Application of Derivatives

Given that, at any instant of time Rate of change in volume w.r.t. time $=$ rate of change in surface area w.r.t. time

i.e.,

$ \begin{equation*} \frac{d v}{d t}=\frac{d s}{d t}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) \end{equation*} $

Volume of sphere of radius $(r), V=\frac{4}{3} \pi r^3$

Differentiating w.r.t. ' $t$ ', we get

$ \begin{align*} & \frac{d v}{d t}=\frac{d}{d t}\left(\frac{4}{3} \pi r^3\right)=\frac{4}{3} \pi \frac{d}{d t}\left(r^3\right) \\ & \frac{d v}{d t}=\frac{4}{3} \pi\left(3 r^2\right) \frac{d r}{d t} \text { or } \frac{d v}{d t}=4 \pi r^2 \frac{d r}{d t} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii) \end{align*} $

Surface area of sphere of radius $(r)$,

$ S=4 \pi r^2 $

Differentiating w.r.t. ' $t$ ', we get

$ \begin{align*} \frac{d s}{d t} & =\frac{d}{d t}\left(4 \pi r^2\right)=4 \pi \frac{d}{d t}\left(r^2\right) \\ \Rightarrow \quad \frac{d s}{d t} & =8 \pi r \frac{d r}{d t} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii) \end{align*} $

Putting the values from Eqs. (ii) and (iii) in Eq. (i), we get

$ 4 \pi r^2 \frac{d r}{d t}=8 \pi r \frac{d r}{d t} \text { or } r=2 $

Given that,

Rate of increase in, $\frac{d V}{d t}=4 \pi \mathrm{cc} / \mathrm{sec}$

Volume, $V=288 \pi \mathrm{cc}$

As we know, volume of a sphere of radius $(r)$,

$ \begin{equation*} V=\frac{4}{3} \pi r^3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) \end{equation*} $

Differentiating above expression w.r.t. ' $t^{\prime}$ on both sides, we get

$ \begin{align*} & \frac{d V}{d t}=\frac{d}{d t}\left(\frac{4}{3} \pi r^3\right)=\frac{4}{3} \pi 3 r^2 \frac{d r}{d t} \\ & \frac{d V}{d t}=4 \pi r^2 \frac{d r}{d t} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii) \end{align*} $

From Eq. (i) $(288 \pi)=\frac{4}{3} \pi r^3$

or  $ r=6 \mathrm{~cm} $

Putting this value in Eq. (ii), we get

$ 4 \pi=4 \pi(6)^2 \frac{d r}{d t} \text { or } \frac{d r}{d t}=\frac{1}{36} $

Given function,

$ f(x)=x-\log \left(\frac{1+x}{x}\right), x>0 $

Differentiating w.r.t. $x$, we get

$ \begin{aligned} f^{\prime}(x) & =\frac{d}{d x}(x)-\frac{d}{d x} \log \left(\frac{1+x}{x}\right) \\ & =1-\frac{1}{\frac{1+x}{x}} \cdot \frac{d}{d x}\left(\frac{1+x}{x}\right) \end{aligned} $

[from chain rule]

$ \begin{aligned} & =1-\frac{x}{1+x} \cdot\left\{\frac{x \frac{d}{d x}(1+x)-(1+x) \frac{d}{d x}(x)}{x^2}\right\} \\ & =1-\frac{x}{1+x}\left\{\frac{x-(1+x)}{x^2}\right\} \\ & =1-\frac{x}{1+x}\left\{-\frac{1}{x^2}\right\} \\ & f^{\prime}(x)=1+\frac{1}{x(x+1)} \end{aligned} $

Since $x>0$, therefore there is no possibility of $f^{\prime}(x)$ to be zero at any value of $x$.

So, for the given function $f^{\prime}(x) \neq 0$ Also, for $\forall x \in \mathbf{R} f^{\prime}(x)>0$ i.e., function $f(x)$ is strictly increasing its defined domain.

So, reason correctly explains given assertion.

