x) = x3 – 3x2 + 3x - 100 is increasing in R.

Answer:

Given, f(x) = x3 – 3x2 + 3x – 100

So, f’(x) = 3x2 – 6x + 3

=> f’(x) = 3(x2 – 2x + 1)

=> f’(x) = 3(x - 1)2

For any x ∈ R, (x − 1)2 > 0.

Thus, f’(x) is always positive in R.

Hence, the given function (f) is increasing in R.

Question 19:

The interval in which y = x2e-x is increasing is                                                                                        

(A) (-∞, ∞)                      (B) (−2, 0)                          (C) (2, ∞)                              (D) (0, 2)

Answer:

Given, y = x2e-x

So, dy/dx = 2x e-x - x2e-x

=> dy/dx = xe-x(2 - x)

Now, dy/dx = 0

=> xe-x(2 - x) = 0

=> x = 0, 2

The points x = 0 and x = 2 divide the real line into three disjoint intervals i.e. (-∞, 0), (0, 2) and

(2, ∞).

In intervals (-∞, 0) and (2, ∞), f’(x) < 0 since e-x is always positive.

So, f is decreasing on (-∞, 0) and (2, ∞)

In interval (0, 2), f’(x) > 0

So, f is strictly increasing on (0, 2).

Hence, f is strictly increasing in interval (0, 2).

Therefore, the correct option is D.

 

                                                                     Exercise 6.3

Question 1:

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

Answer:

The given curve is y = 3x4 − 4x

Now, dy/dx = 12x3 − 4

Then, the slope of the tangent to the given curve at x = 4 is given by,

dy/dx]x = 4 = 12 * 43 – 4 = 12 * 64 – 4 = 768 – 4 = 764

Question 2:

Find the slope of the tangent to the curve, y = (x - 1)/(x - 2), x ≠ 2 at x = 10.

Answer:

The given curve is y = (x - 1)/(x - 2)

Now, dy/dx = {(x - 2) * d(x - 1)/dx - (x - 1) * d(x - 2)/dx}/(x - 2)2

=> dy/dx = {(x - 2) * 1 - (x - 1) * 1}/(x - 2)2

=> dy/dx = (x - 2 - x + 1)/(x - 2)2

=> dy/dx = -1/(x - 2)2

Thus, the slope of the tangent at x = 10 is given by,

=> dy/dx]x = 10 = -1/(10 - 2)2 = -1/82 = -1/64

Hence, the slope of the tangent at x = 10 is -1/64

Question 3:

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

Answer:

The given curve is y = x3 – x + 1

Now, dy/dx = 3x2 – 1

The slope of the tangent to a curve at (x0, y0) is dy/dx](x0, y0)

It is given that x0 = 2

Then, the slope of the tangent to the given curve at x = 2 is given by,

dy/dx]x = 2 = 3 * 22 – 1 = 12  – 1 = 11

Question 4:

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

Answer:

The given curve is y = x3 – 3x + 2

Now, dy/dx = 3x2 – 3

The slope of the tangent to a curve at (x0, y0) is dy/dx](x0, y0)

It is given that x0 = 3

Then, the slope of the tangent to the given curve at x = 4 is given by,

dy/dx]x = 3 = 3 * 32 – 1 = 27  – 3 = 24

Question 5:

Find the slope of the normal to the curve x = a cos3 θ, y = a sin3 θ at θ = π/4.

Answer:

It is given that x = a cos3 θ, y = a sin3 θ

dx/dθ = 3a cos2 θ * (-sin θ) = -3a cos2 θ * sin θ

dy/dθ = 3a sin2 θ * cos θ

Now, dy/dx = (dy/dθ)/( dy/dθ)

                      = (3a sin2 θ * cos θ)/( -3a cos2 θ * sin θ)

                      = - sin θ /cos θ

                      = - tan θ          

Therefore, the slope of the tangent at θ = π/4 is given by,

dy/dx]θ = π/4 = -tan π/4 = -1

Now, the slope of the normal at θ = π/4 = -1/[dy/dx]θ = π/4

                                                                       = -1/(-1)

                                                                       = 1

Question 6:

Find the slope of the normal to the curve x = 1 − a sin θ, y = b cos2 θ at θ = π/2.

Answer:

It is given that x = 1 − a sin θ, y = b cos2 θ

dx/dθ = -a cos θ

dy/dθ = 2b cos θ * (-sin θ) = -2b sin θ * cos θ

Now, dy/dx = (dy/dθ)/(dx/dθ)

                      = (-2b sin θ * cos θ)/(-a cos θ)

                      = (2b/a) sin θ

Therefore, the slope of the tangent at θ = π/2 is given by,

dy/dx]θ = π/2 = (2b/a) sin π/2 = (2b/a) * 1 = 2b/a

Now, the slope of the normal at θ = π/2 is given as

= -1/[dy/dx]θ = π/2

= -1/(2b/a)

= -a/2b

Question 7:

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

Answer:

The equation of the given curve is y = x3 − 3x2 − 9x + 7

So, dy/dx = 3x2 − 6x − 9

Now, the tangent is parallel to the x-axis if the slope of the tangent is zero.

     dy/dx = 0

=> 3x2 − 6x – 9 = 0

=> 3(x2 − 2x – 3) = 0

=> x2 − 2x – 3 = 0

=> (x - 3)(x + 1) = 0

=> x = 3, -1

When x = 3, y = (3)3 − 3 (3)2 − 9 (3) + 7 = 27 − 27 − 27 + 7 = −20

When x = −1, 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).

Question 8:

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).

Answer:

If a tangent is parallel to the chord joining the points (2, 0) and (4, 4), then the slope of

the tangent = the slope of the chord.

The slope of the chord = (4 - 0)/(4 - 2) = 4/2 = 2

Now, the slope of the tangent to the given curve at a point (x, y) is given by,

dy/dx = 2(x - 2)

Since the slope of the tangent = slope of the chord, we have:

      2(x – 2) = 2

=> x – 2 = 1

=> x = 3

When x = 3, y = (3 - 2)2 = 1

Hence, the required point is (3, 1).

Question 9:

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

Answer:

The equation of the given curve is y = x3 − 11x + 5

The equation of the tangent to the given curve is given as y = x − 11 (which is of the form

y = mx + c).

So, Slope of the tangent = 1

Now, the slope of the tangent to the given curve at the point (x, y) is given by,

dy/dx = 3x2 - 11

Then, we have:

      3x2 – 11 = 1

=> 3x2 = 12

=> x2 = 4

=> x = ±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).

But both the points should satisfy the equation of the tangent as there would be point of

contact between tangent and curve.

So, (2, -9) is the required point since (-2, 19) does not satisfy the given equation of tangent.

Question 10:

Find the equation of all lines having slope −1 that are tangents to the curve: y = 1/(x – 1), x ≠ 1

Answer:

The equation of the given curve is y = 1/(x – 1), x ≠ 1

The slope of the tangents to the given curve at any point (x, y) is given by,

dy/dx = -1/(x - 1)2

If the slope of the tangent is −1, then we have:

     -1/(x - 1)2 = -1

=> (x - 1)2 = 1

=> x – 1 = ±1

=> x = 0, 2

When x = 0, y = −1 and 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).

The equation of the tangent through (0, −1) is given by,

     y – (-1) = −1(x − 0)

=> y + 1 = −x

=> x + y + 1 = 0

The equation of the tangent through (2, 1) is given by,

     y − 1 = −1(x − 2)

=> y − 1 = − x + 2

=> y + x − 3 = 0

Hence, the equations of the required lines are x + y + 1 = 0 and x + y − 3 = 0.

Question 11:

Find the equation of all lines having slope 2 which are tangents to the curve: y = 1/(x – 3), x ≠ 3

Answer:

The equation of the given curve is y = 1/(x – 3), x ≠ 3

The slope of the tangent to the given curve at any point (x, y) is given by,

dy/dx = -1/(x - 3)2

If the slope of the tangent is 2, then we have:

     -1/(x - 3)2 = 2

=> (x - 3)2 = -1/2

This is not possible since LHS is positive and RHS is negative.

Hence, there is no tangent to the given curve having slope 2.

Question 12:

Find the equations of all lines having slope 0 which are tangent to the curve: y = 1/(x2 – 2x + 3)

Answer:

The equation of the given curve is y = 1/(x2 – 2x + 3)

The slope of the tangent to the given curve at any point (x, y) is given by,

dy/dx = -(2x - 2)/(x2 – 2x + 3) = -2(x - 1)/(x2 – 2x + 3)

If the slope of the tangent is 0, then we have:

      -2(x - 1)/(x2 – 2x + 3) = 0

=> -2(x - 1) = 0

=> x – 1 = 0

=> x = 1

When x = 1, y = 1/(1 – 2 + 3) = 1/2

Now, the equation of the tangent through (1, 1/2) is given by,

     y – 1/2 = 0(x - 1)

=> y – 1/2 = 0

=> y = 1/2 

Hence, the equation of the required line is y = 1/2

Question 13:

Find points on the curve x2/9 + y2/16 = 1 at which the tangents are   (i) parallel to x-axis  (ii) parallel to y-axis

Answer:

The equation of the given curve is x2/9 + y2/16 = 1

On differentiating both sides with respect to x, we have:

     2x/9 + (2y/16) * dy/dx = 0

=> dy/dx = -16x/9y

(i).The tangent is parallel to the x-axis if the slope of the tangent -16x/9y = 0 is 0 which is

possible if x = 0.

for x = 0,

      0/9 + y2/16 = 1

=> y2/16 = 1

=> y2 = 16

=> y = ±4

Hence, the points at which the tangents are parallel to the x-axis are (0, 4) and (0, − 4).

(ii). The tangent is parallel to the y-axis if the slope of the normal is 0, which gives

     -1/(-16x/9y) = 0

=>9y/16x = 0

=> 9y = 0

=> y = 0

for y = 0,

       x2/9 + 0/16 = 1

=> x2/9 = 1

=> x2 = 9

=> x = ±3

Hence, the points at which the tangents are parallel to the y-axis are (3, 0) and (− 3, 0).

Question 14:

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 = cos t, y = sin t at t = π/4

Answer:

(i) The equation of the curve is y = x4 − 6x3 + 13x2 − 10x + 5

On differentiating with respect to x, we get:

      dy/dx = 4x3 − 18x2 + 26x – 10

      dy/dx](0, 5) = 4 * 03 – 18 * 02 + 26 * 0 – 10

=> dy/dx](0, 5) = -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)

=> y − 5 = − 10x

=> 10x + y = 5

The slope of the normal at (0, 5) = -1/Slope of the tangent at (0, 5) = -1/(-10) = 1/10

Therefore, the equation of the normal at (0, 5) is given as:

      y − 5 = (1/10)(x − 0)

=> 10y − 50 = x

=> x - 10y + 50 = 0

(ii) The equation of the curve is y = x4 − 6x3 + 13x2 − 10x + 5

On differentiating with respect to x, we get:

      dy/dx = 4x3 − 18x2 + 26x – 10

      dy/dx](1, 3) = 4 * 13 – 18 * 12 + 26 * 1 – 10

=> dy/dx](1, 3) = 4 – 18 + 26 – 10

=> dy/dx](1, 3) = 30 – 28

=> dy/dx](1, 3) = 2

Thus, the slope of the tangent at (1, 3) is 2. The equation of the tangent is given as:

     y − 3 = 2(x − 1)

=> y − 3 = 2x - 2

=> y = 2x + 1

The slope of the normal at (1, 3) = -1/Slope of the tangent at (1, 3) = -1/2

Therefore, the equation of the normal at (1, 3) is given as:

      y − 3 = (-1/2)(x − 1)

=> 2y − 6 = -x + 1

=> x + 2y - 7 = 0

(iii) The equation of the curve is y = x3

On differentiating with respect to x, we get:

dy/dx = 3x2

dy/dx](1, 1) = 3 * 12 = 3

Thus, the slope of the tangent at (1, 1) is 3 and the equation of the tangent is given as:

     y – 1 = 3(x – 1)

=> y – 1 = 3x – 3

=> y = 3x - 2

The slope of the normal at (1, 1) = -1/Slope of the tangent at (1, 3) = -1/3

Therefore, the equation of the normal at (1, 1) is given as:

     y – 1 = (-1/3)(x – 1)

=> 3y – 3 = -x + 1

=> x + 3y – 4 = 0

(iv) The equation of the curve is y = x2

On differentiating with respect to x, we get:

dy/dx = 2x

dy/dx](0, 0) = 2 * 0 = 0

Thus, the slope of the tangent at (0, 0) is 0 and the equation of the tangent is given as:

     y – 0 = 0(x – 0)

=> y = 0

The slope of the normal at (0, 0) = -1/Slope of the tangent at (0, 0) = -1/0

Which is not defined.

Therefore, the equation of the normal at (x0, y0) = (0, 0) is given as:

x = x0 = 0

(v) The equation of the curve is x = cos t, y = sin t

So, dx/dt = -sin t and dy/dt = cos t

Now, dy/dx = (dy/dt)/(dx/dt) = -cos t /sin t = -cot t

dy/dx]t = π/4 = -cot π/4 = -1

Thus, the slope of the tangent at t = π/4 is -1

When t = π/4,

x = cos π/4 = 1/√2

y = sin π/4 = 1/√2

Thus, the equation of the tangent at t = π/4 i.e. (1/√2, 1/√2) is:

     y – 1/√2 = -1(x – 1/√2)

=> x + y - 1/√2 - 1/√2 = 0

=> x + y - √2 = 0

The slope of the normal at t = π/4 = -1/(Slope of the tangent at t = π/4) = -1/(-1) = 1

Thus, the equation of the normal at t = π/4 i.e. (1/√2, 1/√2) is:

     y – 1/√2 = 1(x – 1/√2)

=> x - y - 1/√2 + 1/√2 = 0

=> x - y = 0

Question 15:

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

Answer:

The equation of the given curve is y = x2 − 2x + 7

On differentiating with respect to x, we get:

dy/dx = 2x - 2

(a) The equation of the line is 2x − y + 9 = 0

=> y = 2x + 9

This is of the form y = mx + c

So, the slope of the line = 2

If a tangent is parallel to the line 2x − y + 9 = 0, then the slope of the tangent is equal to the

slope of the line.

Therefore, we have:

      2 = 2x − 2

=> 2x = 4

=> x = 2

Now, x = 2,

       y = 2 * 2 − 4 + 7

=> y = 4 – 4 + 7

=> y = 7

Thus, the equation of the tangent passing through (2, 7) is given by,

      y – 7 = 2(x - 2)

=> y – 7 = 2x – 4

=> y – 2x – 3 = 0

Hence, the equation of the tangent line to the given curve (which is parallel to line

2x − y + 9 = 0) is y – 2x – 3 = 0

(b) The equation of the line is 5y − 15x = 13

     5y = 15x + 13

=> y = 3x + 13/5

This is of the form y = mx + c

So, the slope of the line = 3

If a tangent is perpendicular to the line 5y − 15x = 13, then the slope of the tangent is

= -1/slope of the line = -1/3

=> 2x – 2 = -1/3

=> 2x = -1/3 + 2

=> 2x = 5/3

=> x = 5/6

Now, at x = 5/6

      y = (5/6)2 – 2 * 5/6 + 7

=> y = 25/36 – 10/6 + 7

=> y = (25 – 60 + 252)/36

=> y = 217/36

Thus, the equation of the tangent passing through (5/6, 217/36) is given by,

      y – 217/36 = (-1/3)(x – 5/6)

=> (36y – 217)/36 = (-1/18)(6x – 5)

=> (36y – 217)/2 = -(6x – 5)

=> 36y – 217 = -2(6x - 5)

=> 36y – 217 = -12x + 10

=> 36y + 12x – 227 = 0

Hence, the equation of the tangent line to the given curve (which is perpendicular to line

5y – 15x = 13) is 36y + 12x – 227 = 0

Question 16:

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

Answer:

The equation of the given curve is y = 7x3 + 11

So, dy/dx = 21x2

The slope of the tangent to a curve at (x0, y0) is dy/dx](x0, y0)

Therefore, the slope of the tangent at the point where x = 2 is given by,

dy/dx]x = 2 = 21 * 22 = 21 * 4 = 84

It is observed that the slopes of the tangents at the points where x = 2 and x = −2 are equal.

