Chapter 3 of 10

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Updated 11 June 2026

NCERT Solutions

Differentiation and its Applications

CBSE · Class 12 · Applied Mathematics

NCERT Solutions for Differentiation and its Applications — CBSE Class 12 Applied Mathematics.

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79 Questions Solved · 7 Sections

Exercise 3.1

1(i)Find

\frac{dy}{dx}

from

x^3 + y^3 = 3axy

.Show solution

Given:

x^3 + y^3 = 3axy

Concept: Implicit differentiation — differentiate both sides with respect to

x

.

Working:

Differentiating both sides w.r.t.

x

3x^2 + 3y^2\frac{dy}{dx} = 3a\left(y + x\frac{dy}{dx}\right)

3x^2 + 3y^2\frac{dy}{dx} = 3ay + 3ax\frac{dy}{dx}

3y^2\frac{dy}{dx} - 3ax\frac{dy}{dx} = 3ay - 3x^2

\frac{dy}{dx}(3y^2 - 3ax) = 3ay - 3x^2

\boxed{\frac{dy}{dx} = \frac{ay - x^2}{y^2 - ax}}

1(ii)Find

\frac{dy}{dx}

from

e^{xy} - axy = a

.Show solution

Given:

e^{xy} - axy = a

Concept: Implicit differentiation using chain rule and product rule.

Working:

Differentiating both sides w.r.t.

x

e^{xy}\cdot\frac{d}{dx}(xy) - a\frac{d}{dx}(xy) = 0

e^{xy}\left(y + x\frac{dy}{dx}\right) - a\left(y + x\frac{dy}{dx}\right) = 0

(e^{xy} - a)\left(y + x\frac{dy}{dx}\right) = 0

Since

e^{xy} - a \neq 0

in general:

y + x\frac{dy}{dx} = 0

\boxed{\frac{dy}{dx} = -\frac{y}{x}}

1(iii)Find

\frac{dy}{dx}

from

3x^3 - 5x^2y + 2xy^2 + 4y^3 = 0

.Show solution

Given:

3x^3 - 5x^2y + 2xy^2 + 4y^3 = 0

Concept: Implicit differentiation with product rule.

Working:

Differentiating both sides w.r.t.

x

9x^2 - 5\left(2xy + x^2\frac{dy}{dx}\right) + 2\left(y^2 + 2xy\frac{dy}{dx}\right) + 12y^2\frac{dy}{dx} = 0

9x^2 - 10xy - 5x^2\frac{dy}{dx} + 2y^2 + 4xy\frac{dy}{dx} + 12y^2\frac{dy}{dx} = 0

\frac{dy}{dx}(-5x^2 + 4xy + 12y^2) = -9x^2 + 10xy - 2y^2

\boxed{\frac{dy}{dx} = \frac{10xy - 9x^2 - 2y^2}{4xy - 5x^2 + 12y^2}}

1(iv)Find

\frac{dy}{dx}

from

x^{1/3} + y^{1/3} = a^{2/3}

.Show solution

Given:

x^{1/3} + y^{1/3} = a^{2/3}

Concept: Implicit differentiation.

Working:

Differentiating both sides w.r.t.

x

\frac{1}{3}x^{-2/3} + \frac{1}{3}y^{-2/3}\frac{dy}{dx} = 0

\frac{1}{3y^{2/3}}\frac{dy}{dx} = -\frac{1}{3x^{2/3}}

\frac{dy}{dx} = -\frac{y^{2/3}}{x^{2/3}} = -\left(\frac{y}{x}\right)^{2/3}

\boxed{\frac{dy}{dx} = -\sqrt[3]{\frac{y}{x}}}

1(v)Find

\frac{dy}{dx}

from

x = y\log(xy)

.Show solution

Given:

x = y\log(xy)

Concept: Implicit differentiation using product rule and chain rule.

Working:

Differentiating both sides w.r.t.

x

1 = \frac{dy}{dx}\cdot\log(xy) + y\cdot\frac{1}{xy}\cdot\frac{d}{dx}(xy)

1 = \frac{dy}{dx}\log(xy) + \frac{1}{x}\left(y + x\frac{dy}{dx}\right)

1 = \frac{dy}{dx}\log(xy) + \frac{y}{x} + \frac{dy}{dx}

1 - \frac{y}{x} = \frac{dy}{dx}(\log(xy) + 1)

\frac{x - y}{x} = \frac{dy}{dx}(1 + \log(xy))

\boxed{\frac{dy}{dx} = \frac{x - y}{x(1 + \log(xy))}}

2(i)Find

\frac{dy}{dx}

from the parametric equations

x = at

y = \frac{a}{t}

.Show solution

Given:

x = at

y = \dfrac{a}{t}

Concept:

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

Working:

\frac{dx}{dt} = a, \quad \frac{dy}{dt} = -\frac{a}{t^2}

\frac{dy}{dx} = \frac{-a/t^2}{a} = -\frac{1}{t^2}

\boxed{\frac{dy}{dx} = -\frac{1}{t^2}}

2(ii)Find

\frac{dy}{dx}

from the parametric equations

x = t\log t

y = \frac{\log t}{t}

.Show solution

Given:

x = t\log t

y = \dfrac{\log t}{t}

Concept:

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

Working:

\frac{dx}{dt} = \log t + t\cdot\frac{1}{t} = \log t + 1 = 1 + \log t

\frac{dy}{dt} = \frac{\frac{1}{t}\cdot t - \log t\cdot 1}{t^2} = \frac{1 - \log t}{t^2}

\frac{dy}{dx} = \frac{(1-\log t)/t^2}{1 + \log t}

\boxed{\frac{dy}{dx} = \frac{1 - \log t}{t^2(1 + \log t)}}

2(iii)Find

\frac{dy}{dx}

from the parametric equations

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

y = \frac{2bt}{1+t^2}

.Show solution

Given:

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

y = \dfrac{2bt}{1+t^2}

Concept:

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

Working:

\frac{dx}{dt} = a\cdot\frac{-2t(1+t^2) - (1-t^2)\cdot 2t}{(1+t^2)^2} = a\cdot\frac{-2t - 2t^3 - 2t + 2t^3}{(1+t^2)^2} = a\cdot\frac{-4t}{(1+t^2)^2}

\frac{dy}{dt} = \frac{2b(1+t^2) - 2bt\cdot 2t}{(1+t^2)^2} = \frac{2b(1 - t^2)}{(1+t^2)^2}

\frac{dy}{dx} = \frac{2b(1-t^2)/(1+t^2)^2}{-4at/(1+t^2)^2} = \frac{2b(1-t^2)}{-4at} = -\frac{b(1-t^2)}{2at}

\boxed{\frac{dy}{dx} = -\frac{b}{2a}\cdot\frac{1-t^2}{t}}

3(i)Find

\frac{dy}{dx}

from

x^y = y^x

.Show solution

Given:

x^y = y^x

Concept: Take logarithm on both sides, then differentiate implicitly.

Working:

Taking

\log

on both sides:

y\log x = x\log y

Differentiating w.r.t.

x

\frac{dy}{dx}\log x + y\cdot\frac{1}{x} = \log y + x\cdot\frac{1}{y}\cdot\frac{dy}{dx}

\frac{dy}{dx}\log x - \frac{x}{y}\frac{dy}{dx} = \log y - \frac{y}{x}

\frac{dy}{dx}\left(\log x - \frac{x}{y}\right) = \log y - \frac{y}{x}

\frac{dy}{dx}\cdot\frac{y\log x - x}{y} = \frac{x\log y - y}{x}

\boxed{\frac{dy}{dx} = \frac{y(x\log y - y)}{x(y\log x - x)}}

3(ii)Find

\frac{dy}{dx}

from

x^y = e^{x-y}

.Show solution

Given:

x^y = e^{x-y}

Concept: Take logarithm on both sides, then differentiate implicitly.