We have,

$ x^5-8 x^4+25 x^3-38 x^2+28 x-8=0 $

Let

$ \begin{aligned} f(x)= & x^5-8 x^4+25 x^3-38 x^2+28 x-8 \\ & f(2)=32-128+200-152+56-8 \\ & f(2)=0 \\ \Rightarrow & f^{\prime}(x)=5 x^4-32 x^3+75 x^2-76 x+28 \\ & f^{\prime}(2)=80-256+300-152+28 \\ \Rightarrow & f^{\prime}(2)=0 \\ & f^{\prime \prime}(x)=20 x^3-96 x^2+150 x-76 \\ & f^{\prime \prime}(2)=160-384+300-76 \\ \Rightarrow & f^{\prime \prime}(2)=0 \\ & f^{\prime \prime \prime}(x)=60 x^2-192 x+150 \\ & f^{\prime \prime \prime}(2)=240-384+150=6 \\ \Rightarrow & f^{\prime \prime \prime}(2) \neq 0 \end{aligned} $

∴ Given, $\alpha$ is a root of multiplicity 3

$ \begin{array}{lrl} \therefore & \alpha & =2 \\ \Rightarrow & \alpha^2-5 \alpha+6 & =4-10+6=0 \end{array} $

Given, $d \theta=9 \mathrm{~min}$

In radian $d \theta=\frac{9}{60} \times \frac{\pi}{180}=\frac{\pi}{1200}$

$ \begin{aligned} A & =\frac{1}{2} b c \sin \theta \Rightarrow \frac{d A}{d \theta}=\frac{1}{2} b c \cos \theta \\ \Rightarrow \quad d A & =\frac{1}{2} b c \cos \theta d \theta \\ \frac{d A}{A} \times 100 & =\frac{\frac{1}{2} b c \cos \theta d \theta}{\frac{1}{2} b c \sin \theta} \times 100 \\ & =\cot \theta d \theta \times 100 \\ & =\cot \frac{3 \pi}{8} \times \frac{\pi}{1200} \times 100 \\&\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, {\left[\because \theta=67^{\circ}=\frac{3 \pi}{8}\right] } & \\ & =(\sqrt{2}-1) \frac{\pi}{12}\left[\because \cos \frac{3 \pi}{8}=\sqrt{2}-1\right] \end{aligned} $

Given, $y=f(x)$

Equation of tangent of curve $y=f(x)$ at $P(\alpha, 1)$ is

$ y-1=f(\alpha)(x-\alpha) $

Equation of normal of curve at $P(\alpha, 1)$ is

$ y-1=-\frac{1}{f^{\prime}(\alpha)}(x-a) $

∴ Tangent cut of $X$-axis at $A$

$ \therefore \quad A=\left(\alpha-\frac{1}{f^{\prime} \alpha}, 0\right) $

∴ Normal cut of $X$-axis at $B$

$ \therefore \quad B=\left(\alpha+f^{\prime}(\alpha), 0\right) $

Point $C=(\alpha, 0)$

$ \begin{aligned} & A C+B C=\alpha-\alpha+\frac{1}{f^{\prime}(\alpha)}+\alpha+f^{\prime}(\alpha)-\alpha \\ & A C-B C=\frac{1}{f^{\prime}(\alpha)}+f^{\prime}(\alpha) \end{aligned} $

$A C+B C$ is minimum

When, $f^{\prime}(\alpha)=1$

∴ Equation of tangent of curve

$ y-1=1(x-\alpha) \Rightarrow x-y=\alpha+1 $

∴ Equation of tangent of curve is parallel to $x-y=0$.

$ \text { } \begin{aligned}Given,\,\, y & =3 x^5+15 x-8 \\ \frac{d x}{d t} & =\frac{1}{5} \mathrm{unit} / \mathrm{sec} \\ \Rightarrow \quad \frac{d y}{d t} & =6 \mathrm{unit} / \mathrm{sec} \\ \frac{d y}{d t} & =\left(15 x^4+15\right) \frac{d x}{d t} \end{aligned} $