Hence, the two tangents are parallel.

Question 17:

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

Answer:

The equation of the given curve is y = x3

So, dy/dx = 3x2

The slope of the tangent at the point (x, y) is given by,

dy/dx](x, y) = 3x2

When the slope of the tangent is equal to the y-coordinate of the point, then y = 3x2

Also, we have y = x3

So, 3x2 = x3

=> x2(x − 3) = 0

=> x = 0, 3

When x = 0, then y = 0 and

when x = 3, then y = 3 * 32 = 3 * 9 = 27

Hence, the required points are (0, 0) and (3, 27).

Question 18:

For the curve y = 4x3 − 2x5, find all the points at which the tangents passes through the origin.

Answer:

The equation of the given curve is y = 4x3 − 2x5

So, dy/dx = 12x2 − 10x4

Therefore, the slope of the tangent at a point (x, y) is 12x2 − 10x4

The equation of the tangent at (x, y) is given by,

Y – y = (12x2 − 10x4)(X – x)   …………1

When the tangent passes through the origin (0, 0), then X = Y = 0

Therefore, equation 1 reduces to:

     -y = (12x2 − 10x4)(0 – x)

=> y = 12x3 − 10x5

Also, we have y = 4x3 − 2x5

So, 12x3 − 10x5 = 4x3 − 2x5

=> 8x5 − 8x3 = 0

=> x5 − x3 = 0

=> x3(x2 - 1) = 0

=> x = 0, ±1

When x = 1, y = 4 (0)3 − 2 (0)5 = 0

When x = 1, y = 4 (1)3 − 2 (1)5 = 4 – 2 = 2

When x = −1, y = 4 (−1)3 − 2 (−1)5 = -4 + 2 = -2

Hence, the required points are (0, 0), (1, 2), and (−1, −2).

Question 19:

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

Answer:

The equation of the given curve is x2 + y2 − 2x − 3 = 0

On differentiating with respect to x, we have:

     2x + 2y * dy/dx – 2 = 0

=> 2y * dy/dx = 2 – 2x

=> y * dy/dx = 1 – x

=> dy/dx = (1 – x)/y

Now, the tangents are parallel to the x-axis if the slope of the tangent is 0.

So, (1 – x)/y = 0

=> 1 – x = 0

=> x = 1

But, x2 + y2 − 2x − 3 = 0

Now, for x = 1

      12 + y2 – 2 * 1 − 3 = 0

=> 1 + y2 – 2 − 3 = 0

=> y2 – 4 = 0

=> y2 = 4

=> y = ±2

Hence, the points at which the tangents are parallel to the x-axis are (1, 2) and (1, −2).

Question 20:

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

Answer:

The equation of the given curve is ay2 = x3

On differentiating with respect to x, we have:    

2ay * dy/dx = 3x2

=> dy/dx = 3x2/2ay

The slope of the tangent to a curve at (x0, y0) is dy/dx](x0, y0)

=> Slope of the tangent at (am2, am3) is

      dy/dx = 3(am2)2/{2a * am3}

=> dy/dx = 3a2 m4/(2a2 * m3)

=> dy/dx = 3m/2

So, the Slope of normal at (am2, am3) is

= -1/slope of the tangent at (am2, am3)

= -1/(3m/2)

= -2/3m

 Hence, the equation of the normal at (am2, am3) is given by,

      y – am3 = (-2/3m)(x – am2)

=> 3my – 3am4 = -2x + 2am2

=> 2x + 3my – am2 (2 + 3m2) = 0

 

Question 21:

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

Answer:

The equation of the given curve is y = x3 + 2x + 6

The slope of the tangent to the given curve at any point (x, y) is given by,

dy/dx = 3x2 + 2

So, the slope of the normal to the given curve at any point (x, y)

= -1/ slope of the tangent at the point (x, y)

= -1/(3x2 + 2)

The equation of the given line is x + 14y + 4 = 0

=> y = -x/14 – 4/14 (which is of the form y = mx + c)

So, the slope of the given line = -1/14

If the normal is parallel to the line, then we must have the slope of the normal being equal to

the slope of the line.

So, -1/(3x2 + 2) = -1/14

=> 3x2 + 2 = 14

=> 3x2 = 12

=> x2 = 4

=> x = ±2

When x = 2, y = 8 + 4 + 6 = 18

When x = −2, y = − 8 − 4 + 6 = −6

Therefore, there are two normals to the given curve with slope -1/14 and passing through the

points (2, 18) and (−2, −6).

Thus, the equation of the normal through (2, 18) is given by,

      y – 18 = (-1/14)(x - 2)

=> 14y – 252 = -x + 2

=> x + 14y – 254 = 0

And, the equation of the normal through (−2, −6) is given by,

      y – (-6) = (-1/14)[x – (-2)]

=> y + 6 = (-1/14)(x + 2)

=> 14y + 84 = -x - 2

=> x + 14y + 86 = 0

Hence, the equations of the normals to the given curve (which are parallel to the given line)

are x + 14y - 254 = 0 and x + 14y + 86 = 0

Question 22:

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

Answer:

The equation of the given parabola is y2 = 4ax

On differentiating y2 = 4ax with respect to x, we have:

      2y * dy/dx = 4a

=> dy/dx = 2a/y

So, the slope of the tangent at (at2, 2at) is

=> dy/dx = 2a/2at = 1/t

Then, the equation of the tangent at (at2, 2at) is given by

      y – 2at = (1/t)(x - at2)

=> ty – 2at2 = x - at2

=> ty = x + at2

Now, the slope of the normal at (at2, 2at) is

= -1/ Slope of the tangent at (at2, 2at)

= -1/(1/t)

= -t

Then, the equation of the normal at (at2, 2at) is given by

      y – 2at = -t(x - at2)

=> y – 2at = -tx + at3

=> y = -tx + 2at + at3

Question 23:

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

[Hint: Two curves intersect at right angle if the tangents to the curves at the point of intersection are perpendicular to each other.]

Answer:

The equations of the given curves are given as x = y2 and xy = k

Putting x = y2 in xy = k, we get:

      y3 = k

=> y = k1/3

So, x = k2/3

Thus, the point of intersection of the given curves is (k2/3, k1/3).

Differentiating x = y2 with respect to x, we have:

      1 = 2y * dy/dx

=> dy/dx = 1/2y

Therefore, the slope of the tangent to the curve x = y2 at (k2/3, k1/3) is

=> dy/dx = 1/2k1/3

On differentiating xy = k with respect to x, we have:

      x * dy/dx + y = 0

=> dy/dx = -y/x

So, the slope of the tangent to the curve xy = k at (k2/3, k1/3) is

=> dy/dx = - k2/3/k1/3 = -1/ k1/3

We know that two curves intersect at right angles if the tangents to the curves at the point of

intersection i.e., at (k2/3, k1/3) are perpendicular to each other.

This implies that we should have the product of the tangents as − 1.

Thus, the given two curves cut at right angles if the product of the slopes of their respective

tangents at is −1.

i.e. (1/2k1/3) * (-1/k1/3) = -1

=> 2k2/3 = 1

=> (2k2/3)3 = 13

=> 8k2 = 1

Hence, the given two curves cut at right angel if 8k2 = 1.

Question 24:

Find the equations of the tangent and normal to the hyperbola at the x2/a2 – y2/b2 = 1 at the point (x0, y0).

Answer:

Given, equation of hyperbola is x2/a2 – y2/b2 = 1

Differentiate w.r.t. x, we get

     2x/a2 – 2y/b2 * dy/dx = 0

=> 2y/b2 * dy/dx = 2x/a2

=> dy/dx = b2 x/a2 y

Therefore, the slope of the tangent at (x0, y0) is

=> dy/dx = b2 x0/a2 y0

Then, the equation of the tangent at (x0, y0) is

      y – y0 = (b2 x0/a2 y0)(x – x0)

=> a2 yy0 – a2 y02 = b2 xx0 – b2x02

=> b2 xx0 - a2 yy0 - b2x02 + a2 y02 = 0

=> xx0/a2 - yy0/b2 – (x02/a2 - y02 /b2) = 0    [Divide by a2b2 on both side]

=> xx0/a2 - yy0/b2 – 1 = 0                              [Since (x0, y0) lies on the hyperbola x2/a2 – y2/b2 = 1]

=> xx0/a2 - yy0/b2 = 1

Now, the slope of the normal at (x0, y0) is given by,

= -1/ Slope of the tangent at (x0, y0)

= -1/(b2 x0/a2 y0)

= -a2 y0/ b2 x0

Hence, the equation of the normal at (x0, y0) is given by,

      y – y0 = (-a2 y0/ b2 x0)(x – x0)

=> (y – y0)/ a2 y0 = -(x – x0)/b2 x0

=> (y – y0)/ a2 y0 + (x – x0)/b2 x0 = 0

Question 25:

Find the equation of the tangent to the curve y = √(3x - 2) which is parallel to the line  4x − 2y + 5 = 0.

Answer:

The equation of the given curve is y = √(3x - 2)

The slope of the tangent to the given curve at any point (x, y) is given by,

dy/dx = 3/2√(3x - 2)

The equation of the given line is 4x − 2y + 5 = 0  

=> y = 2x + 5/2          (which is of the form y = mx + c)

So, the slope of the line = 2

Now, the tangent to the given curve is parallel to the line 4x − 2y − 5 = 0 if the slope of the

tangent is equal to the slope of the line.

=> 3/2√(3x - 2) = 2

=> √(3x - 2) = 3/4

=> 3x – 2 = 9/16

=> 3x = 9/16 + 2

=> 3x = 41/16

=> x = 41/48

When x = 41/48

      y = √(3 * 41/48 – 2)

=> y = √(41/16 – 2)

=> y = √{(41 – 32)/16}

=> y = √(9/16)

=> y = 3/4

So, equation of tangent at the point (41/48, 3/4) is given by

      y – 3/4 = 2(x – 41/48)

=> (4y – 3)/4 = 2{(48x – 41)/48}

=> (4y – 3)/4 = (48x – 41)/24

=> 4y – 3 = (48x – 41)/6

=> 6(4y – 3) = 48x – 41

=> 24y – 18 = 48x – 41

=> 48x – 24y = 23

Hence, the equation of the required tangent is 48x – 24y = 23

Question 26:

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

Answer:

The equation of the given curve is y = 2x2 + 3 sin x

Slope of the tangent to the given curve at x = 0 is given by,

[dy/dx]x = 0 = [4x + 3 cos x]x = 0 = 0 + 3 * cos 0 = 3

Hence, the slope of the normal to the given curve at x = 0 is

= -1/ Slope of the tangent at x = 0

= -1/3

Hence, the correct answer is option D.

Question 27:

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)

Answer:

The equation of the given curve is y2 = 4x

Differentiating with respect to x, we have:

     2y * dy/dx = 4

=> dy/dx = 2/y

Therefore, the slope of the tangent to the given curve at any point (x, y) is given by,

dy/dx = 2/y

The given line is y = x + 1 (which is of the form y = mx + c)

So, the slope of the line = 1

The line y = x + 1 is a tangent to the given curve if the slope of the line is equal to the slope of

the tangent. Also, the line must intersect the curve.

Thus, we must have:

     2/y = 1

=> y = 2

Now, y = x + 1

=> x = y – 1

=> x = 2 – 1

=> x = 1

So, the line y = x + 1 is a tangent to the given curve at the point (1, 2).

Hence, the correct answer is option A.

                                                                      Exercise 6.4

Question 1:

Using differentials, find the approximate value of each of the following up to 3 places of decimal

(i) √(25.3)                                    (ii) √(49.5)                        (iii) √(0.6)                    (iv) (0.009)1/3         

(v) (0.009)1/10                              (vi) (15)1/4                        (vii) (26)1/3                   (viii) (255)1/4

(ix) (82)1/4                                    (x) (410)1/2                       (xi) (0.0037)1/2             (xii) (26.57)1/3

(xiii) (81.5)1/4                               (xiv) (3.968)3/2                (xv) (32.15)1/5

Answer:

(i) Given √(25.3)

Consider y = √x

Let x = 25 and ∆x = 0.3

Then, ∆y = √(x + ∆x) - √x

                 = √(25.3) - √25

                 = √(25.3) – 5

=> √(25.3) = ∆y + 5  

Now, dy is approximately equal to ∆y and is given by,

dy = (dy/dx) * ∆x

     = 1/2√x * (0.3)                            [Since y = √x]

     = 1/2√25 * (0.3)    

     = 0.3/10

     = 0.03                          

Hence, the approximate value of √(25.3) is 0.03 + 5 = 5.03

 

(ii) Given √(49.5)

Consider y = √x

Let x = 49 and ∆x = 0.5

Then, ∆y = √(x + ∆x) - √x

                 = √(49.5) - √49

                 = √(49.5) – 7

=> √(49.5) = ∆y + 7  

Now, dy is approximately equal to ∆y and is given by,

dy = (dy/dx) * ∆x

     = 1/2√x * (0.5)                            [Since y = √x]

     = 1/2√49 * (0.5)    

     = 0.5/14

     = 0.035                           

Hence, the approximate value of √(49.5) is 7 + 0.035 = 7.035

(iii) Given √(0.6)

Consider y = √x

Let x = 1 and ∆x = -0.4

Then, ∆y = √(x + ∆x) - √x

                 = √(0.6) - √1

                 = √(0.6) – 1

=> √(0.6) = ∆y + 1  

Now, dy is approximately equal to ∆y and is given by,

dy = (dy/dx) * ∆x

     = 1/2√x * (-0.4)                            [Since y = √x]

     = 1/2√1 * (-0.4)    

     = -0.2

Hence, the approximate value of √(0.6) is 1 + (−0.2) = 1 − 0.2 = 0.8

(iv) Given (0.009)1/3

Consider y = x1/3

Let x = 0.008 and ∆x = 0.001

Then, ∆y = (x + ∆x)1/3 – x1/3

                 = (0.009)1/3 – (0.008)1/3

                 = (0.009)1/3 – 0.2

=> (0.009)1/3 = ∆y + 0.2  

Now, dy is approximately equal to ∆y and is given by,

dy = (dy/dx) * ∆x

     = 1/3x2/3 * (0.001)                            [Since y = x1/3]

     = 1/(3 * 0.04) * (0.001)    

     = 0.001/0.12

     = 0.008

Hence, the approximate value of (0.009)1/3 is 0.2 + 0.008 = 0.208

(v) Given (0.999)1/10

Consider y = x1/10

Let x = 1 and ∆x = -0.001

Then, ∆y = (x + ∆x)1/10 – x1/10

                 = (0.999)1/10 – 1

=> (0.999)1/10 = ∆y + 1  

Now, dy is approximately equal to ∆y and is given by,

dy = (dy/dx) * ∆x

     = 1/10x9/10 * (-0.001)                            [Since y = x1/10]

     = (-0.001)/10    

     = -0.0001

Hence, the approximate value of (0.999)1/10 is 1 + (−0.0001) = 1 – 0.0001 = 0.9999

