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Updated 11 June 2026

NCERT Solutions

Application of Derivatives

Himachal Pradesh Board · Class 12 · Mathematics

NCERT Solutions for Application of Derivatives — Himachal Pradesh Board Class 12 Mathematics.

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82 Questions Solved · 4 Sections

Exercise 6.1

1Find the rate of change of the area of a circle with respect to its radius

r

when (a)

r = 3\,\text{cm}

(b)

r = 4\,\text{cm}

Show solution

Given: Area of a circle

A = \pi r^2

.

Formula used: Rate of change of area with respect to radius

= \dfrac{dA}{dr}

\frac{dA}{dr} = 2\pi r

(a) When $r = 3$ cm:

\frac{dA}{dr}\bigg|_{r=3} = 2\pi(3) = 6\pi \text{ cm}^2/\text{cm}

(b) When $r = 4$ cm:

\frac{dA}{dr}\bigg|_{r=4} = 2\pi(4) = 8\pi \text{ cm}^2/\text{cm}

2The volume of a cube is increasing at the rate of

8\,\text{cm}^3/\text{s}

. How fast is the surface area increasing when the length of an edge is

12\,\text{cm}

?Show solution

Given:

\dfrac{dV}{dt} = 8\,\text{cm}^3/\text{s}

, edge

x = 12\,\text{cm}

.

Volume of cube:

V = x^3

\frac{dV}{dt} = 3x^2\frac{dx}{dt} \implies 8 = 3(12)^2\frac{dx}{dt} \implies \frac{dx}{dt} = \frac{8}{432} = \frac{1}{54}\,\text{cm/s}

Surface area of cube:

S = 6x^2

\frac{dS}{dt} = 12x\frac{dx}{dt} = 12(12)\cdot\frac{1}{54} = \frac{144}{54} = \frac{8}{3}\,\text{cm}^2/\text{s}

Hence, the surface area is increasing at the rate of $\dfrac{8}{3}\,\text{cm}^2/\text{s}$ .

3The radius of a circle is increasing uniformly at the rate of

3\,\text{cm/s}

. Find the rate at which the area of the circle is increasing when the radius is

10\,\text{cm}

.Show solution

Given:

\dfrac{dr}{dt} = 3\,\text{cm/s}

r = 10\,\text{cm}

.

Area:

A = \pi r^2

\frac{dA}{dt} = 2\pi r\frac{dr}{dt} = 2\pi(10)(3) = 60\pi\,\text{cm}^2/\text{s}

Hence, the area is increasing at the rate of $60\pi\,\text{cm}^2/\text{s}$ .

4An edge of a variable cube is increasing at the rate of

3\,\text{cm/s}

. How fast is the volume of the cube increasing when the edge is

10\,\text{cm}

long?Show solution

Given:

\dfrac{dx}{dt} = 3\,\text{cm/s}

x = 10\,\text{cm}

.

Volume:

V = x^3

\frac{dV}{dt} = 3x^2\frac{dx}{dt} = 3(10)^2(3) = 900\,\text{cm}^3/\text{s}

Hence, the volume is increasing at the rate of $900\,\text{cm}^3/\text{s}$ .

5A stone is dropped into a quiet lake and waves move in circles at the speed of

5\,\text{cm/s}

. At the instant when the radius of the circular wave is

8\,\text{cm}

, how fast is the enclosed area increasing?Show solution

Given:

\dfrac{dr}{dt} = 5\,\text{cm/s}

r = 8\,\text{cm}

.

Area:

A = \pi r^2

\frac{dA}{dt} = 2\pi r\frac{dr}{dt} = 2\pi(8)(5) = 80\pi\,\text{cm}^2/\text{s}

Hence, the enclosed area is increasing at the rate of $80\pi\,\text{cm}^2/\text{s}$ .

6The radius of a circle is increasing at the rate of

0.7\,\text{cm/s}

. What is the rate of increase of its circumference?Show solution

Given:

\dfrac{dr}{dt} = 0.7\,\text{cm/s}

.

Circumference:

C = 2\pi r

\frac{dC}{dt} = 2\pi\frac{dr}{dt} = 2\pi(0.7) = 1.4\pi\,\text{cm/s}

Hence, the circumference is increasing at the rate of $1.4\pi\,\text{cm/s}$ .

7The length

x

of a rectangle is decreasing at the rate of

5\,\text{cm/minute}

and the width

y

is increasing at the rate of

4\,\text{cm/minute}

. When

x = 8\,\text{cm}

and

y = 6\,\text{cm}

, find the rates of change of (a) the perimeter, and (b) the area of the rectangle.Show solution

Given:

\dfrac{dx}{dt} = -5\,\text{cm/min}

(decreasing),

\dfrac{dy}{dt} = 4\,\text{cm/min}

(increasing),

x = 8\,\text{cm}

y = 6\,\text{cm}

.

(a) Perimeter:

P = 2(x + y)

\frac{dP}{dt} = 2\left(\frac{dx}{dt} + \frac{dy}{dt}\right) = 2(-5 + 4) = 2(-1) = -2\,\text{cm/min}

The perimeter is decreasing at the rate of $2\,\text{cm/min}$ .

(b) Area:

A = xy

\frac{dA}{dt} = x\frac{dy}{dt} + y\frac{dx}{dt} = 8(4) + 6(-5) = 32 - 30 = 2\,\text{cm}^2/\text{min}

The area is increasing at the rate of $2\,\text{cm}^2/\text{min}$ .

8A balloon, which always remains spherical on inflation, is being inflated by pumping in 900 cubic centimetres of gas per second. Find the rate at which the radius of the balloon increases when the radius is

15\,\text{cm}

.Show solution

Given:

\dfrac{dV}{dt} = 900\,\text{cm}^3/\text{s}

r = 15\,\text{cm}

.

Volume of sphere:

V = \dfrac{4}{3}\pi r^3

\frac{dV}{dt} = 4\pi r^2\frac{dr}{dt}

900 = 4\pi(15)^2\frac{dr}{dt} = 4\pi(225)\frac{dr}{dt}

\frac{dr}{dt} = \frac{900}{900\pi} = \frac{1}{\pi}\,\text{cm/s}

Hence, the radius is increasing at the rate of $\dfrac{1}{\pi}\,\text{cm/s}$ .

9A balloon, which always remains spherical has a variable radius. Find the rate at which its volume is increasing with the radius when the latter is

10\,\text{cm}

.Show solution

Given: Radius

r = 10\,\text{cm}

.

Volume of sphere:

V = \dfrac{4}{3}\pi r^3

\frac{dV}{dr} = 4\pi r^2

r = 10\,\text{cm}

\frac{dV}{dr}\bigg|_{r=10} = 4\pi(10)^2 = 400\pi\,\text{cm}^3/\text{cm}

Hence, the volume is increasing at the rate of $400\pi\,\text{cm}^3$ per cm increase in radius.

10A ladder

5\,\text{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\,\text{cm/s}

. How fast is its height on the wall decreasing when the foot of the ladder is

4\,\text{m}

away from the wall?Show solution

Given: Length of ladder

= 5\,\text{m}

\dfrac{dx}{dt} = 2\,\text{cm/s} = 0.02\,\text{m/s}

x = 4\,\text{m}

.

Let

x

= distance of foot from wall,

y

= height on wall.

By Pythagoras theorem:

x^2 + y^2 = 25

Differentiating with respect to

t

2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0 \implies \frac{dy}{dt} = -\frac{x}{y}\frac{dx}{dt}

When

x = 4\,\text{m}

y = \sqrt{25 - 16} = 3\,\text{m}

\frac{dy}{dt} = -\frac{4}{3}\times 0.02 = -\frac{0.08}{3}\,\text{m/s} = -\frac{8}{300}\,\text{m/s}

Converting:

\dfrac{dy}{dt} = -\dfrac{8}{3}\,\text{cm/s}

Hence, the height on the wall is decreasing at the rate of $\dfrac{8}{3}\,\text{cm/s}$ .

11A particle moves along the curve

6y = x^3 + 2

. Find the points on the curve at which the

y

-coordinate is changing 8 times as fast as the

x

-coordinate.Show solution

Given:

6y = x^3 + 2

and

\dfrac{dy}{dt} = 8\dfrac{dx}{dt}

.

Differentiating

6y = x^3 + 2

with respect to

t

6\frac{dy}{dt} = 3x^2\frac{dx}{dt}

Substituting

\dfrac{dy}{dt} = 8\dfrac{dx}{dt}

6\cdot 8\frac{dx}{dt} = 3x^2\frac{dx}{dt}

48 = 3x^2 \implies x^2 = 16 \implies x = \pm 4

When $x = 4$ :

6y = 64 + 2 = 66 \implies y = 11

. Point:

(4, 11)

.

When $x = -4$ :

6y = -64 + 2 = -62 \implies y = -\dfrac{31}{3}

. Point:

\left(-4, -\dfrac{31}{3}\right)

.

Hence, the required points are $(4, 11)$ and $\left(-4, -\dfrac{31}{3}\right)$ .

12The radius of an air bubble is increasing at the rate of

\dfrac{1}{2}\,\text{cm/s}

. At what rate is the volume of the bubble increasing when the radius is

1\,\text{cm}

?Show solution

Given:

\dfrac{dr}{dt} = \dfrac{1}{2}\,\text{cm/s}

r = 1\,\text{cm}

.

Volume of sphere:

V = \dfrac{4}{3}\pi r^3

\frac{dV}{dt} = 4\pi r^2\frac{dr}{dt} = 4\pi(1)^2\cdot\frac{1}{2} = 2\pi\,\text{cm}^3/\text{s}

Hence, the volume of the bubble is increasing at the rate of $2\pi\,\text{cm}^3/\text{s}$ .

13A balloon, which always remains spherical, has a variable diameter

\dfrac{3}{2}(2x + 1)

. Find the rate of change of its volume with respect to

x

.Show solution

Given: Diameter

= \dfrac{3}{2}(2x+1)

, so radius

r = \dfrac{3}{4}(2x+1)

.

Volume:

V = \dfrac{4}{3}\pi r^3 = \dfrac{4}{3}\pi\left[\dfrac{3}{4}(2x+1)\right]^3

V = \frac{4}{3}\pi \cdot \frac{27}{64}(2x+1)^3 = \frac{9\pi}{16}(2x+1)^3

\frac{dV}{dx} = \frac{9\pi}{16}\cdot 3(2x+1)^2\cdot 2 = \frac{27\pi}{8}(2x+1)^2

Hence, $\dfrac{dV}{dx} = \dfrac{27\pi}{8}(2x+1)^2$ .

14Sand is pouring from a pipe at the rate of

12\,\text{cm}^3/\text{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\,\text{cm}

?Show solution

Given:

\dfrac{dV}{dt} = 12\,\text{cm}^3/\text{s}

h = \dfrac{r}{6}

, i.e.,

r = 6h

.

Volume of cone:

V = \dfrac{1}{3}\pi r^2 h = \dfrac{1}{3}\pi(6h)^2 h = \dfrac{1}{3}\pi \cdot 36h^3 = 12\pi h^3

\frac{dV}{dt} = 36\pi h^2\frac{dh}{dt}

12 = 36\pi(4)^2\frac{dh}{dt} = 36\pi(16)\frac{dh}{dt}

\frac{dh}{dt} = \frac{12}{576\pi} = \frac{1}{48\pi}\,\text{cm/s}

Hence, the height of the sand cone is increasing at the rate of $\dfrac{1}{48\pi}\,\text{cm/s}$ .

15The total cost

C(x)

in Rupees associated with the production of

x

units of an item is given by

C(x) = 0.007x^3 - 0.003x^2 + 15x + 4000

. Find the marginal cost when 17 units are produced.Show solution

Given:

C(x) = 0.007x^3 - 0.003x^2 + 15x + 4000

.