$ \Rightarrow \quad \,\,\,\,\,\,\,6=15\left(x^4+1\right) \times \frac{1}{5} $

$ \begin{array}{lc} \Rightarrow \mathrm{M} x^4+1=2 \\ \Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, x^4=1 \\ \Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\quad x= \pm 1 \\ \Rightarrow\,\,\,\,\,\,\,\,\,\,\, y(1)=3+15-8=10 \\\,\,\,\,\,\,\,\,\,\,\,\,\,\, y(-1)=-3-15-8=-26 \\ \therefore A(1,10), B(-1,-26) \\ \text { Slope of } A B=\frac{-26-10}{-1-1}=\frac{-36}{-2}=18 \end{array} $

$ \text { Given, } a+b \text { is constant } $

Let    $ a+b=k $

Area of $\triangle A B C$ say $A$

$ \begin{aligned} A & =\frac{1}{2} \times a \times \sqrt{b^2-a^2} \\ \Rightarrow \quad A^2 & =\frac{1}{4} a^2\left(b^2-a^2\right) \\ A^2 & =\frac{a^2}{4}\left((k-a)^2-c^2\right) \\ & =\frac{k^2 a^2-2 k a^3}{4} \end{aligned} $

Differentiating w.r.t. ' $a$ ',

$ 2 A \frac{d A}{d a}=\frac{2 k^2 a-6 k a^2}{4} $

For maximum or minimum $\frac{d A}{d a}=0$

$ \begin{aligned} k^2 a-3 k a^2 & =0 \Rightarrow a=k / 3 \\ \Rightarrow \quad \frac{d^2 A}{d a^2} & =\frac{-k^2}{4 A}<0 \end{aligned} $

$A^{\prime}$ s maximum value, $a=k / 3$

Now, $a=\frac{k}{3}, b=2 \frac{k}{3}$

$ \Rightarrow \quad \cos \theta=\frac{a}{b}=\frac{1}{2} \Rightarrow \theta=\frac{\pi}{3} $

$ \begin{array}{cc} \text { } \text { Given curve, } x^2-a^2=\frac{x^2 y^2}{a^2} \\ \Rightarrow 2 x=\frac{1}{a^2}\left[x^2(2 y) \frac{d y}{d x}+(2 x) y^2\right] \\ \Rightarrow 1=\frac{1}{a^2}\left[x y \frac{d y}{d x}+y^2\right] \end{array} $

$ \begin{aligned} & \Rightarrow \quad x y \frac{d y}{d x}+y^2=a^2 \Rightarrow \quad x y \frac{d y}{d x}=a^2-y^2 \\ & \Rightarrow \quad \frac{d y}{d x}=\frac{a^2-y^2}{x y} \ldots \\ & \Rightarrow \quad\left(\frac{d y}{d x}\right)_{P(\alpha, y)} \quad=\frac{a^2-y^2}{\alpha y} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

$ \begin{aligned} &\text { Now, Since } P(\alpha, y) \text { lies on the curve. }\\ &\begin{aligned} & \therefore \alpha^2-a^2=\frac{\alpha^2 y^2}{a^2} \\ & \Rightarrow \quad y^2=\frac{a^2}{\alpha^2}\left(\alpha^2-a^2\right) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} \end{aligned} $

∴ Length of subnormal at $P(\alpha, y)$ is

$ y_1\left(\frac{d y}{d x}\right)_{\left(x_1, y_1\right)}=y\left(\frac{a^2-y^2}{\alpha y}\right) $

(using Eq. (i))

$ \begin{aligned} & \quad=\frac{a^2-\left(\frac{a^2}{\alpha 2}\left(\alpha^2-a^2\right)\right)}{\alpha} \\ & \quad(\text { using Eq. }(\mathrm{ii})) \\ & \quad=\quad \frac{a^4}{\alpha^3}=\frac{k}{\alpha^3} \quad \text { (given) } \\ & \therefore \quad k=a^4 \end{aligned} $

Volume of rectangular parallelopiped is $27 \mathrm{~m}^3$

Let $A=$ Area of base and $h$ is height

$ \begin{aligned} & \therefore \quad V=A h \\ & \frac{d V}{d t}=A \frac{d h}{d t} \\ & \frac{d V}{d t}=A\left(\frac{3 d V}{d t}\right) \\ & \quad\left[\because \frac{d h}{d t}=\frac{3 d V}{d t}\right] \end{aligned} $