(vi) Given (15)1/4

Consider y = x1/4

Let x = 16 and ∆x = -1

Then, ∆y = (x + ∆x)1/4 – x1/4

                 = (15)1/4 – (16)1/4

                 = (15)1/4 – 2

=> (15)1/4 = ∆y + 2  

Now, dy is approximately equal to ∆y and is given by,

dy = (dy/dx) * ∆x

     = 1/4x3/4 * (-1)                            [Since y = x1/4]

     = {1/(4(16)3/4 }* (-1)

     = -1/(4 * 8)

     = -1/32         

     = -0.03125

Hence, the approximate value of (15)1/4 is 2 + (−0.03125) = 2 - 0.03125 = 1.96875

(vii) Given (26)1/3

Consider y = x1/3

Let x = 27 and ∆x = -1

Then, ∆y = (x + ∆x)1/3 – x1/3

                 = (26)1/3 – (27)1/3

                 = (26)1/4 – 3

=> (26)1/3 = ∆y + 3  

Now, dy is approximately equal to ∆y and is given by,

dy = (dy/dx) * ∆x

     = 1/3x2/3 * (-1)                            [Since y = x1/3]

     = {1/(3(27)2/3}* (-1)

     = -1/(3 * 7)

     = -1/27        

     = -0.0370

Hence, the approximate value of (26)1/3 is 3 + (−0.0370) = 3 - 0.0370 = 2.9629

(viii) Given (255)1/4

Consider y = x1/4

Let x = 256 and ∆x = -1

Then, ∆y = (x + ∆x)1/4 – x1/4

                 = (255)1/4 – (256)1/4

                 = (255)1/4 – 4

=> (255)1/4 = ∆y + 4  

Now, dy is approximately equal to ∆y and is given by,

dy = (dy/dx) * ∆x

     = 1/4x3/4 * (-1)                            [Since y = x1/4]

     = {1/(4(256)3/4 }* (-1)

     = -1/(4 * 64)

     = -1/256        

     = -0.0039

Hence, the approximate value of (255)1/4 is 4 + (−0.0039) = 4 - 0.0039 = 3.9961

(ix) Given (82)1/4

Consider y = x1/4

Let x = 81 and ∆x = 1

Then, ∆y = (x + ∆x)1/4 – x1/4

                 = (82)1/4 – (81)1/4

                 = (82)1/4 – 3

=> (82)1/4 = ∆y + 3  

Now, dy is approximately equal to ∆y and is given by,

dy = (dy/dx) * ∆x

     = 1/4x3/4 * (1)                            [Since y = x1/4]

     = {1/(4(81)3/4 }

     = 1/(4 * 27)

     = 1/108        

     = 0.0009

Hence, the approximate value of (82)1/4 is 3 + 0.009 = 3.009

(x) Given (401)1/2

Consider y = x1/2

Let x = 400 and ∆x = 1

Then, ∆y = (x + ∆x)1/2 – x1/2

                 = (401)1/2 – (400)1/2

                 = (401)1/2 – 20

=> (401)1/2 = ∆y + 20  

Now, dy is approximately equal to ∆y and is given by,

dy = (dy/dx) * ∆x

     = 1/2x1/2 * (1)                            [Since y = x1/2]

     = 1/(2 * 20)

     = 1/40        

     = 0.025

Hence, the approximate value of (401)1/2 is 20 + 0.025 = 20.025

(xi) Given (0.0037)1/2

Consider y = x1/2

Let x = 0.0036 and ∆x = 0.0001

Then, ∆y = (x + ∆x)1/2 – x1/2

                 = (0.0037)1/2 – (0.0036)1/2

                 = (0.0037)1/2 – 0.06

=> (0.0037)1/2 = ∆y + 0.06  

Now, dy is approximately equal to ∆y and is given by,

dy = (dy/dx) * ∆x

     = 1/2x1/2 * (0.0001)                            [Since y = x1/2]

     = 0.0001/(2 * 0.06)

    = 0.0001/0.12

     = 0.00083

Thus, the approximate value of (0.0037)1/2 is 0.06 + 0.00083 = 0.06083

(xii) Given (26.57)1/3

Consider y = x1/3

Let x = 27 and ∆x = -0.43

Then, ∆y = (x + ∆x)1/3 – x1/3

                 = (26.57)1/3 – (27)1/3

                 = (26.57)1/3 – 3

=> (26.57)1/3 = ∆y + 3  

Now, dy is approximately equal to ∆y and is given by,

dy = (dy/dx) * ∆x

     = 1/3x2/3 * (-0.43)                            [Since y = x1/3]

     = {1/(3(27)2/3}* (-0.43)

     = -0.43/(3 * 9)

     = -0.43/27        

     = -0.015

Hence, the approximate value of (26.57)1/3 is 3 + (−0.015) = 3 - 0.015 = 2.984

(xiii) Given (81.5)1/4

Consider y = x1/4

Let x = 81 and ∆x = 0.5

Then, ∆y = (x + ∆x)1/4 – x1/4

                 = (81.5)1/4 – (81)1/4

                 = (81.5)1/4 – 3

=> (81.5)1/4 = ∆y + 3  

Now, dy is approximately equal to ∆y and is given by,

dy = (dy/dx) * ∆x

     = 1/4x3/4 * (0.5)                            [Since y = x1/4]

     = 0.5/{(4(81)3/4 }

     = 0.5/(4 * 27)

     = 0.5/108        

     = 0.0046

Hence, the approximate value of (81.5)1/4 is 3 + 0.0046 = 3.0046

(xiv) Given (3.968)3/2

Consider y = x3/2

Let x = 4 and ∆x = -0.032

Then, ∆y = (x + ∆x)3/2 – x3/2

                 = (3.968)3/2 – (4)3/2

                 = (3.968)3/2 – 8

=> (3.968)3/2 = ∆y + 8  

Now, dy is approximately equal to ∆y and is given by,

dy = (dy/dx) * ∆x

     = (3/2)x1/2 * (-0.032)                            [Since y = x3/2]

     = (3/2) * 41/2 * (-0.032)

     = -(3/2) * 2 * (0.032)

     = -3 * 0.032

      = -0.096           

Hence, the approximate value of (3.968)3/2 is 8 + (−0.096) = 8 - 0.096 = 7.904

(xv) Given (32.15)1/5

Consider y = x1/5

Let x = 32 and ∆x = 0.15

Then, ∆y = (x + ∆x)1/5 – x1/5

                 = (32.15)1/5 – (32)1/5

                 = (32.15)1/5 – 2

=> (32.15)1/5 = ∆y + 2  

Now, dy is approximately equal to ∆y and is given by,

dy = (dy/dx) * ∆x

     = 1/5x4/5 * (0.15)                            [Since y = x1/5]

     = 0.15/{(5(32)4/5 }

     = 0.15/(5 * 16)

     = 0.15/80        

     = 0.00187

Hence, the approximate value of (32.15)1/5 is 2 + 0.00187 = 2.00187

Question 2:

Find the approximate value of f (2.01), where f (x) = 4x2 + 5x + 2

Answer:

Let x = 2 and ∆x = 0.01. Then, we have:

f(2.01) = f(x + ∆x) = 4(x + ∆x)2 + 5(x + ∆x) + 2

Now, ∆y = f(x + ∆x) − f(x)

f(x + ∆x) = f(x) + ∆y

                 ≈ f(x) + f’(x) * ∆x                                                       [Since dx = ∆x]

Now, f(2.01) ≈ (4x2 + 5x + 2) + (8x + 5) ∆x

                        = (4 * 22 + 5 * 2 + 2) + (8 * 2 + 5)(0.01)           [as x = 2 and ∆x = 0.01]  

                        = (16 + 10 + 2) + (16 + 5)(0.01)

                        = 28 + 21 * 0.01

                        = 28 + 0.21

                        = 28.21     

Hence, the approximate value of f (2.01) is 28.21

Question 3:

Find the approximate value of f (5.001), where f (x) = x3 - 7x2 + 15

Answer:

Let x = 5 and ∆x = 0.001. Then, we have:

f(5.001) = f(x + ∆x) = (x + ∆x)3 - 7(x + ∆x)2 + 15

Now, ∆y = f(x + ∆x) − f(x)

So, f(x + ∆x) = f(x) + ∆y

                 ≈ f(x) + f’(x) * ∆x                                                  [Since dx = ∆x]

Now, f(5.001) ≈ (x3 - 7x2 + 15) + (3x2 – 14x) ∆x

                           = (53 – 7 * 52 + 15) + (3 * 52 – 14 * 5)(0.001)           [as x = 5 and ∆x = 0.001]  

                           = (125 - 175 + 15) + (75 - 70)(0.001)

                            = -35 + 5 * 0.01

                            = -35 + 0.005

                            = -34.995     

Hence, the approximate value of f (5.001) is -34.995

Question 4:

Find the approximate change in the volume V of a cube of side x metres caused by increasing side by 1%.

Answer

The volume of a cube (V) of side x is given by V = x3

So, dV = (dV/dx) * Δx

            = (3x2) * Δx

            = (3x2) * 0.01x    

            = 0.03x3               [Since 1% of x = 0.01x]

Hence, the approximate change in the volume of the cube is 0.03x3 m3

Question 5:

Find the approximate change in the surface area of a cube of side x metres caused by decreasing the side by 1%.

Answer:

The surface area of a cube (V) of side x is given by S = 6x2

So, dS = (dS/dx) * Δx

            = (12x) * Δx

            = (12x) * 0.01x    

            = 0.12x2               [Since 1% of x = 0.01x]

Hence, the approximate change in the surface area of the cube is 0.12x2 m3

Question 6:

If the radius of a sphere is measured as 7 m with an error of 0.02m, then find the approximate error in calculating its volume.

Answer:

Let r be the radius of the sphere and ∆r be the error in measuring the radius.

Then, r = 7 m and ∆r = 0.02 m

Now, the volume V of the sphere is given by,

V = 4πr3/3

dV/dr = 4πr2

So, dV = (dV/dr) * ∆r

             = (4πr2) * ∆r

             = (4π * 72) * 0.02

              = 3.92π

Hence, the approximate error in calculating the volume is 3.92 π m3

Question 7:

If the radius of a sphere is measured as 9 m with an error of 0.03 m, then find the approximate error in calculating in surface area.

Answer:

Let r be the radius of the sphere and ∆r be the error in measuring the radius.

Then, r = 9 m and ∆r = 0.03 m

Now, the surface area of the sphere (S) is given by,

S = 4πr2

So, dS/dr = 8πr

Now, dS = (dS/dr) * ∆r

                = (8πr) * ∆r

                = (8π * 9) * 0.03

                = 2.16 π      

Hence, the approximate error in calculating the surface area is 2.16π m2

Question 8:

If f(x) = 3x2 + 15x + 5, then the approximate value of f (3.02) is                                                        

A. 47.66                              B. 57.66                             C. 67.66                          D. 77.66

Answer:

Let x = 3 and ∆x = 0.02. Then, we have:

f(5.001) = f(x + ∆x) = 3(x + ∆x)2 + 15(x + ∆x) + 5

Now, ∆y = f(x + ∆x) − f(x)

So, f(x + ∆x) = f(x) + ∆y

                 ≈ f(x) + f’(x) * ∆x                                                                     [Since dx = ∆x]

Now, f(3.02) ≈ (3x2 + 15x + 5) + (6x + 15) ∆x

                        = (3 * 32 + 15 * 3 + 5) + (6 * 3 + 15)(0.002)           [as x = 3 and ∆x = 0.02]  

                        = (27 + 45 + 5) + (18 + 15)(0.02)

                        = 77 + 33 * 0.02

                        = 77 + 0.66

                        = 77.66     

Hence, the approximate value of f (3.02) is 77.66

So, the correct answer is option D.

Question 9:

The approximate change in the volume of a cube of side x metres caused by increasing the side by 3% is                                                                                                                                                

A. 0.06 x3 m3                             B. 0.6 x3 m3                          C. 0.09 x3 m3                     D. 0.9 x3 m3

Answer:

The volume of a cube (V) of side x is given by V = x3

So, dV = (dV/dx) * ∆x

            = (3x2) * ∆x

            = (3x2) * 0.03x                            [Since 3% of x is 0.03x] 

            = 0.09x2

Hence, the approximate change in the volume of the cube is 0.09x3 m3

So, the correct answer is option C.

 

                                                                        Exercise 6.5

Question 1:

Find the maximum and minimum values, if any, of the following functions given by

(i) f(x) = (2x − 1)2 + 3                                         (ii) f(x) = 9x2 + 12x + 2

(iii) f(x) = −(x − 1)2 + 10                                     (iv) g(x) = x3 + 1

Answer:

(i) The given function is f(x) = (2x − 1)2 + 3

It can be observed that (2x − 1)2 ≥ 0 for every x ∈ R.

Therefore, f(x) = (2x − 1)2 + 3 ≥ 3 for every x ∈ R.

The minimum value of f is attained when 2x − 1 = 0

=> 2x − 1 = 0

=> x = 1/2

So, minimum value of f(1/2) = (2 * 1/2 − 1)2 + 3 = 3

Hence, function f does not have a maximum value.

(ii) The given function is f(x) = 9x2 + 12x + 2 = (3x + 2)2 − 2

It can be observed that (3x + 2)2 ≥ 0 for every x ∈ R.

Therefore, f(x) = (3x + 2)2 − 2 ≥ −2 for every x ∈ R.

The minimum value of f is attained when 3x + 2 = 0

=> 3x + 2 = 0

=> x = -2/3

So, minimum value of f = {3 * (-2/3) + 2}2 – 2 = -2

Hence, function f does not have a maximum value.

(iii) The given function is f(x) = − (x − 1)2 + 10

It can be observed that (x − 1)2 ≥ 0 for every x ∈ R.

Therefore, f(x) = − (x − 1)2 + 10 ≤ 10 for every x ∈ R.

The maximum value of f is attained when (x − 1) = 0

=> x − 1 = 0

=> x = 1

So, maximum value of f = f(1) = − (1 − 1)2 + 10 = 10

Hence, function f does not have a minimum value.

(iv) The given function is g(x) = x3 + 1

Hence, function g neither has a maximum value nor a minimum value.

Question 2:

Find the maximum and minimum values, if any, of the following functions given by                   

(i) f(x) = |x + 2| − 1                (ii) g(x) = −|x + 1| + 3                  (iii) h(x) = sin(2x) + 5

(iv) f(x) = |sin 4x + 3|            (v) h(x) = x + 4, x ∈ (−1, 1)

Answer:

(i) Given, f(x) = |x + 2| - 1

We know that |x + 2| ≥ 0 for every x ∈ R.

Therefore, f(x) = |x + 2| - 1 ≥ -1 for every x ∈ R.

The minimum value of f is attained when |x + 2| = 0

=> |x + 2| = 0

=> x = -2

So, minimum value of f = f(−2) = |-2 + 2| - 1 = -1

Hence, function f does not have a maximum value.

(ii) Given, g(x) = -|x + 1| + 3

We know that -|x + 1| ≤ 0 for every x ∈ R.

Therefore, g(x) = -|x + 1| + 3 ≤ 3 for every x ∈ R.

The maximum value of g is attained when |x + 1| = 0

=> |x + 1| = 0

=> x = -1

So, maximum value of g = g(−1) = -|-1 + 1| + 3 = 3

Hence, function g does not have a minimum value.

(iii) Given, h(x) = sin2x + 5

We know that − 1 ≤ sin 2x ≤ 1

=> − 1 + 5 ≤ sin 2x + 5 ≤ 1 + 5

=> 4 ≤ sin 2x + 5 ≤ 6

Hence, the maximum and minimum values of h are 6 and 4 respectively.