Working:

Taking

\log

on both sides:

y\log x = (x - y)

y\log x = x - y \implies y\log x + y = x \implies y(1 + \log x) = x

y = \frac{x}{1 + \log x}

Differentiating w.r.t.

x

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

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

3(iii)Find

\frac{dy}{dx}

from

(x-y)e^{x/(x-y)} = 7

.Show solution

Given:

(x-y)e^{x/(x-y)} = 7

Concept: Take logarithm on both sides, then differentiate implicitly.

Working:

Taking

\log

on both sides:

\log(x-y) + \frac{x}{x-y} = \log 7

Differentiating w.r.t.

x

\frac{1-\frac{dy}{dx}}{x-y} + \frac{(x-y)\cdot 1 - x\left(1-\frac{dy}{dx}\right)}{(x-y)^2} = 0

Multiplying through by

(x-y)^2

(x-y)\left(1-\frac{dy}{dx}\right) + (x-y) - x\left(1-\frac{dy}{dx}\right) = 0

\left(1-\frac{dy}{dx}\right)(x-y-x) + (x-y) = 0

\left(1-\frac{dy}{dx}\right)(-y) + (x-y) = 0

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

y\frac{dy}{dx} = 2y - x

\boxed{\frac{dy}{dx} = \frac{2y - x}{y}}

3(iv)Find

\frac{dy}{dx}

from

y = x^{\log x}

.Show solution

Given:

y = x^{\log x}

Concept: Take logarithm on both sides, then differentiate.

Working:

Taking

\log

on both sides:

\log y = \log x \cdot \log x = (\log x)^2

Differentiating w.r.t.

x

\frac{1}{y}\frac{dy}{dx} = 2\log x \cdot \frac{1}{x}

\frac{dy}{dx} = y\cdot\frac{2\log x}{x} = x^{\log x}\cdot\frac{2\log x}{x} = 2x^{\log x - 1}\cdot\log x

\boxed{\frac{dy}{dx} = 2x^{\log x - 1}\cdot\log x}

4(i)Find

\frac{d^2y}{dx^2}

from

y = x\log x

.Show solution

Given:

y = x\log x

Working:

\frac{dy}{dx} = \log x + x\cdot\frac{1}{x} = \log x + 1

\frac{d^2y}{dx^2} = \frac{1}{x}

\boxed{\frac{d^2y}{dx^2} = \frac{1}{x}}

4(ii)Find

\frac{d^2y}{dx^2}

from

y = x^2 e^x

.Show solution

Given:

y = x^2 e^x

Working:

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

\frac{d^2y}{dx^2} = e^x(x^2 + 2x) + e^x(2x + 2) = e^x(x^2 + 2x + 2x + 2)

\boxed{\frac{d^2y}{dx^2} = e^x(x^2 + 4x + 2)}

4(iii)Find

\frac{d^2y}{dx^2}

from

y = \log(\log x)

.Show solution

Given:

y = \log(\log x)

Working:

\frac{dy}{dx} = \frac{1}{\log x}\cdot\frac{1}{x} = \frac{1}{x\log x}

\frac{d^2y}{dx^2} = \frac{d}{dx}\left[(x\log x)^{-1}\right] = -\frac{1}{(x\log x)^2}\cdot\frac{d}{dx}(x\log x)

\frac{d}{dx}(x\log x) = \log x + 1

\frac{d^2y}{dx^2} = -\frac{\log x + 1}{(x\log x)^2}

\boxed{\frac{d^2y}{dx^2} = -\frac{1 + \log x}{(x\log x)^2}}

4(iv)Find

\frac{d^2y}{dx^2}

from

y = 3e^{2x} + 2e^{3x}

.Show solution

Given:

y = 3e^{2x} + 2e^{3x}

Working:

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

\frac{d^2y}{dx^2} = 12e^{2x} + 18e^{3x} = 6(2e^{2x} + 3e^{3x})

\boxed{\frac{d^2y}{dx^2} = 6(2e^{2x} + 3e^{3x})}

5If

x\sqrt{1+y} + y\sqrt{1+x} = 0

, show that

(1+x^2)\frac{dy}{dx} + 1 = 0

.Show solution

Given:

x\sqrt{1+y} + y\sqrt{1+x} = 0

Working:

Rearranging:

x\sqrt{1+y} = -y\sqrt{1+x}

Squaring both sides:

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

x^2 + x^2 y = y^2 + y^2 x

x^2 - y^2 = y^2 x - x^2 y = xy(y - x)

(x-y)(x+y) = -xy(x-y)

Since

x \neq y

, divide by

(x-y)

x + y = -xy \implies y + xy = -x \implies y(1+x) = -x

y = \frac{-x}{1+x}

Differentiating w.r.t.

x

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

Now check

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

:

Note: The problem states

(1+x^2)

but the standard result for this equation uses

(1+x)^2

. Using

y = \dfrac{-x}{1+x}

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

(1+x)^2\frac{dy}{dx} = -1 \implies (1+x)^2\frac{dy}{dx} + 1 = 0

Hence proved that

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

, which gives

(1+x)^2\dfrac{dy}{dx} + 1 = 0

\blacksquare

6If

y^{1/m} + y^{-1/m} = 2x

, then prove that

(x^2-1)y_1^2 = m^2 y^2

.Show solution

Given:

y^{1/m} + y^{-1/m} = 2x

Working:

Let

y^{1/m} = t

, so

t + \dfrac{1}{t} = 2x

, i.e.,

t^2 - 2xt + 1 = 0

t = x \pm \sqrt{x^2 - 1} \implies y^{1/m} = x \pm \sqrt{x^2-1}

y = \left(x \pm \sqrt{x^2-1}\right)^m

Differentiating w.r.t.

x

y_1 = m\left(x \pm \sqrt{x^2-1}\right)^{m-1}\cdot\left(1 \pm \frac{x}{\sqrt{x^2-1}}\right)

y_1 = m\left(x \pm \sqrt{x^2-1}\right)^{m-1}\cdot\frac{\sqrt{x^2-1} \pm x}{\sqrt{x^2-1}}

y_1 = \frac{m\left(x \pm \sqrt{x^2-1}\right)^m}{\sqrt{x^2-1}} = \frac{my}{\sqrt{x^2-1}}

Squaring both sides:

y_1^2 = \frac{m^2 y^2}{x^2-1}

\boxed{(x^2-1)y_1^2 = m^2 y^2}

\blacksquare

7If

y = \log\left(x + \sqrt{a^2+x^2}\right)

, show that

(a^2+x^2)y_2 + xy_1 = 0

.Show solution

Given:

y = \log\left(x + \sqrt{a^2+x^2}\right)

Working:

y_1 = \frac{dy}{dx} = \frac{1}{x+\sqrt{a^2+x^2}}\cdot\left(1 + \frac{x}{\sqrt{a^2+x^2}}\right)

= \frac{1}{x+\sqrt{a^2+x^2}}\cdot\frac{\sqrt{a^2+x^2}+x}{\sqrt{a^2+x^2}} = \frac{1}{\sqrt{a^2+x^2}}

y_1 = (a^2+x^2)^{-1/2}

.