Marginal Cost (MC)

= \dfrac{dC}{dx}

\frac{dC}{dx} = 0.007(3x^2) - 0.003(2x) + 15 = 0.021x^2 - 0.006x + 15

At $x = 17$ :

\text{MC} = 0.021(17)^2 - 0.006(17) + 15

= 0.021(289) - 0.102 + 15

= 6.069 - 0.102 + 15 = 20.967

Hence, the marginal cost when 17 units are produced is ₹ 20.967 (approximately ₹ 20.97).

16The total revenue in Rupees received from the sale of

x

units of a product is given by

R(x) = 13x^2 + 26x + 15

. Find the marginal revenue when

x = 7

.Show solution

Given:

R(x) = 13x^2 + 26x + 15

.

Marginal Revenue (MR)

= \dfrac{dR}{dx}

\frac{dR}{dx} = 26x + 26

At $x = 7$ :

\text{MR} = 26(7) + 26 = 182 + 26 = 208

Hence, the marginal revenue when $x = 7$ is ₹ 208.

17The rate of change of the area of a circle with respect to its radius

r

r = 6\,\text{cm}

is\n(A)

10\pi

(B)

12\pi

(C)

8\pi

(D)

11\pi

Show solution

Correct Answer: (B) $12\pi$

Justification: Area

A = \pi r^2 \Rightarrow \dfrac{dA}{dr} = 2\pi r

.

At

r = 6

\dfrac{dA}{dr} = 2\pi(6) = 12\pi

18The total revenue in Rupees received from the sale of

x

units of a product is given by

R(x) = 3x^2 + 36x + 5

. The marginal revenue, when

x = 15

is\n(A) 116 (B) 96 (C) 90 (D) 126Show solution

Correct Answer: (D) 126

Justification:

\text{MR} = \dfrac{dR}{dx} = 6x + 36

.

At

x = 15

\text{MR} = 6(15) + 36 = 90 + 36 = 126

Exercise 6.2

1Show that the function given by

f(x) = 3x + 17

is increasing on

\mathbf{R}

.Show solution

Given:

f(x) = 3x + 17

.

Differentiating:

f'(x) = 3 &gt; 0

for all

x \in \mathbf{R}

.

Since

f'(x) &gt; 0

for all

x \in \mathbf{R}

, the function

f

is strictly increasing on

\mathbf{R}

\blacksquare

2Show that the function given by

f(x) = e^{2x}

is increasing on

\mathbf{R}

.Show solution

Given:

f(x) = e^{2x}

.

Differentiating:

f'(x) = 2e^{2x}

.

Since

e^{2x} &gt; 0

for all

x \in \mathbf{R}

, we have

f'(x) = 2e^{2x} &gt; 0

for all

x \in \mathbf{R}

.

Hence,

f

is strictly increasing on

\mathbf{R}

\blacksquare

3Show that the function given by

f(x) = \sin x

is (a) increasing in

\left(0, \dfrac{\pi}{2}\right)

(b) decreasing in

\left(\dfrac{\pi}{2}, \pi\right)

(0, \pi)

Show solution

Given:

f(x) = \sin x

, so

f'(x) = \cos x

.

(a) In $\left(0, \dfrac{\pi}{2}\right)$ :

\cos x &gt; 0

, so

f'(x) &gt; 0

. Hence

f

is increasing in

\left(0, \dfrac{\pi}{2}\right)

.

(b) In $\left(\dfrac{\pi}{2}, \pi\right)$ :

\cos x &lt; 0

, so

f'(x) &lt; 0

. Hence

f

is decreasing in

\left(\dfrac{\pi}{2}, \pi\right)

.

(c) In $(0, \pi)$ :

f

is increasing on

\left(0, \dfrac{\pi}{2}\right)

and decreasing on

\left(\dfrac{\pi}{2}, \pi\right)

. Hence

f

is neither increasing nor decreasing on the entire interval

(0, \pi)

\blacksquare

4Find the intervals in which the function

f

given by

f(x) = 2x^2 - 3x

is (a) increasing (b) decreasingShow solution

Given:

f(x) = 2x^2 - 3x

f'(x) = 4x - 3

Setting

f'(x) = 0

4x - 3 = 0 \Rightarrow x = \dfrac{3}{4}

.

(a) Increasing:

f'(x) &gt; 0 \Rightarrow 4x - 3 &gt; 0 \Rightarrow x &gt; \dfrac{3}{4}

.

So

f

is increasing on

\left(\dfrac{3}{4}, \infty\right)

.

(b) Decreasing:

f'(x) &lt; 0 \Rightarrow x &lt; \dfrac{3}{4}

.

So

f

is decreasing on

\left(-\infty, \dfrac{3}{4}\right)

5Find the intervals in which the function

f

given by

f(x) = 2x^3 - 3x^2 - 36x + 7

is (a) increasing (b) decreasingShow solution

Given:

f(x) = 2x^3 - 3x^2 - 36x + 7

f'(x) = 6x^2 - 6x - 36 = 6(x^2 - x - 6) = 6(x-3)(x+2)

Critical points:

x = 3

and

x = -2

.

| Interval | Sign of

(x-3)

| Sign of

(x+2)

| Sign of

f'(x)

|
|---|---|---|---|
|

x &lt; -2

-

-

+

|
|

-2 &lt; x &lt; 3

-

+

-

|
|

x &gt; 3

+

+

+

|

(a) Increasing:

f'(x) &gt; 0

when

x \in (-\infty, -2) \cup (3, \infty)

.

(b) Decreasing:

f'(x) &lt; 0

when

x \in (-2, 3)

6Find the intervals in which the following functions are strictly increasing or decreasing: (a)

x^2 + 2x - 5

(b)

10 - 6x - 2x^2

(c)

-2x^3 - 9x^2 - 12x + 1

(d)

6 - 9x - x^2

(e)

(x+1)^3(x-3)^3

Show solution

(a) $f(x) = x^2 + 2x - 5$

f'(x) = 2x + 2 = 2(x+1)

f'(x) = 0 \Rightarrow x = -1

- Strictly increasing:

f'(x) &gt; 0 \Rightarrow x &gt; -1

, i.e.,

x \in (-1, \infty)

.
- Strictly decreasing:

f'(x) &lt; 0 \Rightarrow x &lt; -1

, i.e.,

x \in (-\infty, -1)

.

---

(b) $f(x) = 10 - 6x - 2x^2$

f'(x) = -6 - 4x = -2(3 + 2x)

f'(x) = 0 \Rightarrow x = -\dfrac{3}{2}

- Strictly increasing:

f'(x) &gt; 0 \Rightarrow -2(3+2x) &gt; 0 \Rightarrow x &lt; -\dfrac{3}{2}

, i.e.,

x \in \left(-\infty, -\dfrac{3}{2}\right)

.
- Strictly decreasing:

x \in \left(-\dfrac{3}{2}, \infty\right)

.

---

(c) $f(x) = -2x^3 - 9x^2 - 12x + 1$

f'(x) = -6x^2 - 18x - 12 = -6(x^2 + 3x + 2) = -6(x+1)(x+2)

f'(x) = 0 \Rightarrow x = -1, -2

| Interval | Sign of

f'(x)

|
|---|---|
|

x &lt; -2

-6(-)(-)= -6(+) &lt; 0

|
|

-2 &lt; x &lt; -1

-6(+)(-) &gt; 0

|
|

x &gt; -1

-6(+)(+) &lt; 0

|

- Strictly increasing:

x \in (-2, -1)

.
- Strictly decreasing:

x \in (-\infty, -2) \cup (-1, \infty)

.

---

(d) $f(x) = 6 - 9x - x^2$

f'(x) = -9 - 2x

f'(x) = 0 \Rightarrow x = -\dfrac{9}{2}

- Strictly increasing:

f'(x) &gt; 0 \Rightarrow -9 - 2x &gt; 0 \Rightarrow x &lt; -\dfrac{9}{2}

, i.e.,

x \in \left(-\infty, -\dfrac{9}{2}\right)

.
- Strictly decreasing:

x \in \left(-\dfrac{9}{2}, \infty\right)

.

---

(e) $f(x) = (x+1)^3(x-3)^3$

f'(x) = 3(x+1)^2(x-3)^3 + (x+1)^3 \cdot 3(x-3)^2

= 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)

f'(x) = 0 \Rightarrow x = -1, 1, 3

Note:

(x+1)^2 \geq 0

and

(x-3)^2 \geq 0

always.

- Strictly increasing:

f'(x) &gt; 0 \Rightarrow (x-1) &gt; 0 \Rightarrow x &gt; 1

(and

x \neq 3

), i.e.,

x \in (1, 3) \cup (3, \infty)

.
- Strictly decreasing:

f'(x) &lt; 0 \Rightarrow (x-1) &lt; 0 \Rightarrow x &lt; 1

(and

x \neq -1

), i.e.,

x \in (-\infty, -1) \cup (-1, 1)

7Show that

y = \log(1+x) - \dfrac{2x}{2+x}

x &gt; -1

, is an increasing function of

x

throughout its domain.Show solution

Given:

y = \log(1+x) - \dfrac{2x}{2+x}

x &gt; -1

\frac{dy}{dx} = \frac{1}{1+x} - \frac{(2+x)(2) - 2x(1)}{(2+x)^2}

= \frac{1}{1+x} - \frac{4+2x-2x}{(2+x)^2} = \frac{1}{1+x} - \frac{4}{(2+x)^2}

= \frac{(2+x)^2 - 4(1+x)}{(1+x)(2+x)^2}

= \frac{4 + 4x + x^2 - 4 - 4x}{(1+x)(2+x)^2} = \frac{x^2}{(1+x)(2+x)^2}

For

x &gt; -1

(1+x) &gt; 0

(2+x)^2 &gt; 0

, and

x^2 \geq 0

.

So

\dfrac{dy}{dx} \geq 0

for all

x &gt; -1

, and

\dfrac{dy}{dx} = 0

only at

x = 0

.

Hence,

y

is an increasing function throughout its domain.

\blacksquare

8Find the values of

x

for which

y = [x(x-2)]^2

is an increasing function.Show solution

Given:

y = [x(x-2)]^2 = [x^2 - 2x]^2

\frac{dy}{dx} = 2(x^2-2x)(2x-2) = 4x(x-2)(x-1)

Critical points:

x = 0, 1, 2

.

| Interval | Sign of

\frac{dy}{dx}

|
|---|---|
|

x &lt; 0

4(-)(-)(-) &lt; 0

|
|

0 &lt; x &lt; 1

4(+)(-)(-) &gt; 0

|
|

1 &lt; x &lt; 2

4(+)(-)(+) &lt; 0

|
|

x &gt; 2

4(+)(+)(+) &gt; 0

y

is increasing when

\dfrac{dy}{dx} \geq 0

, i.e., for

x \in [0, 1] \cup [2, \infty)

9Prove that

y = \dfrac{4\sin\theta}{(2+\cos\theta)} - \theta

is an increasing function of

\theta

\left[0, \dfrac{\pi}{2}\right]

.Show solution

Given:

y = \dfrac{4\sin\theta}{2+\cos\theta} - \theta

\frac{dy}{d\theta} = \frac{4\cos\theta(2+\cos\theta) - 4\sin\theta(-\sin\theta)}{(2+\cos\theta)^2} - 1

= \frac{8\cos\theta + 4\cos^2\theta + 4\sin^2\theta}{(2+\cos\theta)^2} - 1

= \frac{8\cos\theta + 4}{(2+\cos\theta)^2} - 1

= \frac{8\cos\theta + 4 - (2+\cos\theta)^2}{(2+\cos\theta)^2}

= \frac{8\cos\theta + 4 - 4 - 4\cos\theta - \cos^2\theta}{(2+\cos\theta)^2}

= \frac{4\cos\theta - \cos^2\theta}{(2+\cos\theta)^2} = \frac{\cos\theta(4 - \cos\theta)}{(2+\cos\theta)^2}

For

\theta \in \left[0, \dfrac{\pi}{2}\right]

:
-

\cos\theta \geq 0

4 - \cos\theta \geq 4 - 1 = 3 &gt; 0

(2+\cos\theta)^2 &gt; 0

Hence

\dfrac{dy}{d\theta} \geq 0

for all

\theta \in \left[0, \dfrac{\pi}{2}\right]

.