(iv) Given, f(x) = |sin 4x + 3|

We know that −1 ≤ sin 4x ≤ 1

=> 2 ≤ sin 4x + 3 ≤ 4

=> 2 ≤ |sin 4x + 3| ≤ 4

Hence, the maximum and minimum values of f are 4 and 2 respectively.

(v) Given, h(x) = x + 1, x ∈ (−1, 1)

Here, if a point x0 is closest to −1, then we find x0/2 + 1 > x0 + 1 for all x0 ∈ (−1, 1).

Also, if x1 is closest to 1, then x1 + 1 < (x1 + 1)/2 + 1 < 1 for all x ∈ (−1, 1).

Hence, function h(x) has neither maximum nor minimum value in (−1, 1).

 

Question 3:

Find the local maxima and local minima, if any, of the following functions.

Find also the local maximum and the local minimum values, as the case may be:

(i) f(x) = x2  (ii) g(x) = x3 − 3x            (iii) h(x) = sin x + cos x, 0 < x < π/2

(iv) f(x) = sin x − cos x, 0 < x < 2π       (v) f(x) = x3 − 6x2 + 9x + 15         (vi) g(x) = x/2 + 2/x, x > 0

(vii) g(x) = 1/(x2 + 2)                             (viii) f(x) = x√(1 - x), x > 0

Answer:

(i) Given, f(x) = x2

So f’(x) = 2x

Now, f’(x) = 0

=> 2x = 0

=> x = 0

Thus, x = 0 is the only critical point which could possibly be the point of local maxima or local

minima of f.

We have, f”(0) = 2 which is positive.

Therefore, by second derivative test, x = 0 is a point of local minima and local minimum value

of f at x = 0 is f(0) = 0.

(ii) g(x) = x3 − 3x

So, g’(x) = 3x2 – 3

Now, g’(x) = 0

=> 3x2 – 3 = 0

=> 3(x2 – 1) = 0

=> x2 – 1 = 0

=> x2 = 1

=> x = ±1

Now, g”(x) = 6x

g”(1) = 6 * 1 = 6 > 0

g”(-1) = 6 * (-1) = -6 < 0

By second derivative test, x = 1 is a point of local minima and local minimum value of g at x = 1

is g(1) = 13 − 3 = 1 − 3 = −2.

However, x = −1 is a point of local maxima and local maximum value of g at x = −1 is

g(1) = (−1)3 − 3 (− 1) = − 1 + 3 = 2

(iii) Given, h(x) = sin x + cos x, 0 < x < π/2

So, h’(x) = cos x – sin x

Now, h’(x) = 0

=> cos x – sin x = 0

=> cos x = sin x

=> tan x = 1

=> x = π/4 є (0, π/2)

Again, h”(x) = -sin x – cos x

=> h”(x) = -(sin x + cos x)

=> h”( π/4) = -(sin π/4 + cos π/4)

=> h”( π/4) = -(1/√2 + 1/√2)

=> h”( π/4) = -2/√2

=> h”( π/4) = -√2 < 0

Therefore, by second derivative test, x = π/4, is a point of local maxima and the local maximum

value of h at x = π/4 is

      h(π/4) = sin π/4 + cos π/4

=> h(π/4) = 1/√2 + 1/√2

=> h(π/4) = 2/√2

=> h(π/4) = √2

(iv) Given, f(x) = sin x - cos x, 0 < x < 2π

So, f’(x) = cos x + sin x

Now, f’(x) = 0

=> cos x + sin x = 0

=> cos x = -sin x

=> tan x = -1

=> x = 3π/4, 7π/4 є (0, 2π)

Again, f”(x) = -sin x + cos x

f”(3π/4) = -sin 3π/4 + cos 3π/4 = -1/√2 - 1/√2) = -2/√2 = -√2 < 0

f”(7π/4) = -sin 7π/4 + cos 7π/4 = 1/√2 + 1/√2 = 2/√2 = √2 > 0

Therefore, by second derivative test, x = 3π/4 is a point of local maxima and the local

maximum value of f at x = 3π/4 is

f(3π/4) = sin 3π/4 - cos 3π/4 = 1/√2 + 1/√2 = 2/√2 = √2

However, at x = 7π/4 is a point of local minima and the local minimum value of f is

f(7π/4) = sin 7π/4 - cos 7π/4 = -1/√2 - 1/√2 = -2/√2 = -√2

(v) f(x) = x3 − 6x2 + 9x + 15

So, f(x) = 3x2 − 12x + 9

Now, f’(x) = 0

=> 3x2 − 12x + 9 = 0

=> 3(x2 − 4x + 3) = 0

=> 3(x - 1)(x - 3) = 0

=> x = 1, 3

Now, f”(x) = 6x – 12

f”(1) = 6 * 1 – 12 = 6 – 12 = -6 < 0

f”(3) = 6 * 3 – 12 = 18 – 12 = 6 > 0

Therefore, by second derivative test, x = 1 is a point of local maxima and the local maximum

value of f at x = 1 is

f(1) = 1 − 6 + 9 + 15 = 19

However, x = 3 is a point of local minima and the local minimum value of f at x = 3 is

f(3) = 27 − 54 + 27 + 15 = 15

(vi) Given, g(x) = x/2 + 2/x, x > 0

g’(x) = 1/2 - 2/x2

Now, g’(x) = 0

=> 1/2 - 2/x2 = 0

=> x2 = 4

=> x = ±2

Since x > 0, we take x = 2

Now, g”(x) = 4/x3

Now, g”(2) = 4/23 = 4/8 = 1/2 > 0

Therefore, by second derivative test, x = 2 is a point of local minima and the local minimum

value of g at x = 2 is

g(2) = 2/2 + 2/2 = 1 + 1 = 2

(vii) Given, g(x) = 1/(x2 + 2)

So, g’(x) = -2x/(x2 + 2)2

Now, g’(x) = 0

=> -2x/(x2 + 2)2 = 0

=> 2x = 0

=> x = 0

Now, for values close to x = 0 and to the left of 0, g’(x) > 0.

Also, for values close to x = 0 and to the right of 0, g’(x) < 0

Therefore, by first derivative test, x = 0 is a point of local maxima and the local maximum value

of g(0) is

g(0) = 1/(0 + 2) = 1/2

 (viii) Given, f(x) = x√(1 - x), x > 0

So, f’(x) = √(1 - x) + 1/2√(1 - x) * (-1)

=> f’(x) = √(1 - x) - x/2√(1 - x)

=> f’(x) = {2(1 - x) – x}/2√(1 - x)

=> f’(x) = (2 - 2x – x)/2√(1 - x)

=> f’(x) = (2 - 3x)/2√(1 - x)

Now, f’(x) = 0

=> (2 - 3x)/2√(1 - x) = 0

=> 2 – 3x = 0

=> x = 2/3

Again, f”(x) = (1/2){√(1 - x) * (-3) – (2 – 3x) * -1/2√(1 - x)}/(1 - x)

=> f”(x) = {√(1 - x) * (-3) + (2 – 3x) * 1/2√(1 - x)}/2(1 - x)

=> f”(x) = {-6(1 - x) + (2 – 3x)}/4(1 - x)3/2

=> f”(x) = (3x – 4)/4(1 - x)3/2

Now, f”(2/3) = (3 * 2/3 – 4)/4(1 – 2/3)3/2

=> f”(2/3) = (2 – 4)/4(1/3)3/2

=> f”(2/3) = -1/2(1/3)3/2 < 0

Therefore, by second derivative test, x = 2/3 is a point of local maxima and the local maximum

value of f at x = 2/3 is

f(2/3) = (2/3)√(1 – 2/3) = (2/3)√(1/3) = 2√3/9

Question 4:

Prove that the following functions do not have maxima or minima:                                                  

(i) f(x) = ex                      (ii) g(x) = log x                            (iii) h(x) = x3 + x2 + x + 1

Answer:

(i). We have, f(x) = ex

Now, f’(x) = ex

Now, if f’(x) = 0, then ex = 0

But, the exponential function can never assume 0 for any value of x.

Therefore, there does not exist c ∈ R such that f’(c) = 0

Hence, function f does not have maxima or minima.

(ii). We have, g(x) = log x

So, g’(x) = 1/x

Since log x is defined for a positive number

g’(x) > 0 for any x

Therefore, there does not exist c ∈ R such that g’(c) = 0

Hence, function g does not have maxima or minima.

(iii). We have, h(x) = x3 + x2 + x + 1

So, h’(x) = 3x2 + 2x + 1

Now, h(x) = 0  

So, 3x2 + 2x + 1 = 0

=> x = (-2 ± 2√2i)/6 = (-1 ± √2i)/3 ∈ R

Therefore, there does not exist c ∈ R such that h’(c) = 0

Hence, function h does not have maxima or minima.

Question 5:

Find the absolute maximum value and the absolute minimum value of the following functions in the given intervals:

(i) f(x) = x2, x є [-2, 2]                           (ii). f(x) = sin x + cos x, x є [0, π]

(iii) f(x) = 4x – x2/2, x є [-2, 9/2]         (iv) f(x) = (x - 1)2 + 3, x є [-3, 2]

Answer:

(i) The given function is f(x) = x3

So, f’(x) = 3x2

Now, f’(x) = 0

=> 3x2 = 0

=> x = 0

Then, we evaluate the value of f at critical point x = 0 and at end points of the interval [−2, 2].

f(0) = 0

f(−2) = (−2)3 = −8

f(2) = (2)3 = 8

Hence, we can conclude that the absolute maximum value of f on [−2, 2] is 8 occurring at x = 2.

Also, the absolute minimum value of f on [−2, 2] is −8 occurring at x = −2.

(ii)The given function is f(x) = sin x + cos x

So, f’(x) = cos x – sin x

Now, f’(x) = 0

=> cos x – sin x = 0

=> sin x = cos x

=> tan x = 1

=> x = π/4

Then, we evaluate the value of f at critical point x = π/4 and at the end points of the interval

[0, π].

f(π/4) = sin π/4 + cos π/4 = 1/√2 + 1/√2 = 2/√2 = √2

f(0) = sin 0 + cos 0 = 0 + 1 = 1

f(π) = sin π + cos π = 0 – 1 = -1

Hence, we can conclude that the absolute maximum value of f on [0, π] is √2 occurring at x =

π/4 and the absolute minimum value of f on [0, π] is −1 occurring at x = π.

 (iii) The given function is f(x) = 4x – x2/2

So, f’(x) = 4 – 2x/2 = 4 – x

Now, f’(x) = 0

=> 4 – x = 0

=> x = 4

Then, we evaluate the value of f at critical point x = 4 and at the end points of the interval

[−2, 9/2].

Now,

f(4) = 16 – 16/2 = 16 – 8 = 8

f(-2) = -8 – 4/2 = -8 – 2 = -10

f(9/2) = 4(9/2) – (1/2)(9/2)2 = 18 – 81/2 = 18 – 10.125 = 7.875

Hence, we can conclude that the absolute maximum value of f on [−2, 9/2] is 8 occurring at x =

4 and the absolute minimum value of f on [−2, 9/2] is -10 occurring at x = -2.

(iv) The given function is f(x) = (x - 1)2 + 3

So, f’(x) = 2(x - 1)

Now, f’(x) = 0

=> 2(x – 1) = 0

=> x = 1

Then, we evaluate the value of f at critical point x = 1 and at the end points of the interval

[−3, 1].

Now,

f(1) = (1 - 1)2 + 3 = 0 + 3 = 3

f(-3) = (-3 - 1)2 + 3 = 16 + 3 = 19

Hence, we can conclude that the absolute maximum value of f on [−3, 1] is 19 occurring at x =

-3 and the absolute minimum value of f on [−3, 1] is 3 occurring at x = 1.

Question 6:

Find the maximum profit that a company can make, if the profit function is given by p(x) = 41 − 24x − 18x2

Answer:

The profit function is given as p(x) = 41 − 24x − 18x2

So, p’(x) = -24 − 36x

and p”(x) = -36

Now, p’(x) = 0

=> -24 − 36x = 0

=> x = -36/24

=> x = -2/3

Also, p”(-2/3) = -36 < 0

By second derivative test, x = -2/3 is the point of local maxima of p.

So, maximum profit = p(-2/3)

                                     = 41 – 24 * (-2/3) – 18 * (-2/3)2

                                     = 41 + 16 – 8

                                     = 49

Hence, the maximum profit that the company can make is 49 units.

Question 7:

Find both the maximum value and the minimum value of 3x4 − 8x3 + 12x2 − 48x + 25 on the interval [0, 3].

Answer:

Given, f(x) = 3x4 − 8x3 + 12x2 − 48x + 25

f′(x) = 12x3 − 24x2 + 24x − 48

       =12(x3 −2x2 +2x −4)

       =12[x2(x − 2) + 2(x − 2)]

       =12(x2 + 2)(x − 2)

For maxima and minima, f′(x) = 0

=> 12(x2 + 2)(x − 2)=0

=> x = 2, x2 = -2

Since x2 = -2 is not possible

So, x = 2 ∈ [0, 3]

Now we evaluate the value of f at critical point x = 2 and at the end points of the interval [0, 3]

f(0) = 25

f(2) = 3 * 24 – 8 * 23 + 12 * 22 – 48 * 2 + 25

        = 48 – 64 + 48 – 96 + 25

        = −39

f(3) = 3 * 34 – 8 * 33 + 12 * 32 – 48 * 3 + 25

       = 243 – 216 + 108 −144 + 25

       = 16

Hence, at x = 0, Maximum value = 25

At x = 2, Minimum value = -39

Question 8:

At what points in the interval [0, 2π], does the function sin 2x attain its maximum value?

Answer:

Let f(x) = sin 2x

So, f’(x) = 2 cos 2x

Now, f’(x) = 0

=> 2 cos 2x = 0

=> cos 2x = 0

=> 2x = π/2, π/2, 3π/2, 5π/2, 7π/2

=> x = π/4, π/4, 3π/4, 5π/4, 7π/4

Then, we evaluate the values of f at critical points and at the x = π/4, π/4, 3π/4, 5π/4, 7π/4 and

at the end points of the interval [0, 2π].

Now,

f(π/4) = sin π/2 = 1

f(3π/4) = sin 3π/2 = -1

f(5π/4) = sin 5π/2 = 1

f(7π/4) = sin 7π/2 = -1

f(0) = sin 0 = 0

f(2π) = sin π = 0

Hence, we can conclude that the absolute maximum value of f on [0, 2π] is occurring at x = π/4

and x = 5π/4

Question 9:

What is the maximum value of the function sin x + cos x?

Answer:

Let f(x) = sin x + cos x

So, f’(x) = cos x - sin x

Now, f’(x) = 0

=> cos x – sin x = 0

=> sin x = cos x

=> tan x = 1

=> x = π/4, 5π/4, ……..

Now, f”(x) = -sin x – cos x = -(sin x + cos x)

Now, f”(x) will be negative when (sin x + cos x) is positive i.e., when sin x and cos x are both

positive. Also, we know that sin x and cos x both are positive in the first quadrant.

Then f”(x) will be negative when x ∈ [0, π/2]

Thus, we consider x = π/4

f”( π/4) = -(sin π/4 + cos π/4) = -(1/√2 + 1/√2) = -2/√2 = -√2 < 0

By second derivative test, f will be the maximum at x = π/4 and the maximum value of f is

f(π/4) = sin π/4 + cos π/4 = (1/√2 + 1/√2) = 2/√2 = √2

Question 10:

Find the maximum value of 2x3 − 24x + 107 in the interval [1, 3]. Find the maximum value of the same function in [−3, −1].