Differentiating again:

y_2 = -\frac{1}{2}(a^2+x^2)^{-3/2}\cdot 2x = \frac{-x}{(a^2+x^2)^{3/2}}

Now:

(a^2+x^2)y_2 + xy_1 = (a^2+x^2)\cdot\frac{-x}{(a^2+x^2)^{3/2}} + x\cdot\frac{1}{\sqrt{a^2+x^2}}

= \frac{-x}{\sqrt{a^2+x^2}} + \frac{x}{\sqrt{a^2+x^2}} = 0

Hence

(a^2+x^2)y_2 + xy_1 = 0

\blacksquare

8If

y = \left(x + \sqrt{x^2+1}\right)^p

, prove that

(x^2+1)y_2 + xy_1 - p^2 y = 0

.Show solution

Given:

y = \left(x + \sqrt{x^2+1}\right)^p

Working:

y_1 = p\left(x+\sqrt{x^2+1}\right)^{p-1}\cdot\left(1+\frac{x}{\sqrt{x^2+1}}\right)

= p\left(x+\sqrt{x^2+1}\right)^{p-1}\cdot\frac{\sqrt{x^2+1}+x}{\sqrt{x^2+1}}

= \frac{p\left(x+\sqrt{x^2+1}\right)^p}{\sqrt{x^2+1}} = \frac{py}{\sqrt{x^2+1}}

y_1\sqrt{x^2+1} = py

. Squaring:

y_1^2(x^2+1) = p^2 y^2 \quad \cdots (*)

Differentiating

(*)

w.r.t.

x

2y_1 y_2(x^2+1) + y_1^2\cdot 2x = p^2\cdot 2y\cdot y_1

Dividing by

2y_1

(assuming

y_1 \neq 0

y_2(x^2+1) + xy_1 = p^2 y

\boxed{(x^2+1)y_2 + xy_1 - p^2 y = 0}

\blacksquare

Exercise 3.2

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

r

.Show solution

Given: Circumference

C = 2\pi r

Working:

\frac{dC}{dr} = 2\pi

Answer: The rate of change of circumference with respect to radius is

\boxed{2\pi}

2Find the rate of change of lateral surface area of a cube with respect to side

x

, when

x = 4

cm.Show solution

Given: Lateral surface area of a cube

S = 4x^2

Working:

\frac{dS}{dx} = 8x

x = 4

cm:

\frac{dS}{dx}\bigg|_{x=4} = 8(4) = 32 \text{ cm}^2/\text{cm}

Answer:

\boxed{32 \text{ cm}^2/\text{cm}}

3If the rate of change of volume of a sphere is equal to the rate of change of its radius, then find its radius. Also find its surface area.Show solution

Given:

\dfrac{dV}{dt} = \dfrac{dr}{dt}

Working:

Volume of sphere:

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

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

Setting

\dfrac{dV}{dt} = \dfrac{dr}{dt}

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

4\pi r^2 = 1 \implies r^2 = \frac{1}{4\pi} \implies r = \frac{1}{2\sqrt{\pi}}

Surface area

= 4\pi r^2 = 4\pi\cdot\dfrac{1}{4\pi} = 1

sq. unit.

Answer:

r = \dfrac{1}{2\sqrt{\pi}}

and surface area

= 1

sq. unit.

4The volume of a cone changes at the rate

40

cm³/sec. If height of the cone is always equal to its diameter, then find the rate of change of radius when its circular base area is

1

m².Show solution

Given:

\dfrac{dV}{dt} = 40

cm³/sec, height

h = 2r

(diameter

= 2r

), base area

= \pi r^2 = 1

m²

= 10^4

cm².

Working:

Volume of cone:

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

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

40 = 2\pi r^2\frac{dr}{dt}

Given

\pi r^2 = 10^4

cm²:

40 = 2\times 10^4\times\frac{dr}{dt}

\frac{dr}{dt} = \frac{40}{2\times 10^4} = \frac{40}{20000} = 0.002 \text{ cm/sec}

Answer: Rate of change of radius

= \boxed{0.002}

cm/sec.

5For what values of

x

is the rate of increase of total cost function

C(x) = x^3 - 5x^2 + 5x + 8

twice the rate of increase of

x

?Show solution

Given:

C(x) = x^3 - 5x^2 + 5x + 8

, and

\dfrac{dC}{dt} = 2\dfrac{dx}{dt}

.

Working:

\frac{dC}{dt} = \frac{dC}{dx}\cdot\frac{dx}{dt} = (3x^2 - 10x + 5)\frac{dx}{dt}

Setting this equal to

2\dfrac{dx}{dt}

3x^2 - 10x + 5 = 2

3x^2 - 10x + 3 = 0

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

x = \frac{1}{3} \quad \text{or} \quad x = 3

Answer:

x = \dfrac{1}{3}

x = 3

6The radius of the base of a cone is increasing at the rate of 3 cm/minute and the altitude is decreasing at the rate of 4 cm/minute. Find the rate of change of lateral surface area when the radius is 7 cm and the altitude 24 cm.Show solution

Given:

\dfrac{dr}{dt} = 3

cm/min,

\dfrac{dh}{dt} = -4

cm/min,

r = 7

cm,

h = 24

cm.

Working:

Slant height:

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

r=7

h=24

l = \sqrt{49 + 576} = \sqrt{625} = 25

cm.

Lateral surface area:

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

\frac{dS}{dt} = \pi\left[\frac{dr}{dt}\cdot l + r\cdot\frac{dl}{dt}\right]

\frac{dl}{dt} = \frac{r\frac{dr}{dt} + h\frac{dh}{dt}}{\sqrt{r^2+h^2}} = \frac{7(3) + 24(-4)}{25} = \frac{21 - 96}{25} = \frac{-75}{25} = -3

\frac{dS}{dt} = \pi\left[3\times 25 + 7\times(-3)\right] = \pi[75 - 21] = 54\pi \text{ cm}^2/\text{min}

Answer: Rate of change of lateral surface area

= \boxed{54\pi}

cm²/min.

7A ladder 10 meters long rests with one end against a vertical wall, the other on the floor. The lower end moves away from the wall at the rate of 2 meters/minute. Find the rate at which the upper end falls when its base is 6 meters away from the wall.Show solution

Given: Ladder length

= 10

\dfrac{dx}{dt} = 2

m/min,

x = 6

m.

Working:

Let

x

= distance of lower end from wall,

y

= height of upper end.

x^2 + y^2 = 100

x = 6

y = \sqrt{100 - 36} = 8

m.

Differentiating w.r.t.

t

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

\frac{dy}{dt} = -\frac{x}{y}\cdot\frac{dx}{dt} = -\frac{6}{8}\times 2 = -\frac{3}{2} \text{ m/min}

Answer: The upper end falls at the rate of

\dfrac{3}{2}

m/min (i.e.,

-\dfrac{3}{2}

m/min).

8A spherical iron ball 10 cm in radius is coated with a layer of ice of uniform thickness that melts at a rate of 50 cm³/min. When the thickness of ice is 5 cm, find the rate at which the thickness of ice decreases.Show solution

Given: Radius of iron ball

= 10

cm, thickness of ice

= x

cm,

\dfrac{dV}{dt} = -50

cm³/min,

x = 5

cm.

Working:

Total radius

= 10 + x

.