Therefore,

y

is an increasing function of

\theta

\left[0, \dfrac{\pi}{2}\right]

\blacksquare

10Prove that the logarithmic function is increasing on

(0, \infty)

.Show solution

Let

f(x) = \log x

, defined for

x &gt; 0

f'(x) = \frac{1}{x}

For all

x \in (0, \infty)

x &gt; 0 \Rightarrow \dfrac{1}{x} &gt; 0 \Rightarrow f'(x) &gt; 0

.

Hence, the logarithmic function is strictly increasing on

(0, \infty)

\blacksquare

11Prove that the function

f

given by

f(x) = x^2 - x + 1

is neither strictly increasing nor decreasing on

(-1, 1)

.Show solution

Given:

f(x) = x^2 - x + 1

f'(x) = 2x - 1

f'(x) = 0 \Rightarrow x = \dfrac{1}{2} \in (-1, 1)

.

- For

x \in \left(-1, \dfrac{1}{2}\right)

f'(x) = 2x - 1 &lt; 0

(decreasing).
- For

x \in \left(\dfrac{1}{2}, 1\right)

f'(x) = 2x - 1 &gt; 0

(increasing).

Since

f

is decreasing on part of

(-1,1)

and increasing on another part,

f

is neither strictly increasing nor strictly decreasing on

(-1, 1)

\blacksquare

12Which of the following functions are decreasing on

\left(0, \dfrac{\pi}{2}\right)

? (A)

\cos x

(B)

\cos 2x

(C)

\cos 3x

(D)

\tan x

Show solution

Correct Answer: (A) $\cos x$

(A) $f(x) = \cos x$ :

f'(x) = -\sin x &lt; 0

for

x \in \left(0, \dfrac{\pi}{2}\right)

. ✓ Decreasing.

(B) $f(x) = \cos 2x$ :

f'(x) = -2\sin 2x

. For

x \in \left(0, \dfrac{\pi}{4}\right)

\sin 2x &gt; 0

f'(x) &lt; 0

; for

x \in \left(\dfrac{\pi}{4}, \dfrac{\pi}{2}\right)

\sin 2x &lt; 0

f'(x) &gt; 0

. Not entirely decreasing.

(C) $f(x) = \cos 3x$ :

f'(x) = -3\sin 3x

. At

x = \dfrac{\pi}{3}

\sin 3x = 0

and changes sign. Not entirely decreasing.

(D) $f(x) = \tan x$ :

f'(x) = \sec^2 x &gt; 0

. Increasing.

Hence, only (A) $\cos x$ is decreasing on $\left(0, \dfrac{\pi}{2}\right)$ .

13On which of the following intervals is the function

f

given by

f(x) = x^{100} + \sin x - 1

decreasing? (A)

(0,1)

(B)

\left(\dfrac{\pi}{2}, \pi\right)

(C)

\left(0, \dfrac{\pi}{2}\right)

(D) None of theseShow solution

Correct Answer: (D) None of these

f'(x) = 100x^{99} + \cos x

(A) $(0,1)$ :

100x^{99} &gt; 0

and

\cos x &gt; 0

, so

f'(x) &gt; 0

. Increasing.

(B) $\left(\dfrac{\pi}{2}, \pi\right)$ :

100x^{99} &gt; 0

(since

x &gt; 0

). Although

\cos x &lt; 0

here,

100x^{99}

dominates (e.g., at

x = \pi/2

100(\pi/2)^{99}

is very large). So

f'(x) &gt; 0

. Increasing.

(C) $\left(0, \dfrac{\pi}{2}\right)$ : Both

100x^{99} &gt; 0

and

\cos x &gt; 0

, so

f'(x) &gt; 0

. Increasing.

Hence, $f$ is not decreasing on any of the given intervals. Answer: (D).

14For what values of

a

the function

f

given by

f(x) = x^2 + ax + 1

is increasing on

[1, 2]

?Show solution

Given:

f(x) = x^2 + ax + 1

f'(x) = 2x + a

For

f

to be increasing on

[1, 2]

, we need

f'(x) \geq 0

for all

x \in [1, 2]

.

The minimum value of

f'(x) = 2x + a

[1,2]

occurs at

x = 1

f'(1) = 2 + a \geq 0 \implies a \geq -2

Hence, $f$ is increasing on $[1, 2]$ for all $a \geq -2$ .

15Let I be any interval disjoint from

[-1, 1]

. Prove that the function

f

given by

f(x) = x + \dfrac{1}{x}

is increasing on I.Show solution

Given:

f(x) = x + \dfrac{1}{x}

f'(x) = 1 - \frac{1}{x^2} = \frac{x^2 - 1}{x^2}

For any interval

I

disjoint from

[-1, 1]

, we have

|x| &gt; 1

, i.e.,

x^2 &gt; 1

.

So

x^2 - 1 &gt; 0

and

x^2 &gt; 0

, giving

f'(x) = \dfrac{x^2-1}{x^2} &gt; 0

.

Hence,

f

is strictly increasing on

I

\blacksquare

16Prove that the function

f

given by

f(x) = \log \sin x

is increasing on

\left(0, \dfrac{\pi}{2}\right)

and decreasing on

\left(\dfrac{\pi}{2}, \pi\right)

.Show solution

Given:

f(x) = \log \sin x

f'(x) = \frac{1}{\sin x}\cdot \cos x = \cot x

- For

x \in \left(0, \dfrac{\pi}{2}\right)

\cot x &gt; 0

, so

f'(x) &gt; 0

. Hence

f

is increasing.
- For

x \in \left(\dfrac{\pi}{2}, \pi\right)

\cot x &lt; 0

, so

f'(x) &lt; 0

. Hence

f

is decreasing.

\blacksquare

17Prove that the function

f

given by

f(x) = \log|\cos x|

is decreasing on

\left(0, \dfrac{\pi}{2}\right)

and increasing on

\left(\dfrac{3\pi}{2}, 2\pi\right)

.Show solution

Given:

f(x) = \log|\cos x|

f'(x) = \frac{1}{|\cos x|}\cdot \frac{d}{dx}|\cos x|

For

x \in \left(0, \dfrac{\pi}{2}\right)

\cos x &gt; 0

, so

|\cos x| = \cos x

f'(x) = \frac{-\sin x}{\cos x} = -\tan x

Since

\tan x &gt; 0

\left(0, \dfrac{\pi}{2}\right)

f'(x) &lt; 0

. Hence

f

is decreasing on

\left(0, \dfrac{\pi}{2}\right)

.

For

x \in \left(\dfrac{3\pi}{2}, 2\pi\right)

\cos x &gt; 0

, so

|\cos x| = \cos x

f'(x) = -\tan x

Since

\tan x &lt; 0

\left(\dfrac{3\pi}{2}, 2\pi\right)

f'(x) = -\tan x &gt; 0

. Hence

f

is increasing on

\left(\dfrac{3\pi}{2}, 2\pi\right)

\blacksquare

18Prove that the function given by

f(x) = x^3 - 3x^2 + 3x - 100

is increasing in

\mathbf{R}

.Show solution

Given:

f(x) = x^3 - 3x^2 + 3x - 100

f'(x) = 3x^2 - 6x + 3 = 3(x^2 - 2x + 1) = 3(x-1)^2

Since

(x-1)^2 \geq 0

for all

x \in \mathbf{R}

, we have

f'(x) = 3(x-1)^2 \geq 0

for all

x \in \mathbf{R}

f'(x) = 0

only at

x = 1

(a single point), so

f

is strictly increasing on

\mathbf{R}

\blacksquare

19The interval in which

y = x^2 e^{-x}

is increasing is (A)

(-\infty, \infty)

(B)

(-2, 0)

(C)

(2, \infty)

(D)

(0, 2)

Show solution

Correct Answer: (D) $(0, 2)$

\dfrac{dy}{dx} = 2xe^{-x} + x^2(-e^{-x}) = xe^{-x}(2 - x)

Since

e^{-x} &gt; 0

always,

\dfrac{dy}{dx} &gt; 0

when

x(2-x) &gt; 0

, i.e.,

0 &lt; x &lt; 2

.

Hence

y

is increasing on

(0, 2)

Exercise 6.3

1Find the maximum and minimum values, if any, of the following functions given by (i)

f(x) = (2x-1)^2 + 3

(ii)

f(x) = 9x^2 + 12x + 2

(iii)

f(x) = -(x-1)^2 + 10

(iv)

g(x) = x^3 + 1

Show solution

(i) $f(x) = (2x-1)^2 + 3$

Since

(2x-1)^2 \geq 0

for all

x

, we have

f(x) \geq 3

f(x) = 3

when

2x - 1 = 0 \Rightarrow x = \dfrac{1}{2}

.

Minimum value = 3 at

x = \dfrac{1}{2}

. No maximum value (as

f(x) \to \infty

).

---

(ii) $f(x) = 9x^2 + 12x + 2 = (3x+2)^2 - 2$

Since

(3x+2)^2 \geq 0

f(x) \geq -2

f(x) = -2

when

x = -\dfrac{2}{3}

.

Minimum value = $-2$ at

x = -\dfrac{2}{3}

. No maximum value.

---

(iii) $f(x) = -(x-1)^2 + 10$

Since

-(x-1)^2 \leq 0

f(x) \leq 10

f(x) = 10

when

x = 1

.

Maximum value = 10 at

x = 1

. No minimum value.

---

(iv) $g(x) = x^3 + 1$

g'(x) = 3x^2 \geq 0

for all

x

, and

g'(x) = 0

only at

x = 0

.

Since

g'(x)

does not change sign,

x = 0

is neither a maximum nor a minimum.

Neither maximum nor minimum value exists.

2Find 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+1,\, x \in (-1,1)

Show solution

(i) $f(x) = |x+2| - 1$

Since

|x+2| \geq 0

f(x) \geq -1

f(x) = -1

when

x = -2

.

Minimum value = $-1$ at

x = -2

. No maximum value.

---

(ii) $g(x) = -|x+1| + 3$

Since

-|x+1| \leq 0

g(x) \leq 3

g(x) = 3

when

x = -1

.

Maximum value = 3 at

x = -1

. No minimum value.

---

(iii) $h(x) = \sin(2x) + 5$

Since

-1 \leq \sin(2x) \leq 1

:

Maximum value = $1 + 5 = 6$ (when

\sin 2x = 1

).

Minimum value = $-1 + 5 = 4$ (when

\sin 2x = -1

).

---

(iv) $f(x) = |\sin 4x + 3|$

Since

-1 \leq \sin 4x \leq 1

, we have

2 \leq \sin 4x + 3 \leq 4

.

So

\sin 4x + 3 &gt; 0

always, hence

|\sin 4x + 3| = \sin 4x + 3

.

Maximum value = 4, Minimum value = 2.

---

(v) $h(x) = x + 1$ , $x \in (-1, 1)$

h

is strictly increasing on the open interval

(-1, 1)

.

As

x \to -1^+

h \to 0

; as

x \to 1^-

h \to 2

. The endpoints are not attained.

Neither maximum nor minimum value exists (open interval).