Answer:

Let f(x) = 2x3 − 24x + 107

So, f’(x) = 6x2 – 24 = 6(x2 – 4)

Now, f’(x) = 0

=> 6(x2 – 4) = 0

=> x2 – 4 = 0

=> x2 = 4

=> x = ±2

We first consider the interval [1, 3].

Then, we evaluate the value of f at the critical point x = 2 ∈ [1, 3] and at the end points of the

interval [1, 3].

f(2) = 2 * 23 – 24 * 2 + 107 = 16 − 48 + 107 = 75

f(1) = 2 * 13 – 24 * 1 + 107 = 2 − 24 + 107 = 85

f(3) = 2 * 33 – 24 * 3 + 107 = 54 − 72 + 107 = 89

Hence, the absolute maximum value of f(x) in the interval [1, 3] is 89 occurring at x = 3.

Next, we consider the interval [−3, −1].

Evaluate the value of f at the critical point x = −2 ∈ [−3, −1] and at the end points of the

interval [-3, -1].

f(−3) = 2 (−3)3 − 24(−3) + 107 = −54 + 72 + 107 = 125

f(−1) = 2(−1)3 − 24 (−1) + 107 = −2 + 24 + 107 = 129

f(−2) = 2(−2)3 − 24 (−2) + 107 = −16 + 48 + 107 = 139

Hence, the absolute maximum value of f(x) in the interval [−3, −1] is 139 occurring at x = −2.

Question 11:

It is given that at x = 1, the function x4− 62x2 + ax + 9 attains its maximum value, on the interval [0, 2]. Find the value of a.

Answer:

Let f(x) = x4− 62x2 + ax + 9

So, f’(x) = 4x3 − 124x + a

It is given that function f attains its maximum value on the interval [0, 2] at x = 1.

So, f’(1) = 0

=> 4 * 13 − 124 * 1 + a = 0

=> 4 − 124 + a = 0

=> a = 120

Hence, the value of a is 120.

Question 12:

Find the maximum and minimum values of x + sin 2x on [0, 2π].

Answer:

Let f(x) = x + sin 2x

So, f’(x) = 1 + 2 cos 2x

Now, f’(x) = 0

=> 1 + 2 cos 2x = 0

=> cos 2x = -1/2

=> cos 2x = -cos π/3

=> cos 2x = cos (π - π/3)

=> cos 2x = cos 2π/3

=> 2x = 2nπ ± 2π/3, n є Z

=> x = nπ ± π/3, n є Z

=> x = π/3, 2π/3, 4π/3, 5π/3 є [0, 2π]

Now,

f(π/3) = π/3 + sin 2π/3 = π/3 + √3/2

f(2π/3) = 2π/3 + sin 4π/3 = 2π/3 - √3/2

f(4π/3) = 4π/3 + sin 8π/3 = 4π/3 + √3/2

f(5π/3) = 5π/3 + sin 10π/3 = 5π/3 - √3/2

f(0) = 0 + sin 0 = 0

f(2π) = 2π + sin 2π = 2π + 0 = 2π

Hence, we can conclude that the absolute maximum value of f(x) in the interval [0, 2π] is 2π

occurring at x = 2π and the absolute minimum value of f(x) in the interval [0, 2π] is 0 occurring

at x = 0.

Question 13:

Find two numbers whose sum is 24 and whose product is as large as possible.

Answer:

Let one number be x. Then, the other number is (24 − x).

Let P(x) denote the product of the two numbers. Thus, we have

      P(x) = x(24 - x)

=> P(x) = 24x – x2

So, P’(x) = 24 – 2x

and P”(x) = -2

Now, P’(x) = 0

=> 24 – 2x = 0

=> x = 12

and P”(12) = -2 < 0

By second derivative test, x = 12 is the point of local maxima of P.

Hence, the product of the numbers is the maximum when the numbers are 12

and 24 − 12 = 12

Question 14:

Find two positive numbers x and y such that x + y = 60 and xy3 is maximum.

Answer:

The two numbers are x and y such that x + y = 60

=> y = 60 − x

Let f(x) = xy3

=> f(x) = x(60 - x)3

So, f’(x) = (60 - x)3 – 3x(60 - x)2

=> f’(x) = (60 - x)2(60 - x – 3x)

=> f’(x) = (60 - x)2(60 - 4x)

Also, f”(x) = -2(60 - x)(60 - 4x) - 4(60 - x)2

=> f”(x) = -2(60 - x)[60 - 4x - 2(60 – x)]

Now, f’(x) = 0

=> (60 - x)2(60 - 4x) = 0

=> x = 60, 4x = 60

=> x = 60, 15

When x = 60, f”(x) = 0

When x = 15, f”(x) = -12(60 - 15)(30 - 15) = -12 * 45 * 15 < 0

By second derivative test, x = 15 is a point of local maxima of f.

Thus, function xy3 is maximum when x = 15 and y = 60 − 15 = 45

Hence, the required numbers are 15 and 45.

Question 15:

Find two positive numbers x and y such that their sum is 35 and the product x2y5 is a maximum.

Answer:

Let one number be x. Then, the other number is y = (35 − x).

Let f(x) = x2y5

=> f(x) =  x2(35 - x)5

So, f’(x) = 2x(35 - x)5 - 5x2(35 - x)4

=> f’(x) = x(35 - x)4[2(35 - x) – 5x]

=> f’(x) = x(35 - x)4[70 - 2x – 5x]

=> f’(x) = x(35 - x)4(70 - 7x)

Now, f’(x) = 0

=> x(35 - x)4(70 - 7x) = 0

=> x = 0, 35, 10

Again f”(x) = 7(35 - x)4(10 - x) + 7x[-(35 - x)4 – 4(35 - x)3(10 - x)]

=> f”(x) = 7(35 - x)4(10 - x) - 7x * (35 - x)4 – 28x(35 - x)3(10 - x)

=> f”(x) = 7(35 - x)3[(35 - x)(10 - x) - x(35 - x) – 4x (10 - x)]

=> f”(x) = 7(35 - x)3[350 – 45x + x2 – 35x + x2 – 40x + 4x2]

=> f”(x) = 7(35 - x)3(6x2 – 120x + 350)

When x = 35, f’(x) = f(x) = 0 and y = 35 – 35 = 0 and the product x2y5 will be equal to 0.

When x = 0, y = 35 − 0 = 35 and the product x2y5 will be 0.

So, x = 0 and x = 35 cannot be the possible values of x.

When x = 10, we have

f”(x) = 7(35 - 10)3(6  * 102 – 120 * 10 + 350)

         = 7(25)3(-250)

         < 0  

So, By second derivative test, P(x) will be the maximum when x = 10 and y = 35 − 10 = 25

Hence, the required numbers are 10 and 25.

Question 16:

Find two positive numbers whose sum is 16 and the sum of whose cubes is minimum.

Answer:

Let one number be x. Then, the other number is (16 − x).

Let the sum of the cubes of these numbers be denoted by S(x). Then,

S(x) = x3 + (16 - x)3

So, S’(x) = 3x2 - 3(16 - x)2

and S”(x) = 6x + 6(16 – x)

Now, S’(x) = 0

=> 3x2 - 3(16 - x)2 = 0

=> x2 - (16 - x)2 = 0

=> x2 - (256 + x2 - 32x) = 0

=> x2 - 256 - x2 + 32x = 0

=> 32x = 256

=> x = 8

Now, S”(8) = 6 * 8 + 6(16 – 8) = 48 + 48 = 96 > 0

By second derivative test, x = 8 is the point of local minima of S.

Hence, the sum of the cubes of the numbers is the minimum when the numbers are 8 and

16 − 8 = 8.

Question 17:

A square piece of tin of side 18 cm is to made into a box without top, by

cutting a square from each corner and folding up the flaps to form the box.

What should be the side of the square to be cut off so that the volume of the box is the maximum possible?

Answer:

Let the side of the square to be cut off be x cm. Then, the length and the breadth of the box

will be (18 − 2x) cm each and the height of the box is x cm.

Therefore, the volume V(x) of the box is given by,

V(x) = x(18 − 2x)2

So, V’(x) = (18 − 2x)2 – 4x(18 – 2x)

=> V’(x) = (18 − 2x)[18 – 2x – 4x]

=> V’(x) = (18 − 2x)(18 – 6x)

=> V’(x) = 2 * 6 * (9 − x)(3 – x)

=> V’(x) = 12(9 − x)(3 – x)

And V”(x) = 12[-(9 - x) – (3 - x)]

=> V”(x) = -12[9 - x + 3 - x]

=> V”(x) = -12(12 – 2x)

=> V”(x) = -24(6 – x)

Now, V’(x) = 0

=> 12(9 − x)(3 – x) = 0

=> x = 3, 9

If x = 9, then the length and the breadth will become 0.

So, x ≠ 9

Hence, x = 3

Now, V”(x) = -24(6 – 3) = -24 * 3 = -72 < 0

By second derivative test, x = 3 is the point of maxima of V.

Hence, if we remove a square of side 3 cm from each corner of the square tin and make a box

from the remaining sheet, then the volume of the box obtained is the largest possible.

Question 18:

A rectangular sheet of tin 45 cm by 24 cm is to be made into a box without top, by cutting off square from each corner and folding up the flaps.

What should be the side of the square to be cut off so that the volume of the box is the maximum possible?

Answer:

Let the side of the square to be cut off be x cm. Then, the height of the box is x, the

length is 45 − 2x, and the breadth is 24 − 2x.

Therefore, the volume V(x) of the box is given by,

V(x) = x(45 - x)(24 – 2x)

         = x(1080 – 90x – 48x + 4x2

         = x(1080 – 138x + 4x2)

         = 4x3 – 138x2 + 1080x

Now, V’(x) = 12x2 – 276x + 1080

                    = 12(x2 – 23x + 90)

                    = 12(x - 18)(x - 5)

And V”(x) = 24x – 276 = 12(2x - 23)

Now, V’(x) = 0

=> 12(x - 18)(x - 5) = 0

=> x = 5, 18

It is not possible to cut off a square of side 18 cm from each corner of the rectangular sheet.

Thus, x cannot be equal to 18.

So, x = 5

Now, V”(5) = 12(10 - 23) = 12 * (-13) = -156 < 0

So, by second derivative test, x = 5 is the point of maxima.

Hence, the side of the square to be cut off to make the volume of the box maximum possible is

5 cm.

Question 19:

Show that of all the rectangles inscribed in a given fixed circle, the square has the maximum area.

Answer:

Class_12_Maths_Applications_Of_Derivatives_Figure11

Let a rectangle of length l and breadth b be inscribed in the given circle of radius a.

Then, the diagonal passes through the centre and is of length 2a cm.

Now, by applying the Pythagoras theorem, we have:

      (2a)2 = l2 + b2

=> b2 = 4a2 – l2

=> b = √(4a2 – l2)

Area of rectangle A = l√(4a2 – l2)

So, dA/dl = √(4a2 – l2) + l * (-2l) * 1/2√(4a2 – l2)

=> dA/dl = √(4a2 – l2) - l2/√(4a2 – l2)

=> dA/dl = {(4a2 – l2) - l2}/√(4a2 – l2)

=> dA/dl = (4a2 – 2l2)/√(4a2 – l2)

Again, d2A/dl2 = {√(4a2 – l2)(-4l) - (4a2 – 2l2) * (-2l)/2√(4a2 – l2)}/(4a2 – l2)

=> d2A/dl2 = {(4a2 – l2)(-4l) + l(4a2 – 2l2)}/(4a2 – l2)3/2

=> d2A/dl2 = (-12a2 l + 2l3)/(4a2 – l2)3/2

=> d2A/dl2 = -2l(6a2 - l2)/(4a2 – l2)3/2

Now, dA/dt = 0

=> (4a2 – 2l2)/√(4a2 – l2) = 0

=> 4a2 – 2l2 = 0

=> 2a2 – l2 = 0

=> l = a√2

And b = √(4a2 – 2a2) = √(2a2) = a√2

Now, when l = a√2

=> d2A/dl2 = -2(a√2)(6a2 – 2a2)/2√2a3

=> d2A/dl2 = -8√2 a3/2√2a3

=> d2A/dl2 = -4 < 0

By the second derivative test, when l = a√2

then the area of the rectangle is the maximum.

Since l = b = a√2, the rectangle is a square.

Hence, it has been proved that of all the rectangles inscribed in the given fixed circle, the

square has the maximum area.

Question 20:

Show that the right circular cylinder of given surface and maximum volume is such that is

heights is equal to the diameter of the base.

Answer:

Let r and h be the radius and height of the cylinder respectively.

Then, the surface area (S) of the cylinder is given by,

      S = 2πr2 + 2πrh

=> h = (S - 2πr2)/2πr

=> h = (S/2π) * (1/r) - r

Let V be the volume of the cylinder. Then,

      V = πr2h

=> V = πr2[(S/2π) * (1/r) - r]

=> V = Sr/2 – πr3

Now, dV/dr = S/2 - 2πr2

And d2V/dr2 = -6πr

Now, dV/dr = 0

=> S/2 – 3πr2 = 0

=> r2 = S/6π

When r2 = S/6π

Then d2V/dr2 = -6π * √(S/6π) < 0

By second derivative test, the volume is the maximum when r2 = S/6π

Now, when r2 = S/6π, then

h = (6πr2/2π) * (1/r) – r = 3r – r = 2r

Hence, the volume is the maximum when the height is twice the radius i.e., when the height is

equal to the diameter.

Question 21:

Of all the closed cylindrical cans (right circular), of a given volume of 100 cubic centimeters,

find the dimensions of the can which has the minimum surface area?

Answer:

Let r and h be the radius and height of the cylinder respectively.

Then, volume (V) of the cylinder is given by,

      V = πr2h = 100

=> h = 100/ πr2

Surface area (S) of the cylinder is given by,

       S = 2πr2 + 2πrh

=> S = 2πr2 + 200/r

So, dS/dr = 4πr - 200/r2

And d2S/dr2 = 4π + 400/r3

Now, dS/dr = 0

=> 4πr - 200/r2 = 0

=> 4πr = 200/r2

=> r3 = 200/4π

=> r3 = 50/π

=> r = (50/π)1/3

Now, it is observed that when r = (50/π)1/3, d2S/dr2 > 0

So, by second derivative test,

the surface area is the minimum when the radius of the (50/π)1/3 cm

When r = (50/π)1/3,

      h = 100/{π * (50/π)2/3}

=> h = (2 * 50)/{π * (50/π)2/3}

=> h = 2(50/π)1/3 cm

Hence, the required dimensions of the can which has the minimum surface area is given

by radius = (50/π)1/3 cm and height = 2(50/π)1/3 cm

Question 22:

A wire of length 28 m is to be cut into two pieces. One of the pieces is to be made into a square and the other into a circle.

What should be the length of the two pieces so that the combined area of the square and the circle is minimum?

Answer:

Let a piece of length l be cut from the given wire to make a square.

Then, the other piece of wire to be made into a circle is of length (28 − l) m.

Now, side of square = l/4

Let r be the radius of the circle. Then,

      2πr = 28 – l

=> r = (28 - l)/2π

The combined areas of the square and the circle (A) is given by,

      A = (side of the square)2 + πr2

=> A = l2/16 + π[(28 - l)/2π]2

=> A = l2/16 + (28 - l)2/4π

So, dA/dl = 2l/16 + (2/4π) * (28 - l) * (-1)

=> dA/dl = l/8 - (1/2π) * (28 - l)

And d2A/dl2 = l/8 + 1/2π > 0

Now, dA/dl = 0

=> l/8 - (1/2π) * (28 - l) = 0

=> {πl - 4(28 - l)}/π = 0

=> πl - 4(28 - l) = 0

=> πl - 112 + 4l = 0

=> (π + 4)l - 112 = 0

=> l = 112/(π + 4)

When l = 112/(π + 4), d2A/dl2 > 0

By second derivative test, the area (A) is the minimum when l = 112/(π + 4).