$ \begin{aligned} & A=1 / 3 \\ & \because \quad V=\frac{1}{3} h \\ & 27=\frac{h}{3} \Rightarrow h=81 \\ & \therefore \text { Height of tank }=81 \mathrm{~m} \end{aligned} $

$ \begin{aligned} & \text { } \begin{aligned} A.\,\,\, f(x) & =3 x^4-2 x^3-6 x^2+6 x+1 \\ f^{\prime}(x) & =12 x^3-6 x^2-12 x+6 \\ f^{\prime}(x) & =6\left(2 x^3-x^2-2 x+1\right) \\ f^{\prime}(x) & =6(x-1)(2 x-1)(x+1) \\ \text { Put, } \quad f^{\prime}(x) & =0 x=-1, \frac{1}{2}, 1 \end{aligned} \text { } \end{aligned} $

$f^{\prime}(x)$ is increasing in $\left(-1, \frac{1}{2}\right) \cup(1, \infty)$ decreasing value in $(\infty,-1) \cup(1 / 2,1)$ minimum value at $x=1$ or -1 maximum value of $x=\frac{1}{2}$

$ \begin{aligned} & \text { B. } \quad f(x)=x+\frac{1}{x} \\ & \quad f^{\prime}(x)=1-\frac{1}{x^2} \text { put } f^{\prime}(x)=0 \\ & \because \quad x^2=1 \Rightarrow x= \pm 1 \\ & \quad f^{\prime \prime}(x)=\frac{2}{x^3} \\ & \text { Maximum at } x=-1 \end{aligned} $

C. $f(x)=x^4(7-x)^3$

$ f^{\prime}(x)=4 x^3(7-x)^3-3 x^4(7-x)^2 $

Put

$ \begin{aligned} & f^{\prime}(x)=0 \Rightarrow x^3(7-x)^2(28-7 x)=0 \\ & \qquad x=4 \\ & \text { Maximum at } x=4 \end{aligned} $

D.

$ \begin{aligned} f(x) & =x^4+(8-x)^4 \\ f^{\prime}(x) & =4 x^3-4(8-x)^3, \text { put } \end{aligned} $

$ f^{\prime}(x)=0 $

$x^3=(8-x)^3 \Rightarrow x=8-x$

$2 x=8 \Rightarrow x=4$

$f^{\prime \prime}(x)=12 x^2+12(8-x)^2$

$f^{\prime \prime}(4) \geq 0, \quad \therefore$ Minimum of $x=4$

Let $A$ and $P$ are area and perimeter of circle.

We have,

$ \begin{aligned} \frac{d A}{d t} & =\frac{1}{\sqrt{\pi}} \Rightarrow \frac{d}{d t}\left(\pi r^2\right)=\frac{1}{\sqrt{\pi}} \\ \Rightarrow \quad 2 \pi r \frac{d r}{d t} & =\frac{1}{\sqrt{\pi}} \Rightarrow \frac{d r}{d t}=\frac{1}{2 \pi r \sqrt{\pi}} \end{aligned} $

Now, $P=2 \pi r$

$ \begin{aligned} & \Rightarrow \frac{d P}{d r}=2 \pi \times \frac{d r}{d t}=2 \pi \times \frac{1}{2 \pi r \sqrt{\pi}}=\frac{1}{r \sqrt{\pi}} \\ & =\frac{1}{\frac{1}{2 \sqrt{\pi}} \times \sqrt{\pi}} \\ & {\left[\because P=\sqrt{\pi} \Rightarrow 2 \pi r=\sqrt{\pi} \Rightarrow r=\frac{1}{2 \sqrt{\pi}}\right]} \\ & =2 \text { unit } / \text { sec. } \end{aligned} $

$ \begin{aligned} &\text { We have, }\\ &y=\frac{x^n}{a^{n-1}} \quad \therefore \frac{d y}{d x}=\frac{n x^{n-1}}{a^{n-1}} \end{aligned} $