3Find 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) = x^2

(ii)

g(x) = x^3 - 3x

(iii)

h(x) = \sin x + \cos x,\, 0 &lt; x &lt; \dfrac{\pi}{2}

(iv)

f(x) = \sin x - \cos x,\, 0 &lt; x &lt; 2\pi

(v)

f(x) = x^3 - 6x^2 + 9x + 15

(vi)

g(x) = \dfrac{x}{2} + \dfrac{2}{x},\, x &gt; 0

(vii)

g(x) = \dfrac{1}{x^2+2}

(viii)

f(x) = x\sqrt{1-x},\, 0 &lt; x &lt; 1

Show solution

(i) $f(x) = x^2$

f'(x) = 2x = 0 \Rightarrow x = 0

f''(x) = 2 &gt; 0

, so

x = 0

is a point of local minimum.
Local minimum value = $f(0) = 0$ .

---

(ii) $g(x) = x^3 - 3x$

g'(x) = 3x^2 - 3 = 3(x^2-1) = 0 \Rightarrow x = \pm 1

g''(x) = 6x

.

- At

x = 1

g''(1) = 6 &gt; 0

→ local minimum.

g(1) = 1 - 3 = -2

.
- At

x = -1

g''(-1) = -6 &lt; 0

→ local maximum.

g(-1) = -1 + 3 = 2

.

Local maximum value = 2 at

x = -1

; Local minimum value = $-2$ at

x = 1

.

---

(iii) $h(x) = \sin x + \cos x$ , $0 < x < \dfrac{\pi}{2}$

h'(x) = \cos x - \sin x = 0 \Rightarrow \tan x = 1 \Rightarrow x = \dfrac{\pi}{4}

h''(x) = -\sin x - \cos x

.

At

x = \dfrac{\pi}{4}

h''\left(\dfrac{\pi}{4}\right) = -\dfrac{1}{\sqrt{2}} - \dfrac{1}{\sqrt{2}} = -\sqrt{2} &lt; 0

→ local maximum.

h\left(\dfrac{\pi}{4}\right) = \dfrac{1}{\sqrt{2}} + \dfrac{1}{\sqrt{2}} = \sqrt{2}

.

Local maximum value = $\sqrt{2}$ at

x = \dfrac{\pi}{4}

.

---

(iv) $f(x) = \sin x - \cos x$ , $0 < x < 2\pi$

f'(x) = \cos x + \sin x = 0 \Rightarrow \tan x = -1 \Rightarrow x = \dfrac{3\pi}{4}

x = \dfrac{7\pi}{4}

f''(x) = -\sin x + \cos x

.

- At

x = \dfrac{3\pi}{4}

f''\left(\dfrac{3\pi}{4}\right) = -\dfrac{1}{\sqrt{2}} - \dfrac{1}{\sqrt{2}} = -\sqrt{2} &lt; 0

→ local maximum.

f\left(\dfrac{3\pi}{4}\right) = \dfrac{1}{\sqrt{2}} + \dfrac{1}{\sqrt{2}} = \sqrt{2}

.
- At

x = \dfrac{7\pi}{4}

f''\left(\dfrac{7\pi}{4}\right) = \dfrac{1}{\sqrt{2}} + \dfrac{1}{\sqrt{2}} = \sqrt{2} &gt; 0

→ local minimum.

f\left(\dfrac{7\pi}{4}\right) = -\dfrac{1}{\sqrt{2}} - \dfrac{1}{\sqrt{2}} = -\sqrt{2}

.

Local maximum value = $\sqrt{2}$ at

x = \dfrac{3\pi}{4}

; Local minimum value = $-\sqrt{2}$ at

x = \dfrac{7\pi}{4}

.

---

(v) $f(x) = x^3 - 6x^2 + 9x + 15$

f'(x) = 3x^2 - 12x + 9 = 3(x^2 - 4x + 3) = 3(x-1)(x-3) = 0 \Rightarrow x = 1, 3

f''(x) = 6x - 12

.

- At

x = 1

f''(1) = -6 &lt; 0

→ local maximum.

f(1) = 1 - 6 + 9 + 15 = 19

.
- At

x = 3

f''(3) = 6 &gt; 0

→ local minimum.

f(3) = 27 - 54 + 27 + 15 = 15

.

Local maximum value = 19 at

x = 1

; Local minimum value = 15 at

x = 3

.

---

(vi) $g(x) = \dfrac{x}{2} + \dfrac{2}{x}$ , $x > 0$

g'(x) = \dfrac{1}{2} - \dfrac{2}{x^2} = 0 \Rightarrow x^2 = 4 \Rightarrow x = 2

(since

x &gt; 0

g''(x) = \dfrac{4}{x^3}

. At

x = 2

g''(2) = \dfrac{4}{8} = \dfrac{1}{2} &gt; 0

→ local minimum.

g(2) = 1 + 1 = 2

.

Local minimum value = 2 at

x = 2

.

---

(vii) $g(x) = \dfrac{1}{x^2+2}$

g'(x) = \dfrac{-2x}{(x^2+2)^2} = 0 \Rightarrow x = 0

g''(x)

: At

x = 0

, use first derivative test:

g'(x) &gt; 0

for

x &lt; 0

and

g'(x) &lt; 0

for

x &gt; 0

→ local maximum.

g(0) = \dfrac{1}{2}

.

Local maximum value = $\dfrac{1}{2}$ at

x = 0

.

---

(viii) $f(x) = x\sqrt{1-x}$ , $0 < x < 1$

f'(x) = \sqrt{1-x} + x\cdot\dfrac{-1}{2\sqrt{1-x}} = \dfrac{2(1-x) - x}{2\sqrt{1-x}} = \dfrac{2 - 3x}{2\sqrt{1-x}}

f'(x) = 0 \Rightarrow 2 - 3x = 0 \Rightarrow x = \dfrac{2}{3}

.

For

x &lt; \dfrac{2}{3}

f'(x) &gt; 0

; for

x &gt; \dfrac{2}{3}

f'(x) &lt; 0

→ local maximum at

x = \dfrac{2}{3}

f\left(\dfrac{2}{3}\right) = \dfrac{2}{3}\sqrt{1 - \dfrac{2}{3}} = \dfrac{2}{3}\cdot\dfrac{1}{\sqrt{3}} = \dfrac{2}{3\sqrt{3}} = \dfrac{2\sqrt{3}}{9}

.

Local maximum value = $\dfrac{2\sqrt{3}}{9}$ at

x = \dfrac{2}{3}

4Prove that the following functions do not have maxima or minima: (i)

f(x) = e^x

(ii)

g(x) = \log x

(iii)

h(x) = x^3 + x^2 + x + 1

Show solution

(i) $f(x) = e^x$

f'(x) = e^x &gt; 0

for all

x \in \mathbf{R}

.

Since

f'(x)

never equals zero, there are no critical points. Hence

f

has no maxima or minima.

\blacksquare

---

(ii) $g(x) = \log x$ , $x > 0$

g'(x) = \dfrac{1}{x} &gt; 0

for all

x &gt; 0

.

No critical points exist. Hence

g

has no maxima or minima.

\blacksquare

---

(iii) $h(x) = x^3 + x^2 + x + 1$

h'(x) = 3x^2 + 2x + 1

.

Discriminant

= 4 - 12 = -8 &lt; 0

, and leading coefficient

&gt; 0

, so

h'(x) &gt; 0

for all

x \in \mathbf{R}

.

No critical points. Hence

h

has no maxima or minima.

\blacksquare

5Find the absolute maximum value and the absolute minimum value of the following functions in the given intervals: (i)

f(x) = x^3,\, x \in [-2,2]

(ii)

f(x) = \sin x + \cos x,\, x \in [0,\pi]

(iii)

f(x) = 4x - \dfrac{1}{2}x^2,\, x \in \left[-2, \dfrac{9}{2}\right]

(iv)

f(x) = (x-1)^2 + 3,\, x \in [-3,1]

Show solution

(i) $f(x) = x^3$ , $x \in [-2, 2]$

f'(x) = 3x^2 = 0 \Rightarrow x = 0

.

Values:

f(-2) = -8

f(0) = 0

f(2) = 8

.

Absolute maximum = 8 at

x = 2

; Absolute minimum = $-8$ at

x = -2

.

---

(ii) $f(x) = \sin x + \cos x$ , $x \in [0, \pi]$

f'(x) = \cos x - \sin x = 0 \Rightarrow \tan x = 1 \Rightarrow x = \dfrac{\pi}{4}

.

Values:

f(0) = 1

f\left(\dfrac{\pi}{4}\right) = \sqrt{2}

f(\pi) = -1

.

Absolute maximum = $\sqrt{2}$ at

x = \dfrac{\pi}{4}

; Absolute minimum = $-1$ at

x = \pi

.

---

(iii) $f(x) = 4x - \dfrac{1}{2}x^2$ , $x \in \left[-2, \dfrac{9}{2}\right]$

f'(x) = 4 - x = 0 \Rightarrow x = 4

.

Values:

f(-2) = -8 - 2 = -10

f(4) = 16 - 8 = 8

f\left(\dfrac{9}{2}\right) = 18 - \dfrac{81}{8} = \dfrac{144-81}{8} = \dfrac{63}{8}

.

Absolute maximum = 8 at

x = 4

; Absolute minimum = $-10$ at

x = -2

.

---

(iv) $f(x) = (x-1)^2 + 3$ , $x \in [-3, 1]$

f'(x) = 2(x-1) = 0 \Rightarrow x = 1

.

Values:

f(-3) = 16 + 3 = 19

f(1) = 0 + 3 = 3

.

Absolute maximum = 19 at

x = -3

; Absolute minimum = 3 at

x = 1

6Find the maximum profit that a company can make, if the profit function is given by

p(x) = 41 - 72x - 18x^2

.Show solution

Given:

p(x) = 41 - 72x - 18x^2

p'(x) = -72 - 36x = 0 \Rightarrow x = -2

p''(x) = -36 &lt; 0

, so

x = -2

is a point of local (and absolute) maximum.

p(-2) = 41 - 72(-2) - 18(-2)^2 = 41 + 144 - 72 = 113

Hence, the maximum profit is 113 units.

7Find both the maximum value and the minimum value of

3x^4 - 8x^3 + 12x^2 - 48x + 25

on the interval

[0, 3]

.Show solution

Given:

f(x) = 3x^4 - 8x^3 + 12x^2 - 48x + 25

x \in [0, 3]

f'(x) = 12x^3 - 24x^2 + 24x - 48 = 12(x^3 - 2x^2 + 2x - 4)

= 12[x^2(x-2) + 2(x-2)] = 12(x-2)(x^2+2)

f'(x) = 0 \Rightarrow x = 2

(since

x^2 + 2 &gt; 0

always).

Values:
-

f(0) = 25

f(2) = 3(16) - 8(8) + 12(4) - 48(2) + 25 = 48 - 64 + 48 - 96 + 25 = -39

f(3) = 3(81) - 8(27) + 12(9) - 48(3) + 25 = 243 - 216 + 108 - 144 + 25 = 16

Maximum value = 25 at

x = 0

; Minimum value = $-39$ at

x = 2

8At what points in the interval

[0, 2\pi]

, does the function

\sin 2x

attain its maximum value?Show solution

Given:

f(x) = \sin 2x

x \in [0, 2\pi]

f'(x) = 2\cos 2x = 0 \Rightarrow 2x = \dfrac{\pi}{2}, \dfrac{3\pi}{2}, \dfrac{5\pi}{2}, \dfrac{7\pi}{2}

\Rightarrow x = \dfrac{\pi}{4}, \dfrac{3\pi}{4}, \dfrac{5\pi}{4}, \dfrac{7\pi}{4}

.

Values:
-

f\left(\dfrac{\pi}{4}\right) = \sin\dfrac{\pi}{2} = 1

f\left(\dfrac{3\pi}{4}\right) = \sin\dfrac{3\pi}{2} = -1

f\left(\dfrac{5\pi}{4}\right) = \sin\dfrac{5\pi}{2} = 1

f\left(\dfrac{7\pi}{4}\right) = \sin\dfrac{7\pi}{2} = -1

f(0) = 0

f(2\pi) = 0

.