Hence, the combined area is the minimum when the length of the wire in making the square is

112/(π + 4) cm while the length of the wire in making the circle is

28 - l = 28 - 112/(π + 4) = 28 π/(π + 4) cm.

Question 23:

Prove that the volume of the largest cone that can be inscribed in a sphere of radius R is 8/27 of the volume of the sphere.

Answer:

Class_12_Maths_Applications_Of_Derivatives_Figure10

Let r and h be the radius and height of the cone respectively inscribed in a sphere of radius R.

Let V be the volume of the cone.

Then, V = πr2h/3

Height of the cone is given by,

      h = R + AB

=> h = R + √(R2 – r2)                              [Since ABC is a right angle triangle]    

So, V = [πr2{R + √(R2 – r2)}]/3

=> V = πr2R/3 + πr2√(R2 – r2)/3

So, dV/dr = 2πrR/3 + 2πr√(R2 – r2)/3 + (πr2/3) * (-2r)/2√(R2 – r2)

=> dV/dr = 2πrR/3 + 2πr√(R2 – r2)/3 - (πr3/3) * √(R2 – r2)

=> dV/dr = 2πrR/3 + {2πr(R2 – r2) - πr3}/3√(R2 – r2)

=> dV/dr = 2πrR/3 + (2πrR2 – 3πr3)/3√(R2 – r2)

and d2V/dr2 = 2πR/3 + {3√(R2 – r2)*(2πR2 – 9πr2) - (2πrR2 – 3πr3) * (-2r)√6(R2 – r2)}/9(R2 – r2)

=> d2V/dr2 = 2πR/3 + {9(R2 – r2)*(2πR2 – 9πr2) + 2πr2R2 + 3πr4}/27(R2 – r2)3/2

Now, dV/dr = 0

=> 2πrR/3 + (2πrR2 – 3πr3)/3√(R2 – r2) = 0

=> 2rR/3 = -(2rR2 – 3r3)/3√(R2 – r2) = 0

=> 2R = (3r2 - 2R2)/√(R2 – r2)

=> 2R√(R2 – r2) = (3r2 - 2R2)

=> 4R2(R2 – r2) = (3r2 - 2R2)2

=> 4R4 – 4R2r2 = 9r4 + 4R4 - 12r2R2

=> 9r4 = 8R2r2

=> 9r2 = 8R2

=> r2 = 8R2/9

When r2 = 8R2/9 then d2V/dr2 < 0

So, by second derivative test, the volume of the cone is the maximum when r2 = 8R2/9

When r2 = 8R2/9,

h = R + √(R2 – 8R2/9) = R + √(R2/9) = R + R/3 = 4R/3

Now, V1 = {π * 8R2/9 * (4R/3)}/3

                = (8/27) * (4πR3/3)

                = (8/27) * V

                = (8/27) * Volume of the sphere

Hence, the volume of the largest cone that can be inscribed in the sphere is 8/27 the volume

of the sphere.

Question 24:

Show that the right circular cone of least curved surface and given volume has an altitude equal to √2 time the radius of the base.

Answer:

Let r and h be the radius and the height (altitude) of the cone respectively.

Then, the volume (V) of the cone is given as:

      V = πr2h/3   ………….1

=> h = 3V/πr2    ………….2

The surface area (S) of the cone is given by,

     S = πrl        (where l is the slant height)

Squaring on both sides, we get

=> S2 = π2r2l2

=> S2 = π2r2(h2 + r2)                            [Since l2 = h2 + r2]

=> S2 = π2r2{(3V/πr2)2 + r2}

=> S2 = π2r2{9V22r4 + r2}                [From equation 2]    

=> S2 = π2r4 + 9V2/r2

Differentiating w.r.t r, we get

=> 2S * dS/dr = 4π2r3 - 18V2/r3

For maxima and minima, we have

      dS/dr = 0

=> 4π2r3 - 18V2/r3 = 0

=> 4π2r6 - 18V2 = 0

=> 2π2r6 - 9V2 = 0

=> 2π2r6 = 9V2

=> 2π2r6 = 9 * π2r4h2/9                     [From equation 1]

=> 2π2r6 = π2r4h2

=> 2r2 = h2

=> h = √2 * r

Clearly, d2S/dr2 = 12π2r2 + 54V2/r4 > 0

Hence, for a given volume, the right circular cone of the least curved surface has an altitude

equal to √2 times the radius of the base.

Question 25:

Show that the semi-vertical angle of the cone of the maximum volume and of given slant height is tan-1 √2.

Answer:

Class_12_Maths_Applications_Of_Derivatives_Figure9

Let θ be the semi-vertical angle of the cone.

It is clear that θ є [0, π/2]

Let r, h and l be the radius, height and the slant height of the cone respectively.

The slant height of the cone is given as constant.

Now, r = l sin θ and h = l cos θ

The volume (V) of the cone is given by,

      V = πr2h/3  

=> V = (π * l2 * sin2 θ * l * cos θ)/3

=> V = (π * l3 * sin2 θ * cos θ)/3

=> dV/dθ = πl3{sin2 θ * (-sin θ) + cos θ(2 sin θ cos θ)}/3

=> dV/dθ = πl3(-sin3 θ + 2 sin θ cos2 θ)/3

And d2V/dθ2 = πl3(-3sin2 θ cos θ + 2 cos2 θ – 4 sin2 θ cos θ)/3

=> d2V/dθ2 = πl3(2 cos3 θ – 7 sin2 θ cos θ)/3

Now, dV/dθ = 0

=> πl3(-sin3 θ + 2 sin θ cos2 θ)/3 = 0

=> -sin3 θ + 2 sin θ cos2 θ = 0

=> sin3 θ = 2 sin θ cos2 θ

=> sin2 θ = 2 cos2 θ

=> tan2 θ = 2

=> tan θ = √2

=> θ = tan-1 √2

When θ = tan-1 √2, then

     tan θ = √2

=> tan2 θ = 2

=> sin2 θ = 2 cos2 θ

Now, d2V/dθ2 = πl3(2 cos2 θ – 14 cos3 θ)/3 = - 4πl3 cos3 θ < 0 for θ є [0, π/2]

By second derivative test, the volume (V) is the maximum when θ = tan-1 √2

Hence, for a given slant height, the semi-vertical angle of the cone of the maximum volume is

θ = tan-1 √2

Question 26:

Show that semi-vertical angle of right circular cone of given surface area and maximum volume is sin-1 (1/3).

Answer:

Class_12_Maths_Applications_Of_Derivatives_Figure8

Let r be the radius, l be the slant height and h be the height of the cone.

Surface are of the cone (S) is defined as

     S = πr2 + πrl

=> l = (S - πr2)/πr

=> l = S/πr – r   ……………..1

Let V be the volume of the cone.

So, V = πr2h/3

=> V = πr2√(l2 - r2)/3                           [Since l2 = h2 + r2]

=> V2 = π2r4(l2 - r2)/9

=> V2 = π2r4{(S/πr – r)2 - r2)/9           [From equation 1]  

=> V2 = π2r4[S22r2 - 2S/π]/9

=> V2 = S2r2/9 - 2πSr4/9      …………2

Differentiate equation 2 w.r.t. r, we get

=> 2V * dV/dr = 2S2r/9 - 8πSr3/9

=> 2V * dV/dr = (2Sr/9)(S - 4πr2)    ………..3

For maxima and minima,

      dV/dr = 0

=> (2Sr/9)(S - 4πr2) = 0

=> r = 0 or S = 4πr2

But r ≠ 0

So, S = 4πr2

=> r2 = S/4π

Again, differentiate equation 3 w.r.t. r, we get

      2(dV/dr)2 + 2V * d2V/dr2 = (S/9)(2S - 24πr2)

=> 2V * d2V/dr2 = (S/9)(2S – 6S)                                 [Since dV/dr = 0 and r2 = S/4π]

=> 2V * d2V/dr2 = (-4S2/9) < 0

=> V is maximum when r2 = S/4π

Also, r2 = S/4π

=> r2 = (πr2 + πrl)/4π

=> 4πr2 = πr2 + πrl

=> 3πr2 = πrl

=> l = 3r

Now, if α be the semi-vertical angle of the cone, then

     sin α = r/l

=> sin α = r/3r

=> sin α = 1/3

=> α = sin-1(1/3)

Choose the correct answer in the Exercises 27 and 29

Question 27:

The point on the curve x2 = 2y which is nearest to the point (0, 5) is

(A) (2√2, 4)                   (B) (2√2, 4)                           (C) (0, 0)                              (D) (2, 2)

Answer:

The given curve is x2 = 2y

For each value of x, the position of the point will be (x, x2/2).

The distance d(x) between the points (x, x2/2) and (0, 5) is given by,

      d(x) = √{(x - 0)2 + (x2/2 - 5)2}

=> d(x) = √(x2 + x4/4 + 25 – 5x2)

=> d(x) = √(x4/4 – 4x2 + 25)

So, d’(x) = (x3 – 8x)/2√(x4/4 – 4x2 + 25)

=> d’(x) = (x3 – 8x)/√(x4 – 16x2 + 100)

Now, d’(x) = 0

=> (x3 – 8x)/√(x4 – 16x2 + 100) = 0

=> x3 – 8x = 0

=> x(x2 – 8) = 0

=> x = 0, ±2√2

Now,

      d”(x) = {√(x4 – 16x2 + 100)(3x2 – 8) - (x3 – 8x)(4x3 – 32x)/2√(x4 – 16x2 + 100)/(x4 – 16x2 + 100)

=> d”(x) = {(x4 – 16x2 + 100)(3x2 – 8) - 2(x3 – 8x)(x3 – 8x)/(x4 – 16x2 + 100)3/2

=> d”(x) = {(x4 – 16x2 + 100)(3x2 – 8) - 2(x3 – 8x)2/(x4 – 16x2 + 100)3/2

When x = 0, then d”(x) = {36 * (-8)}/63 < 0

When, x = ±2√2, then d”(x) > 0

By second derivative test, d(x) is the minimum at x = ±2√2.

When x = ±2√2, y = (±2√2)2/2 = 4

Hence, the point on the curve x2 = 2y which is nearest to the point (0, 5) is (±2√2, 4).

So, the correct answer is option A.

Question 28:

For all real values of x, the minimum value of (1 – x + x2)/(1 + x + x2) is

(A) 0                               (B) 1                            (C) 3                           (D) 1/3

Answer:

Let f(x) = (1 – x + x2)/(1 + x + x2)

So, f’(x) = {(1 + x + x2)/(-1 + 2x) - (1 – x + x2)(1 + 2x)}/(1 + x + x2)2

=> f’(x) = (-1 + 2x – x + 2x2 – x2 + 2x3 – 1 – 2x + x + 2x2 – x2 – 2x3)/(1 + x + x2)2

=> f’(x) = (2x2 – 2)/(1 + x + x2)2

=> f’(x) = 2(x2 – 1)/(1 + x + x2)2

Now, f’(x) = 0

=> 2(x2 – 1)/(1 + x + x2)2 = 0

=> x2 – 1 = 0

=> x2 = 1

=> x = ±1

Now, f”(x) = 2[(1 + x + x2)2 * (2x) – (x2 – 1) (1 + x + x2)(1 + 2x)]/(1 + x + x2)4

=> f”(x) = 4(1 + x + x2)2 [x(1 + x + x2) – (x2 – 1)(1 + 2x)]/(1 + x + x2)4

=> f”(x) = 4(1 + x + x2)2 [x + x2 + x3 – x2 – 2x3 + 1 + 2x)]/(1 + x + x2)4

=> f”(x) = 4(1 + 3x – x3)/(1 + x + x2)3

Now,

f”(1) = 4(1 + 3 – 13)/(1 + 1 + 12)3 = (4 * 3)/33 = 4/9 > 0

f”(-1) = 4(1 - 3 – 13)/(1 - 1 + 12)3 = 4 * (-1) = -4 < 0

So, by second derivative test, f is the minimum at x = 1 and the minimum value is given by

f(1) = (1 – 1 + 1)/(1 + 1 + 1) = 1/3

Hence, the correct answer is option D.

Question 29:

The maximum value of [x(x – 1) + 1]1/3, 0 ≤ x ≤ 1 is                                                                                     

(A) (1/3)1/3                          (B) 1/2                             (C) 1                              (D) 0

Answer:

Let f(x) = [x(x – 1) + 1]1/3

So, f’(x) = (2x - 1)/3[x(x – 1) + 1]2/3

Now, f’(x) = 0

=> (2x - 1)/3[x(x – 1) + 1]2/3 = 0

=> 2x – 1 = 0

=> x = 1/2

Then, we evaluate the value of f at critical point x = 1/2 and at the end points of the interval

[0, 1] {i.e., at x = 0 and x = 1}.

f(0) = [0(0 – 1) + 1]1/3 = 1

f(1) = [1(1 – 1) + 1]1/3 = 1

f(1/2) = [(1/2)(1/2 – 1) + 1]1/3 = (3/4)1/3

So, we can conclude that the maximum value of f in the interval [0, 1] is 1.

Hence, the correct answer is option C.

 

                                                          Miscellaneous Exercise on Chapter 6

Question 1:

Using differentials, find the approximate value of each of the following.

(a) (17/81)1/4    (b) (33)-1/5

Answer:

(a) Consider y = x1/4

Let x = 16/81 and ∆x = 1/81

Then, ∆y = (x + ∆x)1/4 - x1/4

                 = (17/81)1/4 - (16/81)1/4

                 = (17/81)1/4 – 2/3

=> (17/81)1/4 = 2/3 + ∆y

Now, dy is approximately equal to ∆y and is given by,

dy = (dy/dx) * ∆x

      = 1/{4 * x3/4} * (∆x)

      = 1/{4 * (16/81)3/4} * (1/81)              [As y = x1/4]

      = 27/(4 * 8) * (1/81)

      = 1/(32 * 3)

      = 1/96

      = 0.010               

Hence, the approximate value of (17/81)1/4 is 2/3 + 0.010 = 0.667 + 0.010 = 0.677

(b) Consider y = x-1/5

Let x = 32 and ∆x = 1

Then, ∆y = (x + ∆x)-1/5 - x-1/5

                 = (x + ∆x)-1/5 - x-1/5

                 = (33)-1/5 – (32)-1/5

                 = (33)-1/5 – 1/2

=> (33)-1/5 = 1/2 + ∆y

Now, dy is approximately equal to ∆y and is given by,

dy = (dy/dx) * ∆x

      = -1/{5 * (x)6/5} * ∆x                    [As y = x-1/5]

      = -1/{5 * 26} * 1

      = -1/320

      = -0.003               

Hence, the approximate value of (33)-1/5 is 1/2 + (-0.003) = 0.5 – 0.003 = 0.497

Question 2:

Show that the function given by f(x) = log x /x has maximum at x = e.

Answer:

The given function is: f(x) = log x /x

Differentiate w.r.t. x, we get

f’(x) = {x * 1/x – log x}/x2

        = (1 – log x)/x2 

Now, f’(x) = 0

=> (1 – log x)/x2 = 0

=> 1 − log x = 0

=> log x = 1

=> log x = log e

=> x = e

Now, f”(x) = {x2 * (-1/x) – (1 – log x)(2x)}/x4

                 = {-x – (1 – log x)(2x)}/x4

                 = {-3 + 2 log x}/x3

Now, f”(e) = {-3 + 2 log 3}/e3

                    = (-3 + 2)/e3

                  = -1/e3

                  < 0

Therefore, by second derivative test, f is maximum at x = e.