Volume of ice:

V = \dfrac{4}{3}\pi(10+x)^3 - \dfrac{4}{3}\pi(10)^3

\frac{dV}{dt} = 4\pi(10+x)^2\frac{dx}{dt}

x = 5

(10+5)^2 = 225

-50 = 4\pi(225)\frac{dx}{dt}

\frac{dx}{dt} = \frac{-50}{900\pi} = \frac{-1}{18\pi} \text{ cm/min}

Answer: The thickness of ice decreases at the rate of

\dfrac{1}{18\pi}

cm/min.

9A stationery company manufactures

x

units of pen in a given time. The cost of raw material is the square of the pens produced, cost of transportation is twice the number of pens produced and the property tax costs ₹5000. (i) Find the cost function

C(x)

. (ii) Find the cost of producing the 21st pen. (iii) The marginal cost of producing 50 pens.Show solution

Given: Cost of raw material

= x^2

, transportation cost

= 2x

, property tax

= 5000

.

(i) Cost Function:

C(x) = x^2 + 2x + 5000

(ii) Cost of producing the 21st pen:

Marginal cost

= C(21) - C(20)

C(21) = 441 + 42 + 5000 = 5483

C(20) = 400 + 40 + 5000 = 5440

\text{Cost of 21st pen} = 5483 - 5440 = ₹43

Alternatively,

MC = \dfrac{dC}{dx} = 2x + 2

. At

x = 20

MC = 42

. (Using the difference method gives ₹43.)

(iii) Marginal cost at $x = 50$ :

MC = \frac{dC}{dx} = 2x + 2

MC(50) = 2(50) + 2 = 102

Answers: (i)

C(x) = x^2 + 2x + 5000

; (ii) ₹43; (iii) ₹102.

10A firm knows that the price per unit

p

for one of its products is linear. It can sell 1400 units when the price is ₹4 per unit, and 1800 units at a price of ₹2 per unit. Find the price per unit if

x

units are sold. Also find the revenue function and the marginal revenue function.Show solution

Given:

p

is linear in

x

. Points:

(x_1, p_1) = (1400, 4)

and

(x_2, p_2) = (1800, 2)

.

Working:

Slope

= \dfrac{2-4}{1800-1400} = \dfrac{-2}{400} = -\dfrac{1}{200}

Using point-slope form:

p - 4 = -\frac{1}{200}(x - 1400)

p = 4 - \frac{x-1400}{200} = 4 - \frac{x}{200} + 7 = 11 - \frac{x}{200}

Price per unit:

p = 11 - \dfrac{x}{200}

Revenue function:

R(x) = px = 11x - \frac{x^2}{200}

Marginal Revenue:

MR = \frac{dR}{dx} = 11 - \frac{x}{100}

Answers:

p = 11 - \dfrac{x}{200}

;

R(x) = 11x - \dfrac{x^2}{200}

;

MR = 11 - \dfrac{x}{100}

Exercise 3.3

1(i)Find the slopes of the tangent and normal to the curve

y = x^3 - x

x = 1

.Show solution

Given:

y = x^3 - x

, at

x = 1

.

Working:

\frac{dy}{dx} = 3x^2 - 1

x = 1

: slope of tangent

= 3(1)^2 - 1 = 2

.

Slope of normal

= -\dfrac{1}{2}

.

Answer: Slope of tangent

= 2

; slope of normal

= -\dfrac{1}{2}

1(ii)Find the slopes of the tangent and normal to the curve

y = 3x^2 - 6x

x = 2

.Show solution

Given:

y = 3x^2 - 6x

, at

x = 2

.

Working:

\frac{dy}{dx} = 6x - 6

x = 2

: slope of tangent

= 6(2) - 6 = 6

.

Slope of normal

= -\dfrac{1}{6}

.

Answer: Slope of tangent

= 6

; slope of normal

= -\dfrac{1}{6}

1(iii)Find the slopes of the tangent and normal to the curve

y = \frac{x-1}{x-2}

x \neq 2

, at

x = 10

.Show solution

Given:

y = \dfrac{x-1}{x-2}

, at

x = 10

.

Working:

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

x = 10

: slope of tangent

= \dfrac{-1}{(10-2)^2} = \dfrac{-1}{64}

.

Slope of normal

= 64

.

Answer: Slope of tangent

= -\dfrac{1}{64}

; slope of normal

= 64

1(iv)Find the slopes of the tangent and normal to the curve

x^{2/3} + y^{2/3} = 2

(1,1)

.Show solution

Given:

x^{2/3} + y^{2/3} = 2

, at

(1,1)

.

Working:

Differentiating implicitly:

\frac{2}{3}x^{-1/3} + \frac{2}{3}y^{-1/3}\frac{dy}{dx} = 0

\frac{dy}{dx} = -\frac{y^{1/3}}{x^{1/3}} = -\left(\frac{y}{x}\right)^{1/3}

(1,1)

: slope of tangent

= -1

.

Slope of normal

= 1

.

Answer: Slope of tangent

= -1

; slope of normal

= 1

2(i)Find the equations of the tangent and normal to the curve

y = x^3 - 3x + 5

at the point

(2,7)

.Show solution

Given:

y = x^3 - 3x + 5

, point

(2,7)

.

Working:

\frac{dy}{dx} = 3x^2 - 3

x = 2

: slope of tangent

m = 3(4) - 3 = 9

.

Equation of tangent:

y - 7 = 9(x - 2) \implies 9x - y - 11 = 0

Equation of normal (slope

= -\dfrac{1}{9}

y - 7 = -\frac{1}{9}(x-2) \implies 9y - 63 = -(x-2) \implies x + 9y - 65 = 0

Answer: Tangent:

9x - y - 11 = 0

; Normal:

x + 9y - 65 = 0

2(ii)Find the equations of the tangent and normal to the curve

x = at^2

y = 2at

t = 2

.Show solution

Given:

x = at^2

y = 2at

, at

t = 2

.

Working:

\frac{dx}{dt} = 2at, \quad \frac{dy}{dt} = 2a

\frac{dy}{dx} = \frac{2a}{2at} = \frac{1}{t}

t = 2

: slope of tangent

= \dfrac{1}{2}

.

Point:

x = 4a

y = 4a

.

Equation of tangent:

y - 4a = \frac{1}{2}(x - 4a) \implies 2y - 8a = x - 4a \implies x - 2y + 4a = 0

Equation of normal (slope

= -2

y - 4a = -2(x - 4a) \implies y - 4a = -2x + 8a \implies 2x + y - 12a = 0

Answer: Tangent:

x - 2y + 4a = 0

; Normal:

2x + y - 12a = 0

3Find the equations of the tangents to the curve

y = 2x^3 - 15x^2 + 36x - 21

at points where the tangents are parallel to the

x

-axis.Show solution

Given:

y = 2x^3 - 15x^2 + 36x - 21

.

Concept: Tangent parallel to

x

-axis

\Rightarrow \dfrac{dy}{dx} = 0

.

Working:

\frac{dy}{dx} = 6x^2 - 30x + 36 = 6(x^2 - 5x + 6) = 6(x-2)(x-3)

Setting

\dfrac{dy}{dx} = 0

x = 2

x = 3

.

At

x = 2

y = 16 - 60 + 72 - 21 = 7

. Point

(2,7)

. Tangent:

y = 7

.

At

x = 3

y = 54 - 135 + 108 - 21 = 6

. Point

(3,6)

. Tangent:

y = 6

.

Answer: Tangents are

y = 7

and

y = 6

4Find the equation of the tangents to the curve

y = x^3 + 2x - 4

, which is perpendicular to the line

x + 14y + 3 = 0

.Show solution

Given:

y = x^3 + 2x - 4

; line

x + 14y + 3 = 0

.

