Chapter 3 of 10

Updated 5 October 2026

NCERT Solutions

Differentiation and its Applications — NCERT Solutions

CBSE · Class 12 · Applied Mathematics

NCERT Solutions for Differentiation and its Applications, CBSE Class 12 Applied Mathematics: 79 textbook questions solved step by step.

48 questions20 flashcards7 formulas & key relations5 concepts

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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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:

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

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

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

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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:

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

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

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

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

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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:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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4(i)Find

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

from

y = x\log x

.Show solution

Given: $y = x\log x$

Working:

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

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

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

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4(ii)Find

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

from

y = x^2 e^x

.Show solution

Given: $y = x^2 e^x$

Working:

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

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

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

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4(iii)Find

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

from

y = \log(\log x)

.Show solution

Given: $y = \log(\log x)$

Working:

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

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

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

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

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

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4(iv)Find

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

from

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

.Show solution

Given: $y = 3e^{2x} + 2e^{3x}$

Working:

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

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

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

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

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

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

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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:

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

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

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

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 $(a^2+x^2)y_2 + xy_1 = 0$ . $\blacksquare$

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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:

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

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

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

So $y_1\sqrt{x^2+1} = py$ . 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 $y_1 \neq 0$ ):
$y_2(x^2+1) + xy_1 = p^2 y$

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

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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 $\boxed{2\pi}$ .

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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:
$\frac{dS}{dx} = 8x$

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

Answer: $\boxed{32 \text{ cm}^2/\text{cm}}$

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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.

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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.

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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: $x = \dfrac{1}{3}$ or $x = 3$ .

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

At $r=7$ , $h=24$ : $l = \sqrt{49 + 576} = \sqrt{625} = 25$ 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.

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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 $= 10$ m, $\dfrac{dx}{dt} = 2$ m/min, $x = 6$ m.

Working:

Let $x$ = distance of lower end from wall, $y$ = height of upper end.

$x^2 + y^2 = 100$

At $x = 6$ : $y = \sqrt{100 - 36} = 8$ 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).

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

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

At $x = 5$ : $(10+5)^2 = 225$

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

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

Answer: The thickness of ice decreases at the rate of $\dfrac{1}{18\pi}$ cm/min.

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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.

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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: $p = 11 - \dfrac{x}{200}$ ; $R(x) = 11x - \dfrac{x^2}{200}$ ; $MR = 11 - \dfrac{x}{100}$ .

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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:

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

At $x = 1$ : slope of tangent $= 3(1)^2 - 1 = 2$ .

Slope of normal $= -\dfrac{1}{2}$ .

Answer: Slope of tangent $= 2$ ; slope of normal $= -\dfrac{1}{2}$ .

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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:

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

At $x = 2$ : slope of tangent $= 6(2) - 6 = 6$ .

Slope of normal $= -\dfrac{1}{6}$ .

Answer: Slope of tangent $= 6$ ; slope of normal $= -\dfrac{1}{6}$ .

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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:

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

At $x = 10$ : 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 $= 64$ .

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

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

At $(1,1)$ : slope of tangent $= -1$ .

Slope of normal $= 1$ .

Answer: Slope of tangent $= -1$ ; slope of normal $= 1$ .

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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:

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

At $x = 2$ : 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: $x + 9y - 65 = 0$ .

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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:

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

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

At $t = 2$ : 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: $2x + y - 12a = 0$ .

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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 $\dfrac{dy}{dx} = 0$ : $x = 2$ or $x = 3$ .

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 $y = 6$ .

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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:

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

At $x = 2$ : $y = 8 + 4 - 4 = 8$ . 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 $14x - y + 12 = 0$ .

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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 $(7, 0)$ .

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

At $x = 7$ : 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: $20x + y - 140 = 0$ .

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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: $x + y - 3 = 0$ .

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7For the curve

y = x^2 + 3x + 4

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

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):
$h^2 + 3h + 4 = 2h^2 + 3h \implies h^2 = 4 \implies h = \pm 2$

At $h = 2$ : $k = 4 + 6 + 4 = 14$ . Point $(2,14)$ .

At $h = -2$ : $k = 4 - 6 + 4 = 2$ . Point $(-2, 2)$ .

Answer: The points are $(2, 14)$ and $(-2, 2)$ .

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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.

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

At $x = 0$ : 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 $(0,b)$ . $\blacksquare$

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9Show that the curves

xy = a^2

and

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

touch each other.