Maximum value = 1 attained at

x = \dfrac{\pi}{4}

and

x = \dfrac{5\pi}{4}

9What is the maximum value of the function

\sin x + \cos x

?Show solution

Given:

f(x) = \sin x + \cos x

f(x) = \sqrt{2}\sin\left(x + \dfrac{\pi}{4}\right)

The maximum value of

\sin\left(x + \dfrac{\pi}{4}\right) = 1

.

Maximum value of $f(x) = \sqrt{2}$ .

*(Alternatively:

f'(x) = \cos x - \sin x = 0 \Rightarrow x = \dfrac{\pi}{4}

;

f\left(\dfrac{\pi}{4}\right) = \dfrac{1}{\sqrt{2}} + \dfrac{1}{\sqrt{2}} = \sqrt{2}

, and

f''\left(\dfrac{\pi}{4}\right) &lt; 0

, confirming maximum.)*

10Find the maximum value of

2x^3 - 24x + 107

in the interval

[1, 3]

. Find the maximum value of the same function in

[-3, -1]

.Show solution

Given:

f(x) = 2x^3 - 24x + 107

f'(x) = 6x^2 - 24 = 6(x^2 - 4) = 0 \Rightarrow x = \pm 2

.

On $[1, 3]$ : Critical point

x = 2

.
-

f(1) = 2 - 24 + 107 = 85

f(2) = 16 - 48 + 107 = 75

f(3) = 54 - 72 + 107 = 89

Maximum value on $[1,3]$ = 89 at

x = 3

.

On $[-3, -1]$ : Critical point

x = -2

.
-

f(-3) = -54 + 72 + 107 = 125

f(-2) = -16 + 48 + 107 = 139

f(-1) = -2 + 24 + 107 = 129

Maximum value on $[-3,-1]$ = 139 at

x = -2

11It is given that at

x = 1

, the function

x^4 - 62x^2 + ax + 9

attains its maximum value, on the interval

[0, 2]

. Find the value of

a

.Show solution

Given:

f(x) = x^4 - 62x^2 + ax + 9

attains maximum at

x = 1

[0, 2]

.

For

x = 1

to be a critical point:

f'(1) = 0

f'(x) = 4x^3 - 124x + a

f'(1) = 4 - 124 + a = 0 \Rightarrow a = 120

Hence, $a = 120$ .

12Find the maximum and minimum values of

x + \sin 2x

[0, 2\pi]

.Show solution

Given:

f(x) = x + \sin 2x

x \in [0, 2\pi]

f'(x) = 1 + 2\cos 2x = 0 \Rightarrow \cos 2x = -\dfrac{1}{2}

2x = \dfrac{2\pi}{3}, \dfrac{4\pi}{3}, \dfrac{8\pi}{3}, \dfrac{10\pi}{3}

x = \dfrac{\pi}{3}, \dfrac{2\pi}{3}, \dfrac{4\pi}{3}, \dfrac{5\pi}{3}

Values:
-

f(0) = 0

f\left(\dfrac{\pi}{3}\right) = \dfrac{\pi}{3} + \sin\dfrac{2\pi}{3} = \dfrac{\pi}{3} + \dfrac{\sqrt{3}}{2}

f\left(\dfrac{2\pi}{3}\right) = \dfrac{2\pi}{3} + \sin\dfrac{4\pi}{3} = \dfrac{2\pi}{3} - \dfrac{\sqrt{3}}{2}

f\left(\dfrac{4\pi}{3}\right) = \dfrac{4\pi}{3} + \sin\dfrac{8\pi}{3} = \dfrac{4\pi}{3} + \dfrac{\sqrt{3}}{2}

f\left(\dfrac{5\pi}{3}\right) = \dfrac{5\pi}{3} + \sin\dfrac{10\pi}{3} = \dfrac{5\pi}{3} - \dfrac{\sqrt{3}}{2}

f(2\pi) = 2\pi

Comparing all values:

Absolute maximum = $2\pi$ at

x = 2\pi

.

Absolute minimum = $0$ at

x = 0

13Find two numbers whose sum is 24 and whose product is as large as possible.Show solution

Let the two numbers be

x

and

24 - x

.

Product:

P = x(24-x) = 24x - x^2

P'(x) = 24 - 2x = 0 \Rightarrow x = 12

P''(x) = -2 &lt; 0

, so

x = 12

gives maximum.

The two numbers are

12

and

24 - 12 = 12

.

Maximum product = $12 \times 12 = 144$ .

14Find two positive numbers

x

and

y

such that

x + y = 60

and

xy^3

is maximum.Show solution

Given:

x + y = 60 \Rightarrow x = 60 - y

.

Let

f(y) = xy^3 = (60-y)y^3 = 60y^3 - y^4

f'(y) = 180y^2 - 4y^3 = 4y^2(45 - y) = 0 \Rightarrow y = 45

(since

y &gt; 0

f''(y) = 360y - 12y^2

. At

y = 45

f''(45) = 360(45) - 12(2025) = 16200 - 24300 = -8100 &lt; 0

.

So

y = 45

gives maximum.

x = 60 - 45 = 15

.

The two numbers are $x = 15$ and $y = 45$ .

15Find two positive numbers

x

and

y

such that their sum is 35 and the product

x^2y^5

is a maximum.Show solution

Given:

x + y = 35 \Rightarrow x = 35 - y

.

Let

f(y) = x^2y^5 = (35-y)^2 y^5

f'(y) = 2(35-y)(-1)y^5 + (35-y)^2 \cdot 5y^4

= y^4(35-y)[-2y + 5(35-y)]

= y^4(35-y)[175 - 7y]

= 7y^4(35-y)(25-y)

f'(y) = 0 \Rightarrow y = 25

(since

y &gt; 0

and

y &lt; 35

).

Sign check: For

y &lt; 25

f'(y) &gt; 0

; for

y &gt; 25

f'(y) &lt; 0

→ maximum at

y = 25

x = 35 - 25 = 10

.

The two numbers are $x = 10$ and $y = 25$ .

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

Let the two numbers be

x

and

16 - x

.

Sum of cubes:

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

S'(x) = 3x^2 - 3(16-x)^2 = 3[x^2 - (16-x)^2] = 3(x - 16 + x)(x + 16 - x) = 3(2x-16)(16) = 96(x-8)

S'(x) = 0 \Rightarrow x = 8

S''(x) = 96 &gt; 0

, so

x = 8

gives minimum.

The two numbers are 8 and 8.

17A square piece of tin of side

18\,\text{cm}

is to be 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?Show solution

Let the side of the square cut from each corner be

x

cm.

Then: Length = Width =

18 - 2x

, Height =

x

.

Volume:

V(x) = x(18-2x)^2

, where

0 &lt; x &lt; 9

V(x) = x(324 - 72x + 4x^2) = 4x^3 - 72x^2 + 324x

V'(x) = 12x^2 - 144x + 324 = 12(x^2 - 12x + 27) = 12(x-3)(x-9)

V'(x) = 0 \Rightarrow x = 3

x = 9

. Since

0 &lt; x &lt; 9

x = 3

V''(x) = 24x - 144

. At

x = 3

V''(3) = 72 - 144 = -72 &lt; 0

→ maximum.

The side of the square to be cut off is $3\,\text{cm}$ .

Maximum volume

= 3(18-6)^2 = 3(144) = 432\,\text{cm}^3

18A rectangular sheet of tin

45\,\text{cm}

24\,\text{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 maximum?Show solution

Let the side of the square cut from each corner be

x

cm.

Length

= 45 - 2x

, Width

= 24 - 2x

, Height

= x

, where

0 &lt; x &lt; 12

.

Volume:

V(x) = x(45-2x)(24-2x)

= x(1080 - 90x - 48x + 4x^2) = x(1080 - 138x + 4x^2)

= 4x^3 - 138x^2 + 1080x

V'(x) = 12x^2 - 276x + 1080 = 12(x^2 - 23x + 90) = 12(x-5)(x-18)

V'(x) = 0 \Rightarrow x = 5

x = 18

. Since

0 &lt; x &lt; 12

x = 5

V''(x) = 24x - 276

. At

x = 5

V''(5) = 120 - 276 = -156 &lt; 0

→ maximum.

The side of the square to be cut off is $5\,\text{cm}$ .

19Show that of all the rectangles inscribed in a given fixed circle, the square has the maximum area.Show solution

Let the circle have radius

r

. Let the rectangle have sides

2a

and

2b

, so

(2a)^2 + (2b)^2 = (2r)^2

, i.e.,

a^2 + b^2 = r^2

.

Area:

A = (2a)(2b) = 4ab

.

Using AM-GM inequality:

ab \leq \dfrac{a^2+b^2}{2} = \dfrac{r^2}{2}

, with equality when

a = b

.

So

A = 4ab \leq 4 \cdot \dfrac{r^2}{2} = 2r^2

, maximum when

a = b

, i.e., the rectangle is a square.

Alternatively (calculus): Let

b = \sqrt{r^2 - a^2}

A = 4a\sqrt{r^2-a^2}

\dfrac{dA}{da} = 4\left[\sqrt{r^2-a^2} + a\cdot\dfrac{-a}{\sqrt{r^2-a^2}}\right] = \dfrac{4(r^2-2a^2)}{\sqrt{r^2-a^2}} = 0 \Rightarrow a = \dfrac{r}{\sqrt{2}}

.

Then

b = \dfrac{r}{\sqrt{2}} = a

, confirming the rectangle is a square.

\blacksquare

20Show that the right circular cylinder of given surface and maximum volume is such that its height is equal to the diameter of the base.Show solution

Let radius

= r

, height

= h

, total surface area

S

(constant).

S = 2\pi r^2 + 2\pi r h \Rightarrow h = \dfrac{S - 2\pi r^2}{2\pi r}

Volume:

V = \pi r^2 h = \pi r^2 \cdot \dfrac{S - 2\pi r^2}{2\pi r} = \dfrac{r(S - 2\pi r^2)}{2} = \dfrac{Sr}{2} - \pi r^3

\dfrac{dV}{dr} = \dfrac{S}{2} - 3\pi r^2 = 0 \Rightarrow r^2 = \dfrac{S}{6\pi}

\dfrac{d^2V}{dr^2} = -6\pi r &lt; 0

→ maximum.

Now

h = \dfrac{S - 2\pi r^2}{2\pi r} = \dfrac{6\pi r^2 - 2\pi r^2}{2\pi r} = \dfrac{4\pi r^2}{2\pi r} = 2r

.

So

h = 2r

= diameter of the base.

\blacksquare

21Of all the closed cylindrical cans (right circular), of a given volume of 100 cubic centimetres, find the dimensions of the can which has the minimum surface area?Show solution

Let radius

= r

, height

= h

.

Given:

V = \pi r^2 h = 100 \Rightarrow h = \dfrac{100}{\pi r^2}

.

Surface area:

S = 2\pi r^2 + 2\pi r h = 2\pi r^2 + 2\pi r \cdot \dfrac{100}{\pi r^2} = 2\pi r^2 + \dfrac{200}{r}

\dfrac{dS}{dr} = 4\pi r - \dfrac{200}{r^2} = 0 \Rightarrow 4\pi r^3 = 200 \Rightarrow r^3 = \dfrac{50}{\pi} \Rightarrow r = \left(\dfrac{50}{\pi}\right)^{1/3}

\dfrac{d^2S}{dr^2} = 4\pi + \dfrac{400}{r^3} &gt; 0

→ minimum.

h = \dfrac{100}{\pi r^2} = \dfrac{100}{\pi \left(\frac{50}{\pi}\right)^{2/3}}

Note that

h = 2r

(can be verified), so height equals diameter.