Concept: Slope of given line

= -\dfrac{1}{14}

. Perpendicular tangent has slope

= 14

.

Working:

\frac{dy}{dx} = 3x^2 + 2 = 14 \implies 3x^2 = 12 \implies x^2 = 4 \implies x = \pm 2

x = 2

y = 8 + 4 - 4 = 8

. Tangent:

y - 8 = 14(x-2) \implies 14x - y - 20 = 0

.

At

x = -2

y = -8 - 4 - 4 = -16

. Tangent:

y + 16 = 14(x+2) \implies 14x - y + 12 = 0

.

Answer:

14x - y - 20 = 0

and

14x - y + 12 = 0

5Find the equation of the tangent and the normal to the curve

y = \frac{x-7}{x^2-5x+6}

at the point where it cuts the

x

-axis.Show solution

Given:

y = \dfrac{x-7}{x^2-5x+6}

.

Working:

The curve cuts the

x

-axis where

y = 0

x - 7 = 0 \Rightarrow x = 7

.

At

x = 7

y = 0

. Point is

(7, 0)

\frac{dy}{dx} = \frac{(x^2-5x+6)(1) - (x-7)(2x-5)}{(x^2-5x+6)^2}

x = 7

: numerator

= (49-35+6)(1) - (0)(9) = 20

; denominator

= (20)^2 = 400

.

Slope of tangent

= \dfrac{20}{400} = \dfrac{1}{20}

.

Equation of tangent:

y - 0 = \frac{1}{20}(x-7) \implies 20y = x - 7 \implies x - 20y - 7 = 0

Equation of normal (slope

= -20

y - 0 = -20(x-7) \implies 20x + y - 140 = 0

Answer: Tangent:

x - 20y - 7 = 0

; Normal:

20x + y - 140 = 0

6Find the equation of the normal to the curve

x^2 = 4y

which passes through the point

(1,2)

.Show solution

Given:

x^2 = 4y

, normal passes through

(1,2)

.

Working:

Let the point of contact be

(h, k)

on the curve, so

h^2 = 4k

.

Differentiating

x^2 = 4y

2x = 4\dfrac{dy}{dx} \Rightarrow \dfrac{dy}{dx} = \dfrac{x}{2}

.

At

(h,k)

: slope of tangent

= \dfrac{h}{2}

; slope of normal

= -\dfrac{2}{h}

.

Equation of normal at

(h,k)

y - k = -\frac{2}{h}(x - h)

Passing through

(1,2)

2 - k = -\frac{2}{h}(1-h) = \frac{2(h-1)}{h}

2 - k = 2 - \frac{2}{h} \implies k = \frac{2}{h}

Also

h^2 = 4k = \dfrac{8}{h} \Rightarrow h^3 = 8 \Rightarrow h = 2

k = 1

.

Slope of normal

= -\dfrac{2}{2} = -1

.

Equation of normal:

y - 1 = -1(x-2) \implies x + y - 3 = 0

Answer:

x + y - 3 = 0

7For the curve

y = x^2 + 3x + 4

, find all points at which the tangent passes through the origin.Show solution

Given:

y = x^2 + 3x + 4

.

Working:

Let

(h,k)

be a point on the curve:

k = h^2 + 3h + 4

... (i)

\frac{dy}{dx} = 2x + 3 \implies \text{slope at }(h,k) = 2h+3

Equation of tangent at

(h,k)

passing through origin

(0,0)

0 - k = (2h+3)(0-h) \implies -k = -h(2h+3)

k = h(2h+3) = 2h^2 + 3h \quad \cdots (ii)

From (i) and (ii):

h^2 + 3h + 4 = 2h^2 + 3h \implies h^2 = 4 \implies h = \pm 2

h = 2

k = 4 + 6 + 4 = 14

. Point

(2,14)

.

At

h = -2

k = 4 - 6 + 4 = 2

. Point

(-2, 2)

.

Answer: The points are

(2, 14)

and

(-2, 2)

8Show that the line

\frac{x}{a} + \frac{y}{b} = 1

touches the curve

y = be^{-x/a}

at the point where it crosses the

y

-axis.Show solution

Given: Line

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

; curve

y = be^{-x/a}

.

Working:

The curve crosses the

y

-axis at

x = 0

y = be^0 = b

. Point is

(0, b)

.

Check if

(0,b)

lies on the line:

\dfrac{0}{a} + \dfrac{b}{b} = 1

. ✓

Slope of line

= -\dfrac{b}{a}

.

Slope of tangent to curve:

\frac{dy}{dx} = be^{-x/a}\cdot\left(-\frac{1}{a}\right) = -\frac{b}{a}e^{-x/a}

x = 0

: slope

= -\dfrac{b}{a}

.

Since the slope of the line equals the slope of the curve at

(0,b)

, and the point lies on both, the line is tangent to the curve at

(0,b)

\blacksquare

9Show that the curves

xy = a^2

and

x^2 + y^2 = 2a^2

touch each other.Show solution

Given:

C_1: xy = a^2

and

C_2: x^2 + y^2 = 2a^2

.

Working:

From

C_1

y = \dfrac{a^2}{x}

. Substituting in

C_2

x^2 + \frac{a^4}{x^2} = 2a^2 \implies x^4 - 2a^2x^2 + a^4 = 0 \implies (x^2-a^2)^2 = 0

x = \pm a

x = a

y = a

. At

x = -a

y = -a

.

At point

(a,a)

:

Slope of

C_1

y = a^2/x \Rightarrow dy/dx = -a^2/x^2 = -1

.

Slope of

C_2

2x + 2y\,dy/dx = 0 \Rightarrow dy/dx = -x/y = -1

.

Both curves have the same slope

(-1)

(a,a)

and pass through the same point. Hence they touch each other.

\blacksquare

10Prove that the curves

xy = 4

and

x^2 + y^2 = 8

touch each other.Show solution

Given:

C_1: xy = 4

and

C_2: x^2 + y^2 = 8

.

Working:

From

C_1

y = 4/x

. Substituting in

C_2

x^2 + \frac{16}{x^2} = 8 \implies x^4 - 8x^2 + 16 = 0 \implies (x^2-4)^2 = 0

x = \pm 2

x = 2

y = 2

. At

x = -2

y = -2

.

At point

(2,2)

:

Slope of

C_1

dy/dx = -y/x = -1

.

Slope of

C_2

dy/dx = -x/y = -1

.

Both curves pass through

(2,2)

with the same slope. Hence they touch each other.

\blacksquare

Exercise 3.4

1(i)Find critical points of

f(x) = x^3 - 6x^2 + 9x - 10

.Show solution

Given:

f(x) = x^3 - 6x^2 + 9x - 10

Working:

f'(x) = 3x^2 - 12x + 9 = 3(x^2 - 4x + 3) = 3(x-1)(x-3)

Setting

f'(x) = 0

x = 1

x = 3

.

Answer: Critical points are

x = 1

and

x = 3

1(ii)Find critical points of

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

x &gt; 0

.Show solution

Given:

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

x &gt; 0

Working:

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

Setting

f'(x) = 0

1 - \log x = 0 \Rightarrow \log x = 1 \Rightarrow x = e

.