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 $C_2$ :
$x^2 + \frac{a^4}{x^2} = 2a^2 \implies x^4 - 2a^2x^2 + a^4 = 0 \implies (x^2-a^2)^2 = 0$
$x = \pm a$

At $x = a$ : $y = a$ . 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 $(-1)$ at $(a,a)$ and pass through the same point. Hence they touch each other. $\blacksquare$

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10Prove that the curves

xy = 4

and

x^2 + y^2 = 8

touch each other.

Given: $C_1: xy = 4$ and $C_2: x^2 + y^2 = 8$ .

Working:

From $C_1$ : $y = 4/x$ . Substituting in $C_2$ :
$x^2 + \frac{16}{x^2} = 8 \implies x^4 - 8x^2 + 16 = 0 \implies (x^2-4)^2 = 0$
$x = \pm 2$

At $x = 2$ : $y = 2$ . 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. $\blacksquare$

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Exercise 3.4

1(i)Find critical points of

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

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1(ii)Find critical points of

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

x > 0

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1(iii)Find critical points of

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

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1(iv)Find critical points of

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

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 $e^{-x^3} > 0$ 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 $x = \dfrac{1}{3^{1/3}}$ .

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2(i)Find the intervals in which

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

is increasing or decreasing.

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:

$x < 1$ : $f'(x) < 0$ (decreasing)
$1 < x < 2$ : $f'(x) > 0$ (increasing)
$2 < x < 3$ : $f'(x) < 0$ (decreasing)
$x > 3$ : $f'(x) > 0$ (increasing)

Answer: Increasing on $(1,2)$ and $(3,\infty)$ ; Decreasing on $(-\infty,1)$ and $(2,3)$ .

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2(ii)Find the intervals in which

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

is increasing or decreasing.

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., $x = 3$ or $x = -2$ ) or $2x-1 = 0$ (i.e., $x = 1/2$ ).

Sign of $f'(x)$ depends on $(2x-1)$ :

$x < \dfrac{1}{2}$ : $2x-1 < 0$ , so $f'(x) \leq 0$ (decreasing)
$x > \dfrac{1}{2}$ : $2x-1 > 0$ , so $f'(x) \geq 0$ (increasing)

Answer: Increasing on $\left(\dfrac{1}{2}, \infty\right)$ ; Decreasing on $\left(-\infty, \dfrac{1}{2}\right)$ .

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2(iii)Find the intervals in which

f(x) = x^2 e^x

is increasing or decreasing.

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 $e^x > 0$ always, sign depends on $x(x+2)$ :

Critical points: $x = 0$ and $x = -2$ .

$x < -2$ : $x < 0$ , $x+2 < 0$ , product $> 0$ : increasing
$-2 < x < 0$ : $x < 0$ , $x+2 > 0$ , product $< 0$ : decreasing
$x > 0$ : $x > 0$ , $x+2 > 0$ , product $> 0$ : increasing

Answer: Increasing on $(-\infty,-2)$ and $(0,\infty)$ ; Decreasing on $(-2,0)$ .

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3Show that the function

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

increases on

(0,\infty)

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4Prove that the function

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

is neither increasing nor decreasing in

(0,1)

Given: $f(x) = x^2 - x + 1$

Working:

$f'(x) = 2x - 1$

In $(0,1)$ :

At $x = \dfrac{1}{2}$ : $f'(x) = 0$
For $0 < x < \dfrac{1}{2}$ : $f'(x) < 0$ (decreasing)
For $\dfrac{1}{2} < x < 1$ : $f'(x) > 0$ (increasing)

Since $f'(x)$ changes sign in $(0,1)$ , the function is neither entirely increasing nor entirely decreasing in $(0,1)$ . $\blacksquare$

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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.

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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.

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:

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

$\dfrac{dR}{dp} > 0$ when $p < 1$ or $p > 3$ : Revenue is increasing.

$\dfrac{dR}{dp} < 0$ when $1 < p < 3$ : Revenue is decreasing.

Answer: Revenue increases for $p \in (-\infty,1)\cup(3,\infty)$ and decreases for $p \in (1,3)$ .

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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.

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

$\frac{d(MC)}{dx} = \frac{2(-6)}{(x+3)^3} = \frac{-12}{(x+3)^3} < 0 \text{ for all } x > 0$

Hence MC falls continuously as $x$ increases. $\blacksquare$

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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.

Given: $p = 29 - x$ , $C(x) = 45 + 11x$ .

Working:

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

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

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

$P'(x) > 0$ when $x < 9$ : Profit is increasing.

$P'(x) < 0$ when $x > 9$ : Profit is decreasing.