$ \begin{aligned} \therefore \text { Length of subnormal } & =y \frac{d y}{d x} \\ & =\frac{x^n}{a^{n-1}} \times n \frac{x^{n-1}}{a^{n-1}} \\ & =n \frac{x^{2 n-1}}{a^{2 n-2}} \end{aligned} $

∴ Length of subnormal at

$ (\alpha, \beta)=\frac{n \alpha}{a^{2 n-2}} $

Now, it is given that

Length of subnormal is proportional to $a^2$

$ \therefore \quad 2 n-1=2 \Rightarrow n=\frac{3}{2} $

Let $P(x)=a x^3+b x^2+c x+d$

$ \therefore \quad P^{\prime}(x)=3 a x^2+2 b x+c $

Since, $P(x)$ has extreme value at $x=1$

$ \therefore \quad P^{\prime}(1)=0 \quad \Rightarrow \quad 3 a+2 b+c=0 $

Now, we have

$ \begin{gathered} \lim _{x \rightarrow 0}\left(\frac{P(x)+4}{x^2}+2\right)=6 \\ \Rightarrow \quad \lim _{x \rightarrow 0} \frac{a x^3+b x^2+c x+d+4}{x^2}=4 \\ \Rightarrow \quad \lim _{x \rightarrow 0} a x+b+\frac{c}{x}+\frac{d+4}{x^2}=4 \end{gathered} $

Since, value of limit is finite

$\therefore \quad c=0$ and $d+4=0 \Rightarrow c=0$,

$d=-4$

$ \therefore \quad \mathop {\lim }\limits_{x \to 0} a x+b=4 \Rightarrow b=4 $

$ \begin{aligned} &\text { Putting } b=4, c=0 \text { in Eq. (i), we get }\\ &\begin{gathered} 3 a+8=0 \\ \Rightarrow \quad a=\frac{-8}{3} \\ \therefore \quad P(x)=\frac{-8}{3} x^3+4 x^2-4 \\ \Rightarrow \quad \frac{d P(x)}{d x}=-8 x^2+8 x \\ \left.\therefore \frac{d P(x)}{d x}\right|_{x=1 / 2}=\frac{-8}{4}+\frac{8}{2}=2 \end{gathered} \end{aligned} $

To determine the absolute maximum (M) and absolute minimum (m) of the function $ f(x) = 2x^3 - 9x^2 + 12x + 5 $ over the interval $[0, 3]$, we need to examine its critical points and endpoints.

First, we find the derivative of the function, $ f'(x) $, to locate the critical points:

$ f'(x) = \frac{d}{dx} (2x^3 - 9x^2 + 12x + 5) = 6x^2 - 18x + 12 $

Next, we set the derivative equal to zero to find the critical points:

$ 6x^2 - 18x + 12 = 0 $

Simplifying this equation by dividing by 6:

$ x^2 - 3x + 2 = 0 $

We solve this quadratic equation using the factorization method:

$ (x - 1)(x - 2) = 0 $

So, the critical points are:

$ x = 1 \quad \text{and} \quad x = 2 $

We now evaluate the function $ f(x) $ at the critical points and at the endpoints of the interval [0, 3]:

1. At $ x = 0 $:

$ f(0) = 2(0)^3 - 9(0)^2 + 12(0) + 5 = 5 $

2. At $ x = 1 $:

$ f(1) = 2(1)^3 - 9(1)^2 + 12(1) + 5 = 2 - 9 + 12 + 5 = 10 $

3. At $ x = 2 $:

$ f(2) = 2(2)^3 - 9(2)^2 + 12(2) + 5 = 16 - 36 + 24 + 5 = 9 $

4. At $ x = 3 $:

$ f(3) = 2(3)^3 - 9(3)^2 + 12(3) + 5 = 54 - 81 + 36 + 5 = 14 $

We now identify the absolute maximum (M) and absolute minimum (m) from the above values:

• Maximum value $ M = 14 $ at $ x = 3 $
• Minimum value $ m = 5 $ at $ x = 0 $

Thus, the difference $ M - m $ is:

$ M - m = 14 - 5 = 9 $

Therefore, the correct answer is:

Option B: 9