Answer: Critical point is

x = e

1(iii)Find critical points of

f(x) = 50\sqrt{x} - 0.5x - 1000

.Show solution

Given:

f(x) = 50\sqrt{x} - 0.5x - 1000

Working:

f'(x) = \frac{50}{2\sqrt{x}} - 0.5 = \frac{25}{\sqrt{x}} - 0.5

Setting

f'(x) = 0

\frac{25}{\sqrt{x}} = 0.5 \implies \sqrt{x} = 50 \implies x = 2500

Answer: Critical point is

x = 2500

1(iv)Find critical points of

f(x) = 5xe^{-x^3}

.Show solution

Given:

f(x) = 5xe^{-x^3}

Working:

f'(x) = 5e^{-x^3} + 5x\cdot e^{-x^3}\cdot(-3x^2) = 5e^{-x^3}(1 - 3x^3)

Since

e^{-x^3} &gt; 0

always, setting

f'(x) = 0

1 - 3x^3 = 0 \implies x^3 = \frac{1}{3} \implies x = \frac{1}{\sqrt[3]{3}}

Answer: Critical point is

x = \dfrac{1}{3^{1/3}}

2(i)Find the intervals in which

f(x) = x^4 - 8x^3 + 22x^2 - 24x + 1

is increasing or decreasing.Show solution

Given:

f(x) = x^4 - 8x^3 + 22x^2 - 24x + 1

Working:

f'(x) = 4x^3 - 24x^2 + 44x - 24 = 4(x^3 - 6x^2 + 11x - 6)

Factoring:

x = 1

is a root. Dividing:

x^3 - 6x^2 + 11x - 6 = (x-1)(x^2-5x+6) = (x-1)(x-2)(x-3)

f'(x) = 4(x-1)(x-2)(x-3)

Sign analysis:
-

x &lt; 1

f'(x) &lt; 0

(decreasing)
-

1 &lt; x &lt; 2

f'(x) &gt; 0

(increasing)
-

2 &lt; x &lt; 3

f'(x) &lt; 0

(decreasing)
-

x &gt; 3

f'(x) &gt; 0

(increasing)

Answer: Increasing on

(1,2)

and

(3,\infty)

; Decreasing on

(-\infty,1)

and

(2,3)

2(ii)Find the intervals in which

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

is increasing or decreasing.Show solution

Given:

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

Working:

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

Since

(x^2-x-6)^2 \geq 0

always:

f'(x) = 0

when

x^2-x-6 = 0

(i.e.,

x = 3

x = -2

) or

2x-1 = 0

(i.e.,

x = 1/2

).

Sign of

f'(x)

depends on

(2x-1)

:
-

x &lt; \dfrac{1}{2}

2x-1 &lt; 0

, so

f'(x) \leq 0

(decreasing)
-

x &gt; \dfrac{1}{2}

2x-1 &gt; 0

, so

f'(x) \geq 0

(increasing)

Answer: Increasing on

\left(\dfrac{1}{2}, \infty\right)

; Decreasing on

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

2(iii)Find the intervals in which

f(x) = x^2 e^x

is increasing or decreasing.Show solution

Given:

f(x) = x^2 e^x

Working:

f'(x) = 2xe^x + x^2 e^x = xe^x(2+x) = x(x+2)e^x

Since

e^x &gt; 0

always, sign depends on

x(x+2)

:

Critical points:

x = 0

and

x = -2

.

-

x &lt; -2

x &lt; 0

x+2 &lt; 0

, product

&gt; 0

: increasing
-

-2 &lt; x &lt; 0

x &lt; 0

x+2 &gt; 0

, product

&lt; 0

: decreasing
-

x &gt; 0

x &gt; 0

x+2 &gt; 0

, product

&gt; 0

: increasing

Answer: Increasing on

(-\infty,-2)

and

(0,\infty)

; Decreasing on

(-2,0)

3Show that the function

f(x) = \log(1+x) + \frac{1}{1+x}

increases on

(0,\infty)

.Show solution

Given:

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

Working:

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

For

x \in (0,\infty)

x &gt; 0

and

(1+x)^2 &gt; 0

, so

f'(x) &gt; 0

.

Hence

f(x)

is increasing on

(0,\infty)

\blacksquare

4Prove that the function

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

is neither increasing nor decreasing in

(0,1)

.Show solution

Given:

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

Working:

f'(x) = 2x - 1

(0,1)

:
- At

x = \dfrac{1}{2}

f'(x) = 0

- For

0 &lt; x &lt; \dfrac{1}{2}

f'(x) &lt; 0

(decreasing)
- For

\dfrac{1}{2} &lt; x &lt; 1

f'(x) &gt; 0

(increasing)

Since

f'(x)

changes sign in

(0,1)

, the function is neither entirely increasing nor entirely decreasing in

(0,1)

\blacksquare

5A company finds that its total revenue may be determined by

R(x) = 240000 - (x-500)^2

. Find when is the revenue function increasing and when decreasing.Show solution

Given:

R(x) = 240000 - (x-500)^2

Working:

R'(x) = -2(x-500)

R'(x) &gt; 0 \Rightarrow -2(x-500) &gt; 0 \Rightarrow x &lt; 500

: Revenue is increasing.

R'(x) &lt; 0 \Rightarrow x &gt; 500

: Revenue is decreasing.

Answer: Revenue is increasing for

x &lt; 500

and decreasing for

x &gt; 500

6The price

p

per unit is given by

x = \frac{1}{3}p^2 - 2p + 3

where

x

is the number of units sold. (i) Find the revenue function

R

. (ii) Find the price interval for which the revenue is increasing and decreasing.Show solution

Given:

x = \dfrac{1}{3}p^2 - 2p + 3

(i) Revenue function:

R = px = p\left(\frac{p^2}{3} - 2p + 3\right) = \frac{p^3}{3} - 2p^2 + 3p

(ii) Intervals of increase/decrease:

\frac{dR}{dp} = p^2 - 4p + 3 = (p-1)(p-3)

\dfrac{dR}{dp} &gt; 0

when

p &lt; 1

p &gt; 3

: Revenue is increasing.

\dfrac{dR}{dp} &lt; 0

when

1 &lt; p &lt; 3

: Revenue is decreasing.

Answer: Revenue increases for

p \in (-\infty,1)\cup(3,\infty)

and decreases for

p \in (1,3)

7The total cost function of a manufacturing company is given by

C(x) = 2x\left(\frac{x+4}{x+3}\right) + 3

. Show that MC (Marginal Cost) falls continuously as the output

x

increases.Show solution

Given:

C(x) = 2x\cdot\dfrac{x+4}{x+3} + 3

Working:

MC = \frac{dC}{dx} = 2\cdot\frac{d}{dx}\left[\frac{x(x+4)}{x+3}\right] = 2\cdot\frac{d}{dx}\left[\frac{x^2+4x}{x+3}\right]

= 2\cdot\frac{(2x+4)(x+3) - (x^2+4x)(1)}{(x+3)^2}

Numerator:

(2x+4)(x+3) - (x^2+4x) = 2x^2+6x+4x+12 - x^2-4x = x^2+6x+12

MC = \frac{2(x^2+6x+12)}{(x+3)^2}

Now differentiate

MC

w.r.t.

x

\frac{d(MC)}{dx} = 2\cdot\frac{(2x+6)(x+3)^2 - (x^2+6x+12)\cdot 2(x+3)}{(x+3)^4}

= \frac{2(x+3)[(2x+6)(x+3) - 2(x^2+6x+12)]}{(x+3)^4} = \frac{2[(2x+6)(x+3) - 2(x^2+6x+12)]}{(x+3)^3}

Numerator of bracket:

2x^2+6x+6x+18 - 2x^2-12x-24 = -6

\frac{d(MC)}{dx} = \frac{2(-6)}{(x+3)^3} = \frac{-12}{(x+3)^3} &lt; 0 \text{ for all } x &gt; 0

Hence MC falls continuously as

x

increases.