Answer: Profit increases for $x \in (0,9)$ and decreases for $x \in (9,\infty)$ .

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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

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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 > 0

Given: $f(x) = \dfrac{\log x}{x}$ , $x > 0$

Working:

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

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

At $x = e$ : $f''(e) = \dfrac{2(1)-3}{e^3} = \dfrac{-1}{e^3} < 0$ .

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 $= \dfrac{1}{e}$ .

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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

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 $e^x > 0$ ):
$-x^2 - 2x + 1 = 0 \implies x^2 + 2x - 1 = 0 \implies x = \frac{-2\pm\sqrt{4+4}}{2} = -1\pm\sqrt{2}$

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

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

$(-1+\sqrt{2})^2 = 3-2\sqrt{2}$ ; $-4(-1+\sqrt{2}) = 4-4\sqrt{2}$

$= e^{-1+\sqrt{2}}(-(3-2\sqrt{2}) + 4 - 4\sqrt{2} - 1) = e^{-1+\sqrt{2}}(-3+2\sqrt{2}+3-4\sqrt{2}) = e^{-1+\sqrt{2}}(-2\sqrt{2}) < 0$

So $x = -1+\sqrt{2}$ is a local maximum.

At $x = -1-\sqrt{2}$ : $f''(-1-\sqrt{2}) = e^{-1-\sqrt{2}}(2\sqrt{2}) > 0$ : local minimum.

Answer: Local maximum at $x = -1+\sqrt{2}$ ; local minimum at $x = -1-\sqrt{2}$ .

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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}

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 $+$ to $-$ (check: for $x<4$ , $(x-4)<0,(x-16)<0$ , product $>0$ ; for $4<x<10$ , $(x-4)>0,(x-16)<0$ , product $<0$ ). 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 $-$ to $+$ . So local minimum at $x=16$ .

Local min value $= f(16) = \dfrac{256-112+6}{6} = \dfrac{150}{6} = 25$ .

Answer: Local maximum value $= 1$ at $x = 4$ ; local minimum value $= 25$ at $x = 16$ .

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

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

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

Working:

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

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

At $x = \dfrac{1}{2}$ : $f''\left(\dfrac{1}{2}\right) = 8 > 0$ : local minimum.

Local min value $= 2\cdot\dfrac{1}{2} + \dfrac{1}{2\cdot\frac{1}{2}} = 1 + 1 = 2$ .

At $x = -\dfrac{1}{2}$ : $f''\left(-\dfrac{1}{2}\right) = -8 < 0$ : 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 $= 2$ at $x = \dfrac{1}{2}$ ; local maximum value $= -2$ at $x = -\dfrac{1}{2}$ .

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

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 $Q > 0$ ).

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

$Q'' = -2 < 0$ : maximum.

So $x = 8$ , $y = 8$ .

Answer: Both numbers are $8$ .

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

Given: Perimeter $= 2(l+b) = 2P$ (constant), so $l + b = P$ .

Working:

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

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

$\frac{d^2A}{dl^2} = -2 < 0 \implies \text{maximum}$

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. $\blacksquare$

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4Show that the function

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

has neither a local maximum nor a local minimum value.

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

Working:

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

$f'(x) = 0$ at $x = 2$ , 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. $\blacksquare$

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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.

Given: $P(x) = (400-x)x - 3500 = 400x - x^2 - 3500$

Working:

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

$P''(x) = -2 < 0 \implies \text{maximum at } x = 200$

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.

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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.

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

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

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

$R''(n) = -80 < 0 \implies \text{maximum}$

Total passengers $= 100 + 10(12) = 220$ .

Answer: 220 passengers will maximise the revenue.

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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?

Given: $p = 75 - 2x$ , $C(x) = 350 + 12x + \dfrac{x^2}{4}$ .

Working:

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

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

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

$P''(x) = -\frac{9}{2} < 0 \implies \text{maximum at } x = 14$

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 $= ₹91$ .

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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.

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

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

$C''(v) = \frac{600}{v^3} > 0 \implies \text{minimum}$

Answer: The most economical speed is $\boxed{40}$ km/hr.

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Case Study-I

1If

x

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

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2The production of entire garden if

x

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

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3The marginal production of the garden when

x

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

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4The critical point of producing

x

more units of trees is:

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

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Case Study-II

1Which of the following is the fixed cost?

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2Total cost

C(x)

of toys for

x

units of production is:

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3If the price

p

per unit of item sold is

p = 5000 - 10x

, then the revenue function

R(x)

is given by:

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

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5If the profit function

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

, then it is given by:

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