Dimensions:

r = \left(\dfrac{50}{\pi}\right)^{1/3}\,\text{cm}

and

h = 2\left(\dfrac{50}{\pi}\right)^{1/3}\,\text{cm}

22A wire of length

28\,\text{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?Show solution

Let the length of wire for the square be

x

m, so the circle gets

(28-x)

m.

Side of square

= \dfrac{x}{4}

; radius of circle

= \dfrac{28-x}{2\pi}

.

Combined area:

A = \left(\frac{x}{4}\right)^2 + \pi\left(\frac{28-x}{2\pi}\right)^2 = \frac{x^2}{16} + \frac{(28-x)^2}{4\pi}

\frac{dA}{dx} = \frac{2x}{16} + \frac{2(28-x)(-1)}{4\pi} = \frac{x}{8} - \frac{28-x}{2\pi}

Setting

\dfrac{dA}{dx} = 0

\frac{x}{8} = \frac{28-x}{2\pi} \Rightarrow 2\pi x = 8(28-x) \Rightarrow 2\pi x + 8x = 224 \Rightarrow x = \frac{224}{2\pi+8} = \frac{112}{\pi+4}

\dfrac{d^2A}{dx^2} = \dfrac{1}{8} + \dfrac{1}{2\pi} &gt; 0

→ minimum.

Length for square

= \dfrac{112}{\pi+4}\,\text{m}

; Length for circle

= 28 - \dfrac{112}{\pi+4} = \dfrac{28\pi}{\pi+4}\,\text{m}

23Prove that the volume of the largest cone that can be inscribed in a sphere of radius

R

\dfrac{8}{27}

of the volume of the sphere.Show solution

Let the cone have height

h

and base radius

r

, inscribed in a sphere of radius

R

.

The centre of the sphere is at distance

h - R

from the base of the cone (taking the apex at top).

By geometry:

r^2 = R^2 - (h-R)^2 = 2Rh - h^2

.

Volume of cone:

V = \dfrac{1}{3}\pi r^2 h = \dfrac{\pi h}{3}(2Rh - h^2) = \dfrac{\pi}{3}(2Rh^2 - h^3)

\dfrac{dV}{dh} = \dfrac{\pi}{3}(4Rh - 3h^2) = \dfrac{\pi h}{3}(4R - 3h) = 0 \Rightarrow h = \dfrac{4R}{3}

\dfrac{d^2V}{dh^2} = \dfrac{\pi}{3}(4R - 6h)

. At

h = \dfrac{4R}{3}

= \dfrac{\pi}{3}(4R - 8R) = -\dfrac{4\pi R}{3} &lt; 0

→ maximum.

V_{\max} = \frac{\pi}{3}\left(2R\cdot\frac{16R^2}{9} - \frac{64R^3}{27}\right) = \frac{\pi}{3}\cdot\frac{R^3}{27}(32 - 64) \cdot\frac{1}{1}

Let me recompute:

V_{\max} = \frac{\pi}{3}\left(2R\cdot\frac{16R^2}{9} - \frac{64R^3}{27}\right) = \frac{\pi}{3}\left(\frac{32R^3}{9} - \frac{64R^3}{27}\right) = \frac{\pi}{3}\cdot\frac{96R^3 - 64R^3}{27} = \frac{\pi}{3}\cdot\frac{32R^3}{27} = \frac{32\pi R^3}{81}

Volume of sphere

= \dfrac{4}{3}\pi R^3

\frac{V_{\max}}{V_{\text{sphere}}} = \frac{\frac{32\pi R^3}{81}}{\frac{4\pi R^3}{3}} = \frac{32\pi R^3}{81}\cdot\frac{3}{4\pi R^3} = \frac{96}{324} = \frac{8}{27}

Hence, the volume of the largest inscribed cone is

\dfrac{8}{27}

of the volume of the sphere.

\blacksquare

24Show that the right circular cone of least curved surface and given volume has an altitude equal to

\sqrt{2}

times the radius of the base.Show solution

Let radius

= r

, height

= h

, slant height

l = \sqrt{r^2+h^2}

.

Given volume:

V = \dfrac{1}{3}\pi r^2 h

(constant)

\Rightarrow h = \dfrac{3V}{\pi r^2}

.

Curved surface area:

S = \pi r l = \pi r\sqrt{r^2+h^2}

Minimise

S^2 = \pi^2 r^2(r^2 + h^2) = \pi^2 r^2\left(r^2 + \dfrac{9V^2}{\pi^2 r^4}\right) = \pi^2 r^4 + \dfrac{9V^2}{r^2}

Let

f(r) = \pi^2 r^4 + 9V^2 r^{-2}

f'(r) = 4\pi^2 r^3 - 18V^2 r^{-3} = 0 \Rightarrow 4\pi^2 r^6 = 18V^2 \Rightarrow r^6 = \dfrac{9V^2}{2\pi^2}

Now

V = \dfrac{\pi r^2 h}{3}

, so

V^2 = \dfrac{\pi^2 r^4 h^2}{9}

r^6 = \dfrac{9}{2\pi^2}\cdot\dfrac{\pi^2 r^4 h^2}{9} = \dfrac{r^4 h^2}{2} \Rightarrow r^2 = \dfrac{h^2}{2} \Rightarrow h = r\sqrt{2}

.

Hence, altitude $h = \sqrt{2}\,r$ .

\blacksquare

25Show that the semi-vertical angle of the cone of the maximum volume and of given slant height is

\tan^{-1}\sqrt{2}

.Show solution

Let slant height

= l

(constant), semi-vertical angle

= \theta

.

Then

r = l\sin\theta

h = l\cos\theta

.

Volume:

V = \dfrac{1}{3}\pi r^2 h = \dfrac{1}{3}\pi l^2\sin^2\theta \cdot l\cos\theta = \dfrac{\pi l^3}{3}\sin^2\theta\cos\theta

\dfrac{dV}{d\theta} = \dfrac{\pi l^3}{3}[2\sin\theta\cos^2\theta - \sin^3\theta] = \dfrac{\pi l^3}{3}\sin\theta[2\cos^2\theta - \sin^2\theta] = 0

Since

\sin\theta \neq 0

2\cos^2\theta = \sin^2\theta \Rightarrow \tan^2\theta = 2 \Rightarrow \tan\theta = \sqrt{2}

\dfrac{d^2V}{d\theta^2} &lt; 0

at this point (can be verified), confirming maximum.

Hence, semi-vertical angle

= \tan^{-1}\sqrt{2}

\blacksquare

26Show that semi-vertical angle of right circular cone of given surface area and maximum volume is

\sin^{-1}\left(\dfrac{1}{3}\right)

.Show solution

Let total surface area

S

(constant), radius

= r

, slant height

= l

S = \pi r l + \pi r^2 \Rightarrow l = \dfrac{S - \pi r^2}{\pi r} = \dfrac{S}{\pi r} - r

h^2 = l^2 - r^2 = \left(\dfrac{S}{\pi r} - r\right)^2 - r^2 = \dfrac{S^2}{\pi^2 r^2} - \dfrac{2S}{\pi} + r^2 - r^2 = \dfrac{S^2}{\pi^2 r^2} - \dfrac{2S}{\pi}

V^2 = \dfrac{\pi^2 r^4 h^2}{9} = \dfrac{\pi^2 r^4}{9}\left(\dfrac{S^2}{\pi^2 r^2} - \dfrac{2S}{\pi}\right) = \dfrac{r^2}{9}\left(S^2 - \dfrac{2S\pi r^2}{1}\cdot\dfrac{\pi}{\pi}\right)

Let

f(r) = V^2 = \dfrac{r^2}{9}\left(S^2 - 2\pi S r^2\right) = \dfrac{S^2 r^2 - 2\pi S r^4}{9}

\dfrac{d(V^2)}{dr} = \dfrac{1}{9}(2S^2 r - 8\pi S r^3) = \dfrac{2Sr}{9}(S - 4\pi r^2) = 0

\Rightarrow S = 4\pi r^2

(since

r \neq 0

).

Now

\sin\theta = \dfrac{r}{l}

and

l = \dfrac{S - \pi r^2}{\pi r} = \dfrac{4\pi r^2 - \pi r^2}{\pi r} = 3r

\sin\theta = \dfrac{r}{3r} = \dfrac{1}{3} \Rightarrow \theta = \sin^{-1}\left(\dfrac{1}{3}\right)

\blacksquare

27The point on the curve

x^2 = 2y

which is nearest to the point

(0, 5)

is (A)

(2\sqrt{2}, 4)

(B)

(2\sqrt{2}, 0)

(C)

(0, 0)

(D)

(2, 2)

Show solution

Correct Answer: (A) $(2\sqrt{2}, 4)$

A point on

x^2 = 2y

(x, x^2/2)

. Distance squared from

(0,5)

D = x^2 + \left(\frac{x^2}{2}-5\right)^2

Let

t = x^2

D = t + \left(\dfrac{t}{2}-5\right)^2 = t + \dfrac{t^2}{4} - 5t + 25 = \dfrac{t^2}{4} - 4t + 25

\dfrac{dD}{dt} = \dfrac{t}{2} - 4 = 0 \Rightarrow t = 8 \Rightarrow x^2 = 8 \Rightarrow x = \pm 2\sqrt{2}

y = 4

.

Nearest point:

(2\sqrt{2}, 4)

28For all real values of

x

, the minimum value of

\dfrac{1-x+x^2}{1+x+x^2}

is (A) 0 (B) 1 (C) 3 (D)

\dfrac{1}{3}

Show solution

Correct Answer: (D) $\dfrac{1}{3}$

Let

f(x) = \dfrac{1-x+x^2}{1+x+x^2}

f'(x) = \dfrac{(-1+2x)(1+x+x^2) - (1-x+x^2)(1+2x)}{(1+x+x^2)^2}

Numerator

= (-1+2x)(1+x+x^2) - (1-x+x^2)(1+2x)

= -1-x-x^2+2x+2x^2+2x^3 - (1+2x-x-2x^2+x^2+2x^3)

= -1+x+x^2+2x^3 - 1 - x - x^2 - 2x^3 + 2x^2 - 2x^2

Simplifying: Numerator

= 2(x^2-1)\cdot\dfrac{1}{1}

... Let me use substitution.

At

x = -1

f(-1) = \dfrac{1+1+1}{1-1+1} = 3

. At

x=1

f(1) = \dfrac{1}{3}

f'(x) = 0

gives

x = \pm 1

f(1) = \dfrac{1}{3}

(minimum),

f(-1) = 3

(maximum).

Minimum value = $\dfrac{1}{3}$ .

29The maximum value of

[x(x-1)+1]^{1/3}

0 \leq x \leq 1

is (A)

\left(\dfrac{1}{3}\right)^{1/3}

(B)

\dfrac{1}{2}

Correct Answer: (C) 1

Let

f(x) = [x(x-1)+1]^{1/3} = [x^2 - x + 1]^{1/3}

.

Let

g(x) = x^2 - x + 1

g'(x) = 2x - 1 = 0 \Rightarrow x = \dfrac{1}{2}

g\left(\dfrac{1}{2}\right) = \dfrac{1}{4} - \dfrac{1}{2} + 1 = \dfrac{3}{4}

(minimum of

g

g(0) = 1

g(1) = 1

.

So maximum of

g

[0,1]

1

, giving

f = 1^{1/3} = 1

.