\blacksquare

8The price

p

per unit at which a company can sell all that it produces is given by

p = 29 - x

, where

x

is the number of units produced. The total cost function

C(x) = 45 + 11x

. If

P(x) = R(x) - C(x)

is the profit function, find the interval in which the profit is increasing and decreasing.Show solution

Given:

p = 29 - x

C(x) = 45 + 11x

.

Working:

R(x) = px = (29-x)x = 29x - x^2

P(x) = R(x) - C(x) = 29x - x^2 - 45 - 11x = -x^2 + 18x - 45

P'(x) = -2x + 18 = 2(9-x)

P'(x) &gt; 0

when

x &lt; 9

: Profit is increasing.

P'(x) &lt; 0

when

x &gt; 9

: Profit is decreasing.

Answer: Profit increases for

x \in (0,9)

and decreases for

x \in (9,\infty)

Exercise 3.5

1(i)Find the local maxima, local minima, local minimum value and local maximum value, if any, of

f(x) = x^2 - 6x + 16

.Show solution

Given:

f(x) = x^2 - 6x + 16

Working:

f'(x) = 2x - 6 = 0 \implies x = 3

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

Since

f''(3) = 2 &gt; 0

x = 3

is a point of local minimum.

Local minimum value

= f(3) = 9 - 18 + 16 = 7

.

No local maximum.

Answer: Local minimum at

x = 3

; local minimum value

= 7

1(ii)Find the local maxima, local minima, local minimum value and local maximum value, if any, of

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

x &gt; 0

.Show solution

Given:

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

x &gt; 0

Working:

f'(x) = \frac{1 - \log x}{x^2} = 0 \implies \log x = 1 \implies 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{2\log x - 3}{x^3}

x = e

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

.

So

x = e

is a point of local maximum.

Local maximum value

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

.

Answer: Local maximum at

x = e

; local maximum value

= \dfrac{1}{e}

1(iii)Find the local maxima, local minima, local minimum value and local maximum value, if any, of

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

.Show solution

Given:

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

Working:

f'(x) = -2xe^x + (1-x^2)e^x = e^x(1-x^2-2x) = e^x(-x^2-2x+1)

Setting

f'(x) = 0

(since

e^x &gt; 0

-x^2 - 2x + 1 = 0 \implies x^2 + 2x - 1 = 0 \implies x = \frac{-2\pm\sqrt{4+4}}{2} = -1\pm\sqrt{2}

f''(x) = e^x(-x^2-2x+1) + e^x(-2x-2) = e^x(-x^2-4x-1)

x = -1+\sqrt{2}

f''(-1+\sqrt{2}) = e^{-1+\sqrt{2}}(-((-1+\sqrt{2})^2) - 4(-1+\sqrt{2}) - 1)

(-1+\sqrt{2})^2 = 3-2\sqrt{2}

;

-4(-1+\sqrt{2}) = 4-4\sqrt{2}

= e^{-1+\sqrt{2}}(-(3-2\sqrt{2}) + 4 - 4\sqrt{2} - 1) = e^{-1+\sqrt{2}}(-3+2\sqrt{2}+3-4\sqrt{2}) = e^{-1+\sqrt{2}}(-2\sqrt{2}) &lt; 0

x = -1+\sqrt{2}

is a local maximum.

At

x = -1-\sqrt{2}

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

: local minimum.

Answer: Local maximum at

x = -1+\sqrt{2}

; local minimum at

x = -1-\sqrt{2}

1(iv)Find the local maxima, local minima, local minimum value and local maximum value, if any, of

f(x) = \frac{x^2-7x+6}{x-10}

.Show solution

Given:

f(x) = \dfrac{x^2-7x+6}{x-10}

Working:

f'(x) = \frac{(2x-7)(x-10) - (x^2-7x+6)(1)}{(x-10)^2}

Numerator:

(2x-7)(x-10) - (x^2-7x+6)

= 2x^2-20x-7x+70 - x^2+7x-6 = x^2-20x+64

f'(x) = \frac{x^2-20x+64}{(x-10)^2} = \frac{(x-4)(x-16)}{(x-10)^2}

Critical points:

x = 4

and

x = 16

(excluding

x = 10

).

At

x = 4

: sign of

f'

changes from

+

-

(check: for

x&lt;4

(x-4)&lt;0,(x-16)&lt;0

, product

&gt;0

; for

4&lt;x&lt;10

(x-4)&gt;0,(x-16)&lt;0

, product

&lt;0

). So local maximum at

x=4

.

Local max value

= f(4) = \dfrac{16-28+6}{4-10} = \dfrac{-6}{-6} = 1

.

At

x = 16

: sign changes from

-

+

. So local minimum at

x=16

.

Local min value

= f(16) = \dfrac{256-112+6}{6} = \dfrac{150}{6} = 25

.

Answer: Local maximum value

= 1

x = 4

; local minimum value

= 25

x = 16

1(v)Find the local maxima, local minima, local minimum value and local maximum value, if any, of

f(x) = 2x + \frac{1}{2x}

.Show solution

Given:

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

Working:

f'(x) = 2 - \frac{1}{2x^2} = 0 \implies 2x^2 = \frac{1}{2} \implies x^2 = \frac{1}{4} \implies x = \pm\frac{1}{2}

f''(x) = \frac{1}{x^3}

x = \dfrac{1}{2}

f''\left(\dfrac{1}{2}\right) = 8 &gt; 0

: local minimum.

Local min value

= 2\cdot\dfrac{1}{2} + \dfrac{1}{2\cdot\frac{1}{2}} = 1 + 1 = 2

.

At

x = -\dfrac{1}{2}

f''\left(-\dfrac{1}{2}\right) = -8 &lt; 0

: local maximum.

Local max value

= 2\cdot\left(-\dfrac{1}{2}\right) + \dfrac{1}{2\cdot(-\frac{1}{2})} = -1 - 1 = -2

.

Answer: Local minimum value

= 2

x = \dfrac{1}{2}

; local maximum value

= -2

x = -\dfrac{1}{2}

2The sum of two positive numbers is 16. Find the numbers, if the product of the squares is to be maximum.Show solution

Given:

x + y = 16

, maximize

P = x^2 y^2

.

Working:

Let

y = 16 - x

. Then

P = x^2(16-x)^2

.

Maximizing

P

is equivalent to maximizing

Q = x(16-x)

(since

P = Q^2

and

Q &gt; 0

Q = 16x - x^2 \implies Q' = 16 - 2x = 0 \implies x = 8

Q'' = -2 &lt; 0

: maximum.

So

x = 8

y = 8

.

Answer: Both numbers are

8

3Show that of all rectangles with a given perimeter, the square has the largest area.Show solution

Given: Perimeter

= 2(l+b) = 2P

(constant), so

l + b = P

.

Working:

Area

A = lb = l(P-l) = Pl - l^2

\frac{dA}{dl} = P - 2l = 0 \implies l = \frac{P}{2}

\frac{d^2A}{dl^2} = -2 &lt; 0 \implies \text{maximum}

When

l = \dfrac{P}{2}

b = P - \dfrac{P}{2} = \dfrac{P}{2} = l

.

So

l = b

, which means the rectangle is a square.

Hence, among all rectangles with a given perimeter, the square has the largest area.

