Application of Derivatives

Given,

$\alpha = \sum\limits_{k = 1}^\alpha {{{\sin }^{2k}}\left( {{\pi \over 6}} \right)} $

$ = {\sin ^2}\left( {{\pi \over 6}} \right) + {\sin ^4}\left( {{\pi \over 6}} \right) + {\sin ^6}\left( {{\pi \over 6}} \right)\, + \,......$

This is a infinite G.P.

$\therefore$ $\alpha = {{{{\sin }^2}\left( {{\pi \over 6}} \right)} \over {1 - {{\sin }^2}\left( {{\pi \over 6}} \right)}}$

$ = {{{1 \over 4}} \over {1 - {1 \over 4}}}$

$ = {1 \over 4} \times {4 \over 3}$

$ = {1 \over 3}$

$\therefore$ $g(x) = {2^{{x \over 3}}} + {2^{{1 \over 3}(1 - x)}}$

Given $x \in [0,1]$

We know,

$AM \ge GM$

For ${2^{{x \over 3}}}$ and ${2^{{1 \over 3}(1 - x)}}$ both are positive in $x \in [0,1]$

${{{x^{{x \over 3}}} + {2^{{1 \over 3}(1 - x)}}} \over 2} \ge {\left( {{2^{{x \over 3}}}\,.\,{2^{{1 \over 3}(1 - x)}}} \right)^{{1 \over 2}}}$

$ \Rightarrow {{{2^{{x \over 3}}} + {2^{{1 \over 3}(1 - x)}}} \over 2} \ge {\left( {{2^{{1 \over 3}}}} \right)^{{1 \over 2}}}$

$ \Rightarrow {2^{{x \over 3}}} + {2^{{1 \over 3}(1 - x)}} \ge 2\,.\,{2^{{1 \over 6}}}$

$ \Rightarrow g(x) \ge {2^{{7 \over 6}}}$

We know, $AM = GM$ when terms are equal.

$\therefore$ $g(x) = {2^{{7 \over 6}}}$ when

${2^{{x \over 3}}} = {2^{{1 \over 3}(1 - x)}}$

$ \Rightarrow {x \over 3} = {{1 - x} \over 3}$

$ \Rightarrow 2x = 1$

$ \Rightarrow x = {1 \over 2}$

$\therefore$ At $x = {1 \over 2}$, ${\left[ {g(x)} \right]_{\min }} = {2^{{7 \over 6}}}$

$\therefore$ Option (A) is correct.

And option (D) is wrong as at only at a single point $x = {1 \over 2},\,g(x)$ is minimum.

Now, $g(x) = {2^{{x \over 3}}} + {2^{{1 \over 3}(1 - x)}}$

$ = {2^{{x \over 3}}} + {{{2^{{1 \over 3}}}} \over {{2^{{x \over 3}}}}}$

$ = {2^{{x \over 3}}} + {2^{{1 \over 3}}}\,.\,{2^{ - {x \over 3}}}$

$\therefore$ $g'(x) = {2^{{x \over 3}}}\,.\,\log 2\,.\,{1 \over 3} + {2^{{1 \over 3}}}\,.\,{2^{ - {x \over 3}}}\,.\,\log 2\,.\,\left( { - {1 \over 3}} \right)$

$ = {1 \over 3}\log 2\left[ {{2^{{x \over 3}}} - {{{2^{{1 \over 3}}}} \over {{2^{{x \over 3}}}}}} \right]$

$ = {1 \over 3}\log 2\left[ {{{{2^{{{2x} \over 3}}} + {2^{{1 \over 3}}}} \over {{2^{{x \over 3}}}}}} \right]$

We already found that at $x = {1 \over 2}$ $g(x)$ is minimum.

$\therefore$ $g'(x) = 0$ at $x = {1 \over 2}$

Now, $g'(x) > 0$ when

${1 \over 3}\log \left[ {{{{2^{{{2x} \over 3}}} - {2^{{1 \over 3}}}} \over {{2^{{x \over 3}}}}}} \right] > 0$

$ \Rightarrow {2^{{{2x} \over 3}}} > {2^{{1 \over 3}}}$

$ \Rightarrow {{2x} \over 3} > {1 \over 3}$

$ \Rightarrow x > {1 \over 2}$

Similarly, $g'(x) < 0$ when $x < {1 \over 2}$

If we put it in number line we get this

We know $g'(x)$ represent slope of curve $g(x)$ and it is negative when $x < {1 \over 2}$ and positive when $x > {1 \over 2}$ and zero when $x = {1 \over 2}$.

$\therefore$ Graph of $g(x)$ is

From graph you can see value of $g(x)$ is maximum either at $x = 0$ or $x = 1$ in the range $x \in [0,1]$.

$\therefore$ $g(0) = {2^0} + {2^{{1 \over 3}}} = 1 + {2^{{1 \over 3}}}$

$g(1) = {2^{{1 \over 3}}} + {2^0} = 1 + {2^{{1 \over 3}}}$

$\therefore$ We get maximum value at $x = 0$ and $x = 1$ both.

$\therefore$ B and C options are correct.