Question 3:

The two equal sides of an isosceles triangle with fixed base b are decreasing at the rate of 3 cm per second.

How fast is the area decreasing when the two equal sides are equal to the base?

Answer:

Class_12_Maths_Applications_Of_Derivatives_Figure7

Let ∆ABC be isosceles where BC is the base of fixed length b.

Let the length of the two equal sides of ∆ABC be a.

Draw AD Ʇ BC

Now, in ∆ADC, by applying the Pythagoras theorem, we have:

AD = √(a2 – b2/4)

Now, area of triangle = (1/2) * b * √(a2 – b2/4)

The rate of change of the area with respect to time (t) is given by,

dA/dt = (1/2) * b * {2a/√(a2 – b2/4)} * da/dt

            = {ab/√(4a2 – b2)} * da/dt 

It is given that the two equal sides of the triangle are decreasing at the rate of 3 cm per

second.

i.e. da/dt = -3 cm/s

So, dA/dt = -3ab/√(4a2 – b2)

Then, when a = b, we have:

dA/dt = -3b2/√(4b2 – b2) = -3b2/√(3b2) = -√(3b)

Hence, if the two equal sides are equal to the base, then the area of the triangle is decreasing

at the rate of √(3b) cm2/s.

Question 4:

Find the equation of the normal to curve y2 = 4x at the point (1, 2).

Answer:

The equation of the given curve is y2 = 4x

Differentiating with respect to x, we have:

      2y * dy/dx = 4

=> dy/dx = 2/y

=> dy/dx](1, 2) = 2/2 = 1

Now, the slope of the normal at point (1, 2) is -1/[dy/dx](1, 2) = -1/1 = -1

So, equation of the normal at (1, 2) is y − 2 = −1(x − 1)

=> y − 2 = − x + 1

=> x + y − 3 = 0

Question 5:

Show that the normal at any point θ to the curve

x = a cos θ + aθ sin θ; y = a sin θ – aθ cos θ is at a constant distance from the origin.

Answer:

We have x = a cos θ + a θ sin θ

So, dx/dθ = -a sin θ + a sin θ + aθ cos θ = aθ cos θ

y = a sin θ – aθ cos θ

So, dy/dθ = a cos θ - a cos θ + aθ sin θ = aθ sin θ

Now, dy/dx = (dy/dθ) * (dθ/dx)

                      = aθ sin θ/aθ cos θ

                      = tan θ   

Slope of the normal at any point θ = -1/tan θ

The equation of the normal at a given point (x, y) is given by,

      y - a sin θ + aθ cos θ = (-1/tan θ)(x – a cos θ – aθ sin θ)

=> y sin θ - a sin2 θ + aθ sin θ cos θ = -x cos θ + a cos2 θ + aθ sin θ cos θ

=> x cos θ  + y sin θ – a(sin2 θ + cos2) = 0

=> x cos θ  + y sin θ – a = 0

Now, the perpendicular distance of the normal from the origin is

|-a|/( sin2 θ + cos2) = |-a| = |-a|, which is independent of θ.

Hence, the perpendicular distance of the normal from the origin is constant.

Question 6:

Find the intervals in which the function f given by

f(x) = (4 * sin x – 2x – x cos x)/(2 + cos x) is 

(i) increasing   (ii) decreasing

Answer:

Given, f(x) = (4 * sin x – 2x – x cos x)/(2 + cos x)

f’(x) = {(2 + cos x) (4 * cos x – 2 – cos x + x sin x) - (4 * sin x – 2x – x cos x)/(-sin x)}/ (2 + cos x)2

        = {(2 + cos x) (3 * cos x – 2 + x sin x) + sin x(4 * sin x – 2x – x cos x)}/ (2 + cos x)2

        = (6 cos x – 4 + 2x sin x + 3cos2 x – 2 cos x + x sin x cos x + 4 sin2 x – 2x sin x – x sin x cos x)}            

           }/(2 + cos x)2

        = (4 cos x – 4 + 3cos2 x + 4 sin2 x)/(2 + cos x)2

        = (4 cos x – 4 + 3cos2 x + 4 – 4 cos2 x)/(2 + cos x)2

        = (4 cos x – cos2 x)/(2 + cos x)2

        = cos x(4 – cos x)/(2 + cos x)2

Now, f’(x) = 0

=> cos x = 0 or cos x = 4

But cos x ≠ 4

So, cos x = 0

=> x = π/2, 3π/2

Now, at x = π/2 and x = 3π/2 divides (0, 2π) into three disjoint intervals i.e.,

(0, π/2), (π/2, 3π/2) and (3π/2, 2π)

In intervals (0, π/2) and (3π/2, 2π), f’(x) > 0

Thus, f(x) is increasing for 0 < x < π/2 and 3π/2 < x < 2π

In the interval (π/2, 3π/2), f’(x) < 0

Thus, f(x) is decreasing for π/2 < x < 3π/2

Question 7:

Find the intervals in which the function f given by is f(x) = x3 + 1/x3, x ≠ 0 is

(i) increasing  (ii) decreasing

Answer:

Given, f(x) = x3 + 1/x3

So, f’(x) = 3x2 - 3/x4

=> f’(x) = (3x6 – 3)/x4

Then f’(x) = 0

=> (3x6 – 3)/x4 = 0

=> 3x6 – 3 = 0

=> x6 = 1

=> x = ±1

Now, the points x = 1 and x = −1 divide the real line into three disjoint intervals

i.e., (-∞, -1), (-1, 1) and (1, ∞)

In intervals i.e., (-∞, -1), and (1, ∞) when x < −1 and x > 1, f’(x) > 0

Thus, when x < −1 and x > 1, f is increasing.

In interval (−1, 1) i.e., when −1 < x < 1, f’(x) < 0

Thus, when −1 < x < 1, f is decreasing.

Question 8:

Find the maximum area of an isosceles triangle inscribed in the ellipse

x2/a2 + y2/b2 = 1 with its vertex at one end of the major axis.

Answer:

Class_12_Maths_Applications_Of_Derivatives_Figure6

The given ellipse is x2/a2 + y2/b2 = 1

Let the major axis be along the x −axis.

Let ABC be the triangle inscribed in the ellipse where vertex C is at (a, 0).

Since the ellipse is symmetrical with respect to the x−axis and y −axis, we can assume the

coordinates of A to be (−x1, y1) and the coordinates of B to be (−x1, −y1).

Now, we have y1 = ±(b/a)√(a2 – x12)

So, the coordinates of A are (−x1, (b/a)√(a2 – x12)) and

the coordinates of B are (−x1, -(b/a)√(a2 – x12))

As the point (x1, y1) lies on the ellipse, the area of triangle ABC (A) is given by,

A = |a{(2b/a)√(a2 – x12)} + (-x1){(-b/a)√(a2 – x12)} + (-x1){(-b/a)√(a2 – x12)}|

=> A = b√(a2 – x12) + (x1)(b/a)√(a2 – x12)   ………..1

Now, dA/dx1 = -2 x1b/2√(a2 – x12) + (b/a)√(a2 – x12) – 2bx12/2a√(a2 – x12)

                        = b/a√(a2 – x12)[- x1a(a2 – x12) – x12]

                        = b[-2x12 – ax1 + a2]/a√(a2 – x12)

Now, dA/dx1 = 0

=> b[-2x12 – ax1 + a2]/a√(a2 – x12) = 0

=> -2x12 – ax1 + a2 = 0

=> x1 = [a ± √{a2 – 4(-2)(a2)}]/2(-2)

=> x1 = [a ± √(9a2)]/(-4)

=> x1 = (a ± 3a)/(-4)

=> x1 = -a, a/2

But, x1 cannot be equal to a.

So, x1 = -a/2

Now, y1 = (b/a)√(a2 – a2/4) = (ba/2a)√3 = b√3/2

Now, d2A/dx12 = (b/a)[√(a2 – x12) (-4x1 – a) – (-2x12 – x1a + a2) * (-2x1)/2√(a2 – x12)]/(a2 – x12)

                           = (b/a)[(a2 – x12) (-4x1 – a) + x1 (-2x12 – x1a + a2)]/(a2 – x12)3/2

                           = (b/a)[2x13 – 3a2x1 – a3)]/(a2 – x12)3/2

Also, when x1 = a/2

d2A/dx12 = (b/a)[2a3/8 – 3a3/2 – a3)]/(3a2/4)3/2

                = (b/a)[a3/4 – 3a3/2 – a3)]/(3a2/4)3/2         

                = -(b/a)[9a3/4]/(3a2/4)3/2

                < 0    

Thus, the area is the maximum when x1 = a/2

So, Maximum area of the triangle is given by,

A = b√(a2 – a2/4) + (a/2)(b/a)√(a2 – a2/4)

    = ab√3/2 + (a/2)(b/a)(a√3/2)

    = ab√3/2 + ab√3/4

    = 3ab√3/4  

Question 9:

A tank with rectangular base and rectangular sides, open at the top is to be constructed so that its depth is 2 m and volume is 8 m3.

If building of tank costs Rs 70 per sq meters for the base and Rs 45 per square metre for sides.

What is the cost of least expensive tank?

Answer

Let l, b, and h represent the length, breadth, and height of the tank respectively.

Then, we have height (h) = 2 m

Volume of the tank = 8m3

Volume of the tank = l * b * h

So, 8 = l * b * 2

=> lb = 4

=> b = 4/l

Now, area of the base = lb = 4

Area of the 4 walls (A) = 2h(l + b)

=> A = 4(l + 4/l)

So, dA/dl = 4(1 – 4/l2)

Now, dA/dl = 0

=> 4(1 – 4/l2) = 0

=> l2 = 4

=> l = ±2

However, the length cannot be negative.

Therefore, we have l = 4

So, b = 4/l = 4/2 = 2

Now, d2A/dl2 = 32/l3

When l = 2, d2A/dl2 = 32/8 = 4 > 0

Thus, by second derivative test, the area is the minimum when l = 2

We have l = b = h = 2

So, cost of building the base = Rs 70 * (lb) = Rs 70 * 4 = Rs 280

And cost of building the walls = Rs 2h(l + b) * 45 = Rs 90 * 2(2 + 2) = Rs 8 * 90 = Rs 720

Required total cost = Rs (280 + 720) = Rs 1000

Hence, the total cost of the tank will be Rs 1000.

Question 10:

The sum of the perimeter of a circle and square is k, where k is some constant.

Prove that the sum of their areas is least when the side of square is double the radius of the circle.

Answer:

Let r be the radius of the circle and a be the side of the square.

Then, we have:

       2πr + 4a = k (Where k is a constant)

=> a = (k - 2πr)/4

The sum of the areas of the circle and the square (A) is given by,

A = πr2 + a2 = πr2 + (k - 2πr)2/16

Now, dA/dr = 2πr + 2(k - 2πr)(- 2π)/16

                      = 2πr - π(k - 2πr)/4

Now, dA/dr = 0

=> 2πr - π(k - 2πr)/4 = 0

=> 2πr = π(k - 2πr)/4

=> 8r = k - 2πr

=> r = k/(8 + 2π)

=> r = k/2(4 + π)

Now, d2A/dr2 = 2π + π2/2 > 0

So, when r = k/2(4 + π), d2A/dr2 > 0

The sum of the areas is least when r = k/2(4 + π)

When r = k/2(4 + π), a = [k - 2πk/2(4 + π)]/4 = 2r

Hence, it has been proved that the sum of their areas is least when the side of the square is

double the radius of the circle.

Question 11:

A window is in the form of rectangle surmounted by a semicircular opening.

The total perimeter of the window is 10 m.

Find the dimensions of the window to admit maximum light through the whole opening.

Answer:

Class_12_Maths_Applications_Of_Derivatives_Figure5

Let x and y be the length and breadth of the rectangular window.

Radius of the semicircular opening = x/2

It is given that the perimeter of the window is 10 m.

=> x + 2y + πx/2 = 10

=> x(1 + π/2) + 2y = 10

=> 2y = 10 - x(1 + π/2)

=> y = 5 - x(1/2 + π/4)

Area of the window (A) is given by,

A = xy + (π/2)(x/2)2

    = x[5 - x(1/2 + π/4)] + (π/2)(x/2)2 

    = 5x – x2(1/2 + π/4)] + πx2/8

Now, dA/dx = 5 – 2x(1/2 + π/4) + πx/4

                      = 5 – x(1 + π/2) + πx/4

d2A/dx2 = -(1 + π/2) + π/4

               = -1 - π/4

Now, dA/dx = 0

=> 5 – x(1 + π/2) + πx/4 = 0

=> 5 - x - πx/4 = 0

=> x(1 + π/4) = 5

=> x = 5/(1 + π/4)

=> x = 20/(π + 4)

Thus, when x = 20/(π + 4), d2A/dx2 < 0

Therefore, by second derivative test, the area is the maximum when length x = 20/(π + 4).

Now,

y = 5 - x = 5 - 20/(π + 4) * (2 + π)/4

               = 5 - 5(2 + π)/(π + 4)

               = 10/(π + 4)      

Hence, the required dimensions of the window to admit maximum light is given by length =

20/(π + 4) m and breadth = 10/(π + 4) m

Question 12:

A point on the hypotenuse of a triangle is at distance a and b from the sides of the triangle.

Show that the minimum length of the hypotenuse is (a2/3 + b2/3)3/2

Answer:

Class_12_Maths_Applications_Of_Derivatives_Figure4

Let ∆ABC be right-angled at B. Let AB = x and BC = y

Let P be a point on the hypotenuse of the triangle such that P is at a distance of a and b from

the sides AB and BC respectively.

Let C = θ

From the figure, we have,

AC = √(x2 + y2)

Now,

PC = b cosec θ

And, AP = a sec θ

So, AC = AP + PC

=> AC = b cosec θ + a sec θ   ...........1

Now, d(AC)/dθ = -b cosec θ cot θ + a sec θ tan θ

      d(AC)/dθ = 0

=> -b cosec θ cot θ + a sec θ tan θ = 0

=> a sec θ tan θ = b cosec θ cot θ

=> (a/cos θ) * (sin θ/cos θ) = (b/sin θ) * (cos θ/sin θ)

=> a sin3 θ = b cos3 θ

=> a1/3 sin θ = b1/3 cos θ

=> tan θ = (b/a)1/3

So, sin θ = b1/3/√(a2/3 + b2/3) and cos θ = a1/3/√(a2/3 + b2/3)   ……….2

It can be clearly shown that d2(AC)/dθ2 < 0 when tan θ = (b/a)1/3

Therefore, by second derivative test, the length of the hypotenuse is the maximum when

tan θ = (b/a)1/3

Now, when tan θ = (b/a)1/3, we have:

AC = b/√(a2/3 + b2/3)/ b1/3 + a/√(a2/3 + b2/3)/a1/3

       = √(a2/3 + b2/3) (b2/3 + a2/3)

       = (a2/3 + b2/3)3/2

Hence, the maximum length of the hypotenuses is (a2/3 + b2/3)3/2

Question 13:

Find the points at which the function f given by f(x) = (x - 2)4(x + 1)3 has

(i) local maxima                          (ii) local minima                           (iii) point of inflexion

Answer:

The given function is f(x) = (x - 2)4(x + 1)3

Now, f’(x) = 4(x - 2)3(x + 1)3 + 3(x - 2)4(x + 1)2

                   = (x - 2)3(x + 1)2[4(x + 1) + 3(x – 2)]

                   = (x - 2)3(x + 1)2(7x - 2)3      

Now, f’(x) = 0

=> (x - 2)3(x + 1)2(7x - 2)3 = 0

=> x = -1, 2/7, 2

Now, for values of x close to 2/7 and to the left of 2/7, f’(x) > 0

Also, for values of x close to 2/7 and to the right of 2/7, f’(x) < 0

Thus, x = 2/7 is the point of local maxima.