\blacksquare

4Show that the function

f(x) = x^3 - 6x^2 + 12x + 50

has neither a local maximum nor a local minimum value.Show solution

Given:

f(x) = x^3 - 6x^2 + 12x + 50

Working:

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

f'(x) = 0

x = 2

, but

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

for all

x

.

Since

f'(x)

does not change sign at

x = 2

(it is

\geq 0

on both sides),

x = 2

is neither a local maximum nor a local minimum (it is a point of inflection).

Hence

f(x)

has neither a local maximum nor a local minimum.

\blacksquare

5The profit function, in rupees, of a firm selling

x

items (

x \geq 0

) per week is given by

P(x) = (400-x)x - 3500

. How many items should the firm sell to make the maximum profit? Also find the maximum profit.Show solution

Given:

P(x) = (400-x)x - 3500 = 400x - x^2 - 3500

Working:

P'(x) = 400 - 2x = 0 \implies x = 200

P''(x) = -2 &lt; 0 \implies \text{maximum at } x = 200

Maximum profit

= P(200) = 400(200) - (200)^2 - 3500 = 80000 - 40000 - 3500 = ₹36500

.

Answer: The firm should sell

200

items per week for maximum profit of ₹36,500.

6A tour operator charges ₹136 per passenger for 100 passengers with a discount of ₹4 for each 10 passengers in excess of 100. Find the number of passengers that will maximise the amount of money the tour operator receives.Show solution

Given: Base: 100 passengers at ₹136 each. For every 10 extra passengers, price reduces by ₹4.

Working:

Let

n

= number of groups of 10 in excess of 100. Total passengers

= 100 + 10n

. Price per passenger

= 136 - 4n

.

Revenue

R(n) = (100+10n)(136-4n)

= 13600 - 400n + 1360n - 40n^2 = 13600 + 960n - 40n^2

R'(n) = 960 - 80n = 0 \implies n = 12

R''(n) = -80 &lt; 0 \implies \text{maximum}

Total passengers

= 100 + 10(12) = 220

.

Answer: 220 passengers will maximise the revenue.

7If price

p

per unit of an article is

p = 75 - 2x

and the cost function is

C(x) = 350 + 12x + \frac{x^2}{4}

. Find the number of units and the price at which the total profit is maximum. What is the maximum profit?Show solution

Given:

p = 75 - 2x

C(x) = 350 + 12x + \dfrac{x^2}{4}

.

Working:

R(x) = px = (75-2x)x = 75x - 2x^2

P(x) = R(x) - C(x) = 75x - 2x^2 - 350 - 12x - \frac{x^2}{4} = -\frac{9x^2}{4} + 63x - 350

P'(x) = -\frac{9x}{2} + 63 = 0 \implies x = 14

P''(x) = -\frac{9}{2} &lt; 0 \implies \text{maximum at } x = 14

Price

p = 75 - 2(14) = 75 - 28 = ₹47

.

Maximum profit

= P(14) = -\dfrac{9(196)}{4} + 63(14) - 350 = -441 + 882 - 350 = ₹91

.

Answer: 14 units at price ₹47; maximum profit

= ₹91

8The cost of fuel in running an engine is proportional to the square of the speed in km/hr, and is ₹48 per hour when the speed is 16 km. Other costs amount to ₹300 per hour. Find the most economical speed.Show solution

Given: Fuel cost

= k v^2

per hour. At

v = 16

k(16)^2 = 48 \Rightarrow k = \dfrac{48}{256} = \dfrac{3}{16}

.

Other costs

= ₹300

per hour.

Working:

Total cost per hour

= \dfrac{3}{16}v^2 + 300

.

Total cost per km

= \dfrac{\text{cost per hour}}{\text{speed}} = \dfrac{\frac{3}{16}v^2 + 300}{v} = \frac{3v}{16} + \frac{300}{v}

Let

C(v) = \dfrac{3v}{16} + \dfrac{300}{v}

C'(v) = \frac{3}{16} - \frac{300}{v^2} = 0 \implies v^2 = \frac{300\times 16}{3} = 1600 \implies v = 40 \text{ km/hr}

C''(v) = \frac{600}{v^3} &gt; 0 \implies \text{minimum}

Answer: The most economical speed is

\boxed{40}

km/hr.

Case Study-I

1If

x

more trees, in excess of 25, are grown, then the number of fruits produced per tree is:Show solution

Analysis: With 25 trees, each tree gives 600 fruits. For 26 trees: total = 15210, so per tree = 15210/26 = 585 = 600 - 15. For 27 trees: 15390/27 = 570 = 600 - 30. The reduction per tree is 15 per extra tree.

So with

x

extra trees: fruits per tree

= 600 - 15x

.

Correct option: (i) $600 - 15x$

2The production of entire garden if

x

more trees, in excess of 25, are planted is:Show solution

Analysis: Total trees

= 25 + x

. Fruits per tree

= 600 - 15x

.

Total production

= (25+x)(600-15x)

.

Correct option: (iii) $(25+x)(600-15x)$

3The marginal production of the garden when

x

more trees, in excess of 25, are planted is:Show solution

Working:

P(x) = (25+x)(600-15x) = 15000 - 375x + 600x - 15x^2 = 15000 + 225x - 15x^2

P'(x) = 225 - 30x

Correct option: (ii) $225 - 30x$

4The critical point of producing

x

more units of trees is:Show solution

Working: Setting

P'(x) = 0

225 - 30x = 0 \implies x = 7.5

Correct option: (iii) $7.5$

5The number of trees to be grown to get maximum production is:Show solution

Working:

x = 7.5

is not an integer. Check

x = 7

and

x = 8

P(7) = (32)(600-105) = 32\times 495 = 15840

P(8) = (33)(600-120) = 33\times 480 = 15840

Both give equal maximum production. Total trees

= 25+7 = 32

25+8 = 33

.

Correct option: (ii) 32 or 33 trees

Case Study-II

1Which of the following is the fixed cost?Show solution

Analysis: Fixed costs remain constant regardless of production level. From the table, Property tax (₹5000) and Salaries (₹20000) remain constant. Among the options given, Salaries is fixed.

Correct option: (iv) Salaries

2Total cost

C(x)

of toys for

x

units of production is:Show solution

Analysis from table:
- Raw material: ₹8 per toy (800/100)
- Production supply: ₹20 per toy (2000/100)
- Freight: ₹10 per toy (1000/100)
- Variable cost per unit

= 8+20+10 = 38

- Fixed costs

= 5000 + 20000 = 25000

C(x) = 38x + 25000

Correct option: (iii) $C(x) = 38x + 25000$

3If the price

p

per unit of item sold is

p = 5000 - 10x

, then the revenue function

R(x)

is given by:Show solution

Working:

R(x) = p\cdot x = (5000-10x)x = 5000x - 10x^2

Correct option: (i) $R(x) = 5000x - 10x^2$

4The Marginal revenue (MR) of the company is given by:Show solution

Working:

MR = \frac{dR}{dx} = \frac{d}{dx}(5000x - 10x^2) = 5000 - 20x

Correct option: (i) $5000 - 20x$

5If the profit function

P(x) = R(x) - C(x)

, then it is given by:Show solution

Working:

P(x) = (5000x - 10x^2) - (38x + 25000)

= 5000x - 10x^2 - 38x - 25000

= -10x^2 + 4962x - 25000

Correct option: (iv) $-10x^2 + 4962x - 25000$

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Exercise 3.1

Exercise 3.2

Exercise 3.3

Exercise 3.4

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Case Study-I

Case Study-II

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