Maximum value = 1 at

x = 0

x = 1

Miscellaneous Exercise on Chapter 6

1Show that the function given by

f(x) = \dfrac{\log x}{x}

has maximum at

x = e

.Show solution

Given:

f(x) = \dfrac{\log x}{x}

x &gt; 0

f'(x) = \frac{\frac{1}{x}\cdot x - \log x}{x^2} = \frac{1 - \log x}{x^2}

f'(x) = 0 \Rightarrow 1 - \log x = 0 \Rightarrow \log x = 1 \Rightarrow x = e

f''(x) = \frac{-\frac{1}{x}\cdot x^2 - (1-\log x)\cdot 2x}{x^4} = \frac{-x - 2x(1-\log x)}{x^4} = \frac{-1-2(1-\log x)}{x^3}

x = e

f''(e) = \dfrac{-1 - 2(0)}{e^3} = \dfrac{-1}{e^3} &lt; 0

.

Hence,

f

has a maximum at $x = e$ .

\blacksquare

2The two equal sides of an isosceles triangle with fixed base

b

are decreasing at the rate of

3\,\text{cm}

per second. How fast is the area decreasing when the two equal sides are equal to the base?Show solution

Let each equal side

= a

, base

= b

(fixed),

\dfrac{da}{dt} = -3\,\text{cm/s}

.

Height of isosceles triangle:

h = \sqrt{a^2 - \dfrac{b^2}{4}}

.

Area:

A = \dfrac{1}{2}\cdot b \cdot h = \dfrac{b}{2}\sqrt{a^2 - \dfrac{b^2}{4}}

\frac{dA}{dt} = \frac{b}{2}\cdot\frac{2a}{2\sqrt{a^2-b^2/4}}\cdot\frac{da}{dt} = \frac{ab}{2\sqrt{a^2-b^2/4}}\cdot\frac{da}{dt}

When

a = b

\frac{dA}{dt} = \frac{b\cdot b}{2\sqrt{b^2-b^2/4}}\cdot(-3) = \frac{b^2}{2\cdot\frac{b\sqrt{3}}{2}}\cdot(-3) = \frac{b^2}{b\sqrt{3}}\cdot(-3) = \frac{-3b}{\sqrt{3}} = -b\sqrt{3}\,\text{cm}^2/\text{s}

The area is decreasing at the rate of $b\sqrt{3}\,\text{cm}^2/\text{s}$ .

3Find the intervals in which the function

f

given by

f(x) = \dfrac{4\sin x - 2x - x\cos x}{2+\cos x}

is (i) increasing (ii) decreasing.Show solution

Given:

f(x) = \dfrac{4\sin x - 2x - x\cos x}{2+\cos x}

f'(x) = \frac{(4\cos x - 2 - \cos x + x\sin x)(2+\cos x) - (4\sin x - 2x - x\cos x)(-\sin x)}{(2+\cos x)^2}

Numerator:

= (3\cos x - 2 + x\sin x)(2+\cos x) + \sin x(4\sin x - 2x - x\cos x)

= 6\cos x - 4 + 2x\sin x + 3\cos^2 x - 2\cos x + x\sin x\cos x + 4\sin^2 x - 2x\sin x - x\sin x\cos x

= 4\cos x - 4 + 3\cos^2 x + 4\sin^2 x

= 4\cos x - 4 + 3\cos^2 x + 4(1-\cos^2 x)

= 4\cos x - 4 + 3\cos^2 x + 4 - 4\cos^2 x

= 4\cos x - \cos^2 x

= \cos x(4 - \cos x)

f'(x) = \dfrac{\cos x(4-\cos x)}{(2+\cos x)^2}

.

Since

4 - \cos x &gt; 0

always and

(2+\cos x)^2 &gt; 0

always, the sign of

f'(x)

depends on

\cos x

.

(i) Increasing:

f'(x) &gt; 0 \Rightarrow \cos x &gt; 0 \Rightarrow x \in \left(0, \dfrac{\pi}{2}\right) \cup \left(\dfrac{3\pi}{2}, 2\pi\right)

(in

[0, 2\pi]

).

In general:

x \in \left(2n\pi - \dfrac{\pi}{2},\, 2n\pi + \dfrac{\pi}{2}\right)

n \in \mathbb{Z}

.

(ii) Decreasing:

f'(x) &lt; 0 \Rightarrow \cos x &lt; 0 \Rightarrow x \in \left(\dfrac{\pi}{2}, \dfrac{3\pi}{2}\right)

(in

[0, 2\pi]

).

In general:

x \in \left(2n\pi + \dfrac{\pi}{2},\, 2n\pi + \dfrac{3\pi}{2}\right)

n \in \mathbb{Z}

4Find the intervals in which the function

f

given by

f(x) = x^3 + \dfrac{1}{x^3}

x \neq 0

is (i) increasing (ii) decreasing.Show solution

Given:

f(x) = x^3 + x^{-3}

f'(x) = 3x^2 - \dfrac{3}{x^4} = \dfrac{3(x^6-1)}{x^4} = \dfrac{3(x^2-1)(x^4+x^2+1)}{x^4}

Since

x^4 + x^2 + 1 &gt; 0

always and

x^4 &gt; 0

for

x \neq 0

, sign depends on

(x^2-1)

f'(x) = 0 \Rightarrow x = \pm 1

.

(i) Increasing:

f'(x) &gt; 0 \Rightarrow x^2 &gt; 1 \Rightarrow |x| &gt; 1

, i.e.,

x \in (-\infty,-1) \cup (1,\infty)

.

(ii) Decreasing:

f'(x) &lt; 0 \Rightarrow x^2 &lt; 1 \Rightarrow -1 &lt; x &lt; 1

(excluding

x = 0

), i.e.,

x \in (-1,0) \cup (0,1)

5Find the maximum area of an isosceles triangle inscribed in the ellipse

\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1

with its vertex at one end of the major axis.Show solution

Let the vertex at one end of the major axis be

A = (a, 0)

.

A point on the ellipse:

P = (a\cos\theta, b\sin\theta)

. By symmetry, the other vertex is

Q = (a\cos\theta, -b\sin\theta)

.

Base

PQ = 2b\sin\theta

.

Height (horizontal distance from

A

to line

PQ

)

= a - a\cos\theta

.

Area:

A = \dfrac{1}{2}\cdot 2b\sin\theta\cdot a(1-\cos\theta) = ab\sin\theta(1-\cos\theta)

\dfrac{dA}{d\theta} = ab[\cos\theta(1-\cos\theta) + \sin\theta\cdot\sin\theta]

= ab[\cos\theta - \cos^2\theta + \sin^2\theta]

= ab[\cos\theta - \cos^2\theta + 1 - \cos^2\theta]

= ab[\cos\theta - 2\cos^2\theta + 1]

= ab(1+2\cos\theta)(1-\cos\theta) \cdot \dfrac{1}{1}

Actually:

\cos\theta - 2\cos^2\theta + 1 = 0 \Rightarrow 2\cos^2\theta - \cos\theta - 1 = 0 \Rightarrow (2\cos\theta+1)(\cos\theta-1) = 0

\Rightarrow \cos\theta = -\dfrac{1}{2}

(since

\cos\theta = 1

gives zero area)

\Rightarrow \theta = \dfrac{2\pi}{3}

\dfrac{d^2A}{d\theta^2} &lt; 0

\theta = \dfrac{2\pi}{3}

(maximum).

A_{\max} = ab\sin\dfrac{2\pi}{3}\left(1-\cos\dfrac{2\pi}{3}\right) = ab\cdot\dfrac{\sqrt{3}}{2}\cdot\dfrac{3}{2} = \dfrac{3\sqrt{3}}{4}ab

Maximum area of the inscribed isosceles triangle $= \dfrac{3\sqrt{3}}{4}ab$ .

6A tank with rectangular base and rectangular sides, open at the top is to be constructed so that its depth is

2\,\text{m}

and volume is

8\,\text{m}^3

. If building of tank costs Rs 70 per sq metres for the base and Rs 45 per square metre for sides. What is the cost of least expensive tank?Show solution

Let the base dimensions be

x\,\text{m}

and

y\,\text{m}

, depth

= 2\,\text{m}

.

Volume:

V = 2xy = 8 \Rightarrow xy = 4 \Rightarrow y = \dfrac{4}{x}

.

Cost:
- Base area

= xy = 4\,\text{m}^2

; Cost

= 70 \times 4 = 280

.
- Side areas:

2(2x) + 2(2y) = 4(x+y)

; Cost

= 45 \times 4(x+y) = 180(x+y)

.

Total cost:

C = 280 + 180\left(x + \dfrac{4}{x}\right)

\dfrac{dC}{dx} = 180\left(1 - \dfrac{4}{x^2}\right) = 0 \Rightarrow x^2 = 4 \Rightarrow x = 2

\dfrac{d^2C}{dx^2} = 180\cdot\dfrac{8}{x^3} &gt; 0

→ minimum.

At

x = 2

y = 2

C = 280 + 180(2+2) = 280 + 720 = 1000

.

The cost of the least expensive tank is Rs 1000.

7The 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.Show solution

Let radius of circle

= r

, side of square

= a

.

Given:

2\pi r + 4a = k \Rightarrow a = \dfrac{k - 2\pi r}{4}

.

Sum of areas:

A = \pi r^2 + a^2 = \pi r^2 + \left(\dfrac{k-2\pi r}{4}\right)^2

\dfrac{dA}{dr} = 2\pi r + 2\cdot\dfrac{k-2\pi r}{4}\cdot\dfrac{-2\pi}{4} = 2\pi r - \dfrac{\pi(k-2\pi r)}{4}

Setting

\dfrac{dA}{dr} = 0

2\pi r = \dfrac{\pi(k-2\pi r)}{4} \Rightarrow 8r = k - 2\pi r \Rightarrow r(8+2\pi) = k \Rightarrow r = \dfrac{k}{2(\pi+4)}

\dfrac{d^2A}{dr^2} = 2\pi + \dfrac{\pi^2}{2} &gt; 0

→ minimum.

At this

r

a = \dfrac{k - 2\pi r}{4} = \dfrac{k - \frac{\pi k}{\pi+4}}{4} = \dfrac{k\cdot\frac{4}{\pi+4}}{4} = \dfrac{k}{\pi+4} = 2r

.

So

a = 2r

, i.e., side of square = double the radius of circle.

\blacksquare

8A window is in the form of a rectangle surmounted by a semicircular opening. The total perimeter of the window is

10\,\text{m}

. Find the dimensions of the window to admit maximum light through the whole opening.Show solution

Let width of rectangle

= 2r

(diameter of semicircle), height of rectangle

= h

.

Perimeter:

2h + 2r + \pi r = 10 \Rightarrow h = \dfrac{10 - 2r - \pi r}{2} = 5 - r - \dfrac{\pi r}{2}

.

Total area (light admitted):

A = 2rh + \dfrac{\pi r^2}{2} = 2r\left(5 - r - \dfrac{\pi r}{2}\right) + \dfrac{\pi r^2}{2}

= 10r - 2r^2 - \pi r^2 + \dfrac{\pi r^2}{2} = 10r - 2r^2 - \dfrac{\pi r^2}{2}

\dfrac{dA}{dr} = 10 - 4r - \pi r = 0 \Rightarrow r = \dfrac{10}{4+\pi}

\dfrac{d^2A}{dr^2} = -(4+\pi) &lt; 0

→ maximum.

h = 5 - r - \dfrac{\pi r}{2} = 5 - \dfrac{10}{4+\pi}\left(1+\dfrac{\pi}{2}\right) = 5 - \dfrac{10(2+\pi)}{2(4+\pi)} = 5 - \dfrac{5(2+\pi)}{4+\pi} = \dfrac{5(4+\pi) - 5(2+\pi)}{4+\pi} = \dfrac{10}{4+\pi}

h = r = \dfrac{10}{4+\pi}

.

Dimensions: Width

= 2r = \dfrac{20}{4+\pi}\,\text{m}

, Height

= h = \dfrac{10}{4+\pi}\,\text{m}

9A 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

\left(a^{2/3} + b^{2/3}\right)^{3/2}

.Show solution

Let the right angle be at

C

, and let

P

be a point on hypotenuse

AB

such that

PL = a

(perpendicular to

BC

) and

PM = b

(perpendicular to

AC

).