Now, for values of x close to 2 and to the left of 2, f’(x) < 0

Also, for values of x close to 2 and to the right of 2, f’(x) > 0

Thus, x = 2 is the point of local minima.

Now, as the value of x varies through −1, f’(x) does not changes its sign.

Thus, x = −1 is the point of inflexion.

Question 14:

Find the absolute maximum and minimum values of the function f given by

f(x) = cos2 x + sin x, x є [0, π]

Answer:

Given, f(x) = cos2 x + sin x

Now, f’(x) = -2 cos x sin x + cos x

Now, f’(x) = 0

=> -2 cos x sin x + cos x = 0

=> cos x(1 - 2 sin x) = 0

=> sin x = 1/2 , cos x = 0

=> x = π/6, π/2 as x є [0, π]

Now, evaluating the value of f at critical points x = π/2 and x = π/6 at the end points of the

interval [0, π] (i.e., at x = 0 and x = π), we have:

f(π/6) = cos2 π/6 + sin π/6 = (√3/2)2 + 1/2 = 3/4 + 1/2 = 5/4

f(0) = cos2 0 + sin 0 = 1

f(π) = cos2 π + sin π = (-1)2 + 0 = 1

f(π/2) = cos2 π/2 + sin π/2 = 0 + 1 = 1

Hence, the absolute maximum value of f is 5/4 occurring at x = π/6 and the absolute minimum

value of f is 1 occurring at x = 0, π/2 and π.

Question 15:

Show that the altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius r is 4r/3.

Answer:

Class_12_Maths_Applications_Of_Derivatives_Figure3

A sphere of fixed radius (r) is given.

Let R and h be the radius and the height of the cone respectively.

The volume (V) of the cone is given by,

V = πR2h/3

Now, from the right triangle BCD, we have:

     BC = √(r2 - R2)

=> h = r + √(r2 - R2)

So, V = πR2 * {r + √(r2 - R2)}/3

=> V = πR2r/3 + πR2√(r2 - R2)/3

Now, dV/dR = 2πRr/3 + 2πR√(r2 - R2)/3 + R2/3 * (-2R)/2√(r2 - R2)

=> dV/dR = 2πRr/3 + 2πR√(r2 - R2)/3 – R3/3√(r2 - R2)

=> dV/dR = 2πRr/3 + {2πR(r2 - R2)/3 – πR3}/3√(r2 - R2)

=> dV/dR = 2πRr/3 + {2πRr2 - 3πR3}/3√(r2 - R2)

Now, dV/dR = 0

=> 2πRr/3 + {2πRr2 - 3πR3}/3√(r2 - R2) = 0

=> 2πRr/3 = {3πR3 - 2πrR3}/3√(r2 - R2)

=> 2πRr√(r2 - R2) = 3πR3 - 2πrR3

=> 2r√(r2 - R2) = 3R2 - 2r2

=> 4r2(r2 - R2) = (3R2 - 2r2)2

=> 4r4 - 4r2R2 = 9R4 + 4r4 - 12 R2 r2

=> 9R4 - 8r2R2 = 0

=> 9R2 = 8r2

=> R2 = 8r2/9

Now, d2V/dR2 = 2πr/3 + [3√(r2 - R2)(2πr2 - 9πR2} – (2πRr2 - 3πR3)(-6R) * 1/2√(r2 - R2)]/9(r2 - R2)

                          = 2πr/3 + [3√(r2 - R2)(2πr2 - 9πR2} + (2πRr2 - 3πR3)(3R) * 1/2√(r2 - R2)]/9(r2 - R2)

Now, R2 = 8r2/9, it can be shown that d2V/dR2 < 0

The volume is the maximum when R2 = 8r2/9

When R2 = 8r2/9,

Height of the cone = r + √(r2 - 8r2/9)

                                   = r + √(r2/9)   

                                   = r + r/3

                                   = 4r/3     

Hence, it can be seen that the altitude of the right circular cone of maximum volume that can

be inscribed in a sphere of radius r is 4r/3.

Question 16:

Let f be a function defined on [a, b] such that f ′(x) > 0, for all x ∈ (a, b).

Then prove that f is an increasing function on (a, b)

Answer:

Let x1, x2 be two numbers in the interval (a, b)

i.e. x1, x2 є (a, b)

and x1 < x2

Let us consider the interval [x1, x2]

f is continuous and differentiable in (x1, x2) as f is continuous and differentiable in (a, b).

By Mean value of theorem,

There exists c in (x1, x2) i.e. c є (x1, x2) such that

f’(c) = {f(x2) – f(x1)}/(x2 – x1)

Given that f’(x) > 0 for all x є (a, b)

So, f’(c) > 0 for all c є (x1, x2)

=> {f(x2) – f(x1)}/(x2 – x1) > 0

=> f(x2) – f(x1) > 0

=> f(x2) > f(x1)

So, for two points x1, x2 in the interval [a, b] where x2 > x1

=> f(x2) > f(x1)

=> f(x1) < f(x2)

So, f is increasing in the interval (a, b).

Question 17:

Show that the height of the cylinder of maximum volume that can be inscribed in a sphere of radius R is 2R/√3.

Also find the maximum volume.

Answer:

Class_12_Maths_Applications_Of_Derivatives_Figure2

A sphere of fixed radius (R) is given.

Let r and h be the radius and the height of the cylinder respectively.

From the given figure, we have

h = 2√(R2 - r2)

The volume (V) of the cylinder is given by,

V = πr2h = 2πr2√(R2 - r2)

Now, dV/dr = 4πr√(R2 - r2) + {2πr2 * (-2r)}/2√(R2 - r2)

                      = 4πr√(R2 - r2) - 2πr3/√(R2 - r2)

                      = {4πr(R2 - r2) - 2πr3}/√(R2 - r2)

                      = {4πrR2 - 6πr3}/√(R2 - r2)       

Now, dV/dr = 0

=> {4πrR2 - 6πr3}/√(R2 - r2) = 0

=> 4πrR2 - 6πr3 = 0

=> r2 = 2R2/3

Now, d2V/dr2 = [√(R2 - r2){4πR2 - 18πr2} - {4πrR2 - 6πr3} * (-2r)}/2√(R2 - r2)]/(R2 - r2)       

                         = [(R2 - r2){4πR2 - 18πr2}  -r{4πrR2 - 6πr3}]/(R2 - r2)3/2    

                         = [4πR4 - 22πr2 R2 + 12πr4 + 4πr2 R2]/(R2 - r2)3/2

Now, it can be observed that at r2 = 2R2/3, d2V/dr2 < 0

So, the volume is the maximum when r2 = 2R2/3

When r2 = 2R2/3,

Height of the cylinder = 2√(R2 – 2R2/3)

                                         = 2√(R2/3)   

                                         = 2R/√3

Hence, the volume of the cylinder is the maximum when the height of the cylinder is 2R/√3.

Question 18:

Show that height of the cylinder of greatest volume which can be inscribed in a right circular cone of height h

and semi vertical angle α is one-third that of the cone and the greatest volume of cylinder is (4/27)πh3 tan3 α.

Answer:

Class_12_Maths_Applications_Of_Derivatives_Figure1

The given right circular cone of fixed height (h) and semi-vertical angle (α) can be drawn as:

Here, a cylinder of radius R and height H is inscribed in the cone.

Then, angle GAO = α, OG = r, OA = h, OE = R, and CE = H.

We have,

r = h tan α

Now, since ∆AOG is similar to ∆CEG, we have:

      AO/OG = CE/EG

=> h/r = H/(r - R)               [Since EG = OG - OE]

=> H = h(r - R)/r

=> H = (h/h tan α) * (h tan α - R)

=> H = (h tan α - R)/tan α

Now, the volume (V) of the cylinder is given by,

     V = πR2 H

=> V = πR2 * (h tan α - R)/tan α

=> V = πR2 h – πR3/tan α

Now, dV/dR = 2πRh – 3πR2/tan α

Now, dV/dR = 0

=> 2πRh – 3πR2/tan α = 0

=> 2πRh = 3πR2/tan α

=> 2h tan α = 3R

=> R = (2h/3)tan α

Now, d2V/dR2 = 2πh – 6πR/tan α

For R = 2h tan α/3, we have

      d2V/dR2 = 2πh – (6π/tan α) * (2h/3 * tan α)

=> d2V/dR2 = 2πh – 4πh = -2πh < 0

By second derivative test, the volume of the cylinder is the greatest when R = 2h/3 * tan α

When R = 2h/3 * tan α,

=> H = (h tan α - 2h/3 * tan α)/tan α

=> H = (h/3 * tan α)/tan α

=> H = h/3

Thus, the height of the cylinder is one-third the height of the cone when the volume of the

cylinder is the greatest.

Now, the maximum volume of the cylinder can be obtained as:

π(2h/3 * tan α)2 (h/3) = π(4h2/9 * tan2 α) (h/3) = (4/27)πh3 tan3 α

Hence, the given result is proved.

Question 19:

A cylindrical tank of radius 10 m is being filled with wheat at the rate of 314 cubic mere per hour.

Then the depth of the wheat is increasing at the rate of

(A) 1 m/h                        (B) 0.1 m/h                   (C) 1.1 m/h                         (D) 0.5 m/h

Answer:

Let r be the radius of the cylinder.

Then, volume (V) of the cylinder is given by,

V = π*(radius)2 * height

    = π(10)2 * h

    = 100 πh    

Differentiating with respect to time t, we have:

dV/dt = 100 π * dh/dt

The tank is being filled with wheat at the rate of 314 cubic metres per hour.

=> dV/dt = 314 m3/h

Thus, we have:

      314 = 100 π * dh/dt

=> dh/dt = 314/(100 * 3.14)

=> dh/dt = 314/314

=> dh/dt = 1

Hence, the depth of wheat is increasing at the rate of 1 m/h.

The correct answer is option A.

Question 20:

The slope of the tangent to the curve x = t2 + 3t – 8, y = 2t2 – 2t – 5 at the point (2, −1) is

(A) 22/7                         (B) 6/7                         (C) 7/6                          (D) -6/7

Answer:

The given curve is:

x = t2 + 3t – 8 and y = 2t2 – 2t – 5

Now, dx/dt = 2t + 3 and dy/dt = 4t – 2

So, dy/dx = (dy/dt) * (dt/dx) = (4t - 2)/(2t + 3)

The given point is (2, −1).

At x = 2, we have:

      t2 + 3t – 8 = 2

=> t2 + 3t – 10 = 0

=> (t - 2)(t + 5) = 0

=> t = 2, -5

At y = -1, we have

      2t2 – 2t – 5 = -1

=> 2t2 – 2t – 4 = 0

=> (t - 2)(t + 1) = 0

=> t = 2, -1

The common value of t is 2.

Hence, the slope of the tangent to the given curve at point (2, −1) is

dy/dx]t = 2 = (4 * 2 - 2)/(2 * 2 + 3) = (8 - 2)/(4 + 3) = 6/7

Hence, the correct answer is option B.

Question 21:

The line y = mx + 1 is a tangent to the curve y2= 4x if the value of m is

(A) 1                                (B) 2                              (C) 3                                (D) 1/2

Answer:

The equation of the tangent to the given curve is y = mx + 1

Now, substituting y = mx + 1 in y2= 4x, we get:

=> (mx + 1)2= 4x

=> m2x2 + 1 + 2mx – 4x = 0

=> m2x2 + x(2m – 4) + 1 = 0   ………..1

Since a tangent touches the curve at one point, the roots of equation 1 must be equal.

Therefore, we have:

      Discriminant = 0

=> (2m - 4)2 – 4 m2 * 1 = 0

=> 4m2 + 16 – 16m - 4m2 = 0

=> 16 – 16m = 0

=> m = 1

So, the required value of m is 1.

Hence, the correct answer is option A.

Question 22:

The normal at the point (1, 1) on the curve 2y + x2= 3 is

(A) x + y = 0                  (B) x − y = 0              (C) x + y + 1 = 0                    (D) x − y = 1

Answer:

The equation of the given curve is 2y + x2= 3

Differentiating with respect to x, we have:

      2dy/dx + 2x = 0

=> dy/dx = -x

=> dy/dx](1, 1) = -1

The slope of the normal to the given curve at point (1, 1) = -1/dy/dx](1, 1) = -1/(-1) = 1

Hence, the equation of the normal to the given curve at (1, 1) is given as:

=> y – 1 = 1(x - 1)

=> y - 1 = x – 1

=> x – y = 0

Hence, the correct answer is option B.

Question 23:

The normal to the curve x2= 4y passing (1, 2) is

(A) x + y = 3                       (B) x − y = 3                      (C) x + y = 1                    (D) x − y = 1

Answer:

The equation of the given curve is x2 = 4y

Differentiating with respect to x, we have:

      2x = 4dy/dx

=> dy/dx = x/2

The slope of the normal to the given curve at point (h, k) is given by, -1/{dy/dx](h, k)} = -2/h

Equation of the normal at point (h, k) is given as:

=> y – k = (-2/h)(x - h)

Now, it is given that the normal passes through the point (1, 2).

Therefore, we have:

=> 2 – k = (-2/h)(x - h)

=> k = 2 + 2(1 - h)/h    …………1

Since (h, k) lies on the curve x2 = 4y, we have h2 = 4k

=> k = h2/4

From equation 1, we have:

      h2 /4 = 2 + 2(1 - h)/h

=> h3/4 = 2h + 2 - 2h

=> h3/4 = 2

=> h3 = 8

=> h = 2

So, k = h2/4 = 22/4 = 1

Hence, the equation of the normal is given as:

=> y – 1 = (-2/2)(x - 2)

=> y – 1 = -(x - 2)

=> y – 1 = -x + 2

=> x + y = 3

Hence, the correct answer is option A.

Question 24:

The points on the curve 9y2 = x3, where the normal to the curve makes equal intercepts with the axes are

(A) (4, ±8/3)                       (B) (4, -8/3)                           (C) (4, ±3/8)                         (D) (±4, 3/8)

Answer:

The equation of the given curve is 9y2 = x3

Differentiating with respect to x, we have:

      9 * 2y * dy/dx = 3x2

=> dy/dx = x2/6y

The slope of the normal to the given curve at point (x1, y1) is

-1/(dy/dx)](x1, y1) = -6y1/x12

The equation of the normal to the curve at (x1, y1) is

      y – y1 = (-6y1/x12)(x – x1)

=> x12y – x12y1 = -6xy1 + 6x1y1

=> 6xy1 + x12y = 6x1y1 + x12 y1

=> 6xy1/(6x1y1 + x12 y1) + x12y/(6x1y1 + x12 y1) = 1

=> x/{x1(6 + x1)/6} + y/{y1(6 + x1)/x1} = 1

It is given that the normal makes equal intercepts with the axes.

Therefore, we have:

x1(6 + x1)/6 = y1(6 + x1)/x1

=> x1/6 = y1/x1

=> x12 = 6y1 …………1

Also, the point (x1, y1) lies on the curve, so we have

9y12 = x13   ……….2

From equation 1 and 2, we have:

      9(x12/6)2 = x13

=> x14/4 = x13

=> x1 = 4

From equation 2, we have:

     9y12 = 43

=> y12 = 64/9

=> y1 = ±8/3

So, the required points are (4, ±8/3).

Hence, the correct answer is option A.

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