Let

\angle PAC = \theta

(angle that

AP

makes with

AC

). Then:
-

AP = \dfrac{b}{\sin\theta}

(since

PM = b

PM = AP\sin\theta

)
-

PB = \dfrac{a}{\cos\theta}

(since

PL = a

PL = PB\cos\theta

... actually

PL = PB\sin(90°-\theta) = PB\cos\theta

)

Wait, let

\angle ABP = \phi

. Then

PL = PB\sin\phi = a

and

PM = AP\sin\theta = b

where

\theta + \phi = 90°

.

Let

\angle CAB = \theta

. Then

\angle ABC = 90° - \theta

AP = \dfrac{b}{\sin\theta}

PB = \dfrac{a}{\sin(90°-\theta)} = \dfrac{a}{\cos\theta}

.

Hypotenuse:

L = AP + PB = \dfrac{b}{\sin\theta} + \dfrac{a}{\cos\theta}

\dfrac{dL}{d\theta} = -\dfrac{b\cos\theta}{\sin^2\theta} + \dfrac{a\sin\theta}{\cos^2\theta} = 0

\Rightarrow a\sin^3\theta = b\cos^3\theta \Rightarrow \tan^3\theta = \dfrac{b}{a} \Rightarrow \tan\theta = \left(\dfrac{b}{a}\right)^{1/3}

\sin\theta = \dfrac{b^{1/3}}{(a^{2/3}+b^{2/3})^{1/2}}

\cos\theta = \dfrac{a^{1/3}}{(a^{2/3}+b^{2/3})^{1/2}}

L_{\min} = \frac{b}{\sin\theta} + \frac{a}{\cos\theta} = b\cdot\frac{(a^{2/3}+b^{2/3})^{1/2}}{b^{1/3}} + a\cdot\frac{(a^{2/3}+b^{2/3})^{1/2}}{a^{1/3}}

= (a^{2/3}+b^{2/3})^{1/2}\left(b^{2/3} + a^{2/3}\right) = (a^{2/3}+b^{2/3})^{3/2}

Hence, minimum length of hypotenuse

= \left(a^{2/3}+b^{2/3}\right)^{3/2}

\blacksquare

10Find 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 inflexionShow solution

Given:

f(x) = (x-2)^4(x+1)^3

f'(x) = 4(x-2)^3(x+1)^3 + (x-2)^4\cdot 3(x+1)^2

= (x-2)^3(x+1)^2[4(x+1) + 3(x-2)]

= (x-2)^3(x+1)^2[4x+4+3x-6]

= (x-2)^3(x+1)^2(7x-2)

f'(x) = 0 \Rightarrow x = 2,\, x = -1,\, x = \dfrac{2}{7}

.

At $x = -1$ :

(x+1)^2

factor → sign of

f'

does not change. Point of inflexion.

At $x = \dfrac{2}{7}$ :
- For

x &lt; \dfrac{2}{7}

(near):

(x-2)^3 &lt; 0

(x+1)^2 &gt; 0

(7x-2) &lt; 0

→

f'(x) = (-)(+)(-) &gt; 0

.
- For

x &gt; \dfrac{2}{7}

(near):

(x-2)^3 &lt; 0

(x+1)^2 &gt; 0

(7x-2) &gt; 0

→

f'(x) = (-)(+)(+) &lt; 0

f'

changes from

+

-

→ local maximum at

x = \dfrac{2}{7}

.

At $x = 2$ :
- For

x &lt; 2

(near):

(x-2)^3 &lt; 0

(x+1)^2 &gt; 0

(7x-2) &gt; 0

→

f'(x) &lt; 0

.
- For

x &gt; 2

(near):

(x-2)^3 &gt; 0

(x+1)^2 &gt; 0

(7x-2) &gt; 0

→

f'(x) &gt; 0

f'

changes from

-

+

→ local minimum at

x = 2

.

Summary:
- (i) Local maxima at

x = \dfrac{2}{7}

.
- (ii) Local minima at

x = 2

.
- (iii) Point of inflexion at

x = -1

11Find the absolute maximum and minimum values of the function

f

given by

f(x) = \cos^2 x + \sin x

x \in [0, \pi]

.Show solution

Given:

f(x) = \cos^2 x + \sin x = 1 - \sin^2 x + \sin x

f'(x) = -2\sin x\cos x + \cos x = \cos x(1 - 2\sin x) = 0

\Rightarrow \cos x = 0

\sin x = \dfrac{1}{2}

.

In

[0, \pi]

x = \dfrac{\pi}{2}

x = \dfrac{\pi}{6}

.

Values:
-

f(0) = 1 + 0 = 1

f\left(\dfrac{\pi}{6}\right) = \cos^2\dfrac{\pi}{6} + \sin\dfrac{\pi}{6} = \dfrac{3}{4} + \dfrac{1}{2} = \dfrac{5}{4}

f\left(\dfrac{\pi}{2}\right) = 0 + 1 = 1

f(\pi) = 1 + 0 = 1

Absolute maximum value = $\dfrac{5}{4}$ at

x = \dfrac{\pi}{6}

.

Absolute minimum value = $1$ at

x = 0, \dfrac{\pi}{2}, \pi

12Show that the altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius

r

\dfrac{4r}{3}

.Show solution

Let the cone have height

h

and base radius

R

, inscribed in a sphere of radius

r

.

The centre of the sphere lies on the axis of the cone. If the apex is at the top of the sphere, the centre is at distance

r

from apex, so the base is at distance

h - r

below the centre.

By Pythagoras:

R^2 = r^2 - (h-r)^2 = 2rh - h^2

.

Volume:

V = \dfrac{1}{3}\pi R^2 h = \dfrac{\pi h}{3}(2rh - h^2) = \dfrac{\pi}{3}(2rh^2 - h^3)

\dfrac{dV}{dh} = \dfrac{\pi}{3}(4rh - 3h^2) = \dfrac{\pi h}{3}(4r - 3h) = 0 \Rightarrow h = \dfrac{4r}{3}

\dfrac{d^2V}{dh^2} = \dfrac{\pi}{3}(4r - 6h)

. At

h = \dfrac{4r}{3}

= \dfrac{\pi}{3}\left(4r - 8r\right) = -\dfrac{4\pi r}{3} &lt; 0

→ maximum.

Hence, altitude of cone of maximum volume

= \dfrac{4r}{3}

\blacksquare

13Let

f

be a function defined on

[a, b]

such that

f'(x) &gt; 0

, for all

x \in (a, b)

. Then prove that

f

is an increasing function on

(a, b)

.Show solution

To prove:

f

is increasing on

(a, b)

, given

f'(x) &gt; 0

for all

x \in (a, b)

.

Proof: Let

x_1, x_2 \in (a, b)

with

x_1 &lt; x_2

.

By the Mean Value Theorem, there exists

c \in (x_1, x_2)

such that:

f'(c) = \frac{f(x_2) - f(x_1)}{x_2 - x_1}

Since

c \in (x_1, x_2) \subset (a, b)

, we have

f'(c) &gt; 0

.

Also

x_2 - x_1 &gt; 0

.

Therefore:

f(x_2) - f(x_1) = f'(c)(x_2 - x_1) &gt; 0

, i.e.,

f(x_1) &lt; f(x_2)

.

Since

x_1 &lt; x_2 \Rightarrow f(x_1) &lt; f(x_2)

for all

x_1, x_2 \in (a, b)

f

is strictly increasing on

(a, b)

\blacksquare

14Show that the height of the cylinder of maximum volume that can be inscribed in a sphere of radius

R

\dfrac{2R}{\sqrt{3}}

. Also find the maximum volume.Show solution

Let the cylinder have radius

r

and height

h

, inscribed in a sphere of radius

R

.

By geometry:

r^2 + \left(\dfrac{h}{2}\right)^2 = R^2 \Rightarrow r^2 = R^2 - \dfrac{h^2}{4}

.

Volume:

V = \pi r^2 h = \pi h\left(R^2 - \dfrac{h^2}{4}\right) = \pi\left(R^2 h - \dfrac{h^3}{4}\right)

\dfrac{dV}{dh} = \pi\left(R^2 - \dfrac{3h^2}{4}\right) = 0 \Rightarrow h^2 = \dfrac{4R^2}{3} \Rightarrow h = \dfrac{2R}{\sqrt{3}}

\dfrac{d^2V}{dh^2} = -\dfrac{3\pi h}{2} &lt; 0

→ maximum.

Maximum volume:

r^2 = R^2 - \dfrac{1}{4}\cdot\dfrac{4R^2}{3} = R^2 - \dfrac{R^2}{3} = \dfrac{2R^2}{3}

V_{\max} = \pi\cdot\dfrac{2R^2}{3}\cdot\dfrac{2R}{\sqrt{3}} = \dfrac{4\pi R^3}{3\sqrt{3}} = \dfrac{4\pi R^3}{3\sqrt{3}} = \dfrac{4\pi R^3\sqrt{3}}{9}

Height $= \dfrac{2R}{\sqrt{3}}$ and Maximum volume $= \dfrac{4\pi R^3}{3\sqrt{3}}$ .

\blacksquare

15Show that height of the cylinder of greatest volume which can be inscribed in a right circular cone of height

h

and semi vertical angle

\alpha

is one-third that of the cone and the greatest volume of cylinder is

\dfrac{4}{27}\pi h^3\tan^2\alpha

.Show solution

Let the cylinder have radius

r

and height

H

, inscribed in a cone of height

h

and semi-vertical angle

\alpha

.

From similar triangles:

\dfrac{h - H}{r} = \dfrac{h}{h\tan\alpha} = \cot\alpha

, so

r = (h-H)\tan\alpha

.

Volume of cylinder:

V = \pi r^2 H = \pi(h-H)^2\tan^2\alpha\cdot H

Let

f(H) = H(h-H)^2

f'(H) = (h-H)^2 + H\cdot 2(h-H)(-1) = (h-H)[(h-H) - 2H] = (h-H)(h-3H)

f'(H) = 0 \Rightarrow H = h

(trivial) or

H = \dfrac{h}{3}

f''(H) = -1\cdot(h-3H) + (h-H)(-3)

. At

H = \dfrac{h}{3}

f''\left(\dfrac{h}{3}\right) = 0 + \left(h - \dfrac{h}{3}\right)(-3) = -2h &lt; 0

→ maximum.

So height of cylinder $= \dfrac{h}{3}$ = one-third of cone's height.

\blacksquare

Greatest volume:

V_{\max} = \pi\tan^2\alpha\cdot\dfrac{h}{3}\left(h - \dfrac{h}{3}\right)^2 = \pi\tan^2\alpha\cdot\dfrac{h}{3}\cdot\dfrac{4h^2}{9} = \dfrac{4\pi h^3\tan^2\alpha}{27}

\blacksquare

16A cylindrical tank of radius

10\,\text{m}

is being filled with wheat at the rate of 314 cubic metre per hour. Then the depth of the wheat is increasing at the rate of (A)

1\,\text{m/h}

(B)

0.1\,\text{m/h}

(C)

1.1\,\text{m/h}

(D)

0.5\,\text{m/h}

Show solution

Correct Answer: (A) $1\,\text{m/h}$

Given: Radius

r = 10\,\text{m}

\dfrac{dV}{dt} = 314\,\text{m}^3/\text{h}

V = \pi r^2 h = \pi(10)^2 h = 100\pi h

\dfrac{dV}{dt} = 100\pi\dfrac{dh}{dt}

\dfrac{dh}{dt} = \dfrac{314}{100\pi} = \dfrac{314}{100 \times 3.14} = \dfrac{314}{314} = 1\,\text{m/h}

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