$x^2+y-7=4 x$

$ y=-x^2+4 x+7 $

On differentiating w.r.t. $x$, we get

$ \begin{aligned} & y^{\prime}=-2 x+4 \\ & m=-2 x+4 \end{aligned} $

At $(1,10)$

$ m=-2 \times 1+4=2 $

Equation of tangent,

$ \begin{aligned} & & y-10 & =2(x-1) \\ \Rightarrow & & y & =2 x+8 \end{aligned} $

Curve I $x^2-y^2=4$

On differentiating w.r.t. $x$, we get

$ \begin{aligned} & 2 x-2 y y^{\prime}=0 \Rightarrow x=y y^{\prime} \\ \Rightarrow & y^{\prime}=\frac{x}{y} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ \Rightarrow & m_1=\frac{x}{y} \end{aligned} $

Curve II $y^2=3 x$

On differentiating w.r.t. $x$, we get

$ \begin{aligned} & & 2 y y^{\prime} & =3 \times 1 \\ \Rightarrow & & y^{\prime} & =\frac{3}{2 y} \\ \Rightarrow & & m_2 & =\frac{3}{2 y} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

To get intersection point,

$ x^2-4=y^2, y^2=3 x $

On putting $x^2-4=3 x$,

$ x^2-3 x-4=0 $

$ \begin{aligned} &\begin{aligned} \Rightarrow & & x^2-4 x+x-4 & =0 \\ \Rightarrow & & x(x-4)+1(x-4) & =0 \\ \Rightarrow & & (x-4)(x+1) & =0 \\ & & x & =4,-1 \end{aligned}\\ &\text { From the graph it is clear that } x=4 \end{aligned} $

$ \therefore \quad y^2=3 \times 4 \Rightarrow y= \pm 2 \sqrt{3} $

At point $(4,2 \sqrt{3})$ from Eqs. (i) and (ii), we get

$ \begin{aligned} m_1 & =\frac{4}{2 \sqrt{3}}=\frac{2}{\sqrt{3}} \\ m_2 & =\frac{3}{2(2 \sqrt{3})}=\frac{\sqrt{3}}{4} \\ \tan \theta & =\left|\frac{m_1-m_2}{1+m_1 m_2}\right| \\ \tan \theta & =\left|\frac{\frac{2}{\sqrt{3}}-\frac{\sqrt{3}}{4}}{1+\frac{2}{\sqrt{3}} \cdot \frac{\sqrt{3}}{4}}\right| \\ & =\left|\frac{\frac{8-3}{4 \sqrt{3}}}{\frac{4 \sqrt{3}+2 \sqrt{3}}{4 \sqrt{3}}}\right|=\frac{5}{6 \sqrt{3}} \end{aligned} $

$f(x)=2 x^3-3 x^2-36 x+9$

$ \begin{aligned} & x \in[-3,3] \\ & f(x)=2 x^3-3 x^2-36 x+9 \end{aligned} $

On differentiating w.r.t. $x$, we get

$ f^{\prime}(x)=6 x^2-6 x-36 $

On putting $f^{\prime}(x)=0$ for critical numbers.

$ \begin{aligned} & \Rightarrow \quad 6 x^2-6 x-36=0 \\ & \Rightarrow \quad x^2-x-6=0 \\ & \Rightarrow \quad x^2-3 x+2 x-6=0 \\ & \Rightarrow \quad x(x-3)+2(x-3)=0 \\ & \Rightarrow \quad(x-3)(x+2)=0 \\ & \quad x=3,-2 \\ & f(3)=2 \times 27-3 \times 9-36 \times 3+9 \\ & =54-27-108+9=-72 \,\,\,\,\text { (min.) }\\ & f(-2)=2(-8)-3(4)-36(-2)+9 \\ & =-16-12+72+9=53 \,\,\,\text { (max.) }\\ & f(-3)=2(-27)-3(9)-36(-3)+9 \\ & =-54-27+108+9=36 \end{aligned} $

The absolute maximum value is 53 .

To find approximate value of

$ (28)^{1 / 3}=(27+1)^{1 / 3} $

Let $x=27, \Delta x=1, y=x^{1 / 3}$

$ \frac{d y}{d x}=\frac{1}{3} x^{-2 / 3}=\frac{1}{3}(27)^{-2 / 3}=\frac{1}{3} 3^{-2}=\frac{1}{27} $

Approximate value of $3 \sqrt{28}$

$ \begin{aligned} & =y+\frac{d y}{d x} \cdot \Delta x=27^{\frac{1}{3}}+\frac{1}{27} \times 1 \\ & =3+0.037=3.037 \end{aligned} $

Given, curve is $y=x^2$

If the curve is dragged along the positive $X$-axis to a distance of ' $a$ ' units, then new curve will be $y=(x-a)^2$

So, point of intersection

$ \begin{aligned} x^2 & =(x-a)^2 \\ \Rightarrow \quad x-a & = \pm x \Rightarrow x=a / 2 \end{aligned} $

At $x=\frac{a}{2}, y=\frac{a^2}{4}$

Point of intersection $\left(\frac{a}{2}, \frac{a^2}{4}\right)$

Curve $1 y=x^2$

$ \left(\frac{d y}{d x}\right)_1=2 x=2 \times \frac{a}{2}=a $

Curve II $y=(x-a)^2$

$ \begin{aligned} \left(\frac{d y}{d x}\right)_2 & =2(x-a)=2\left(\frac{a}{2}-a\right) \\ & =2 \times-\frac{a}{2}=-a \end{aligned} $

$ \begin{aligned} &\text { Angle between two curve is } \theta \text {. }\\ &\begin{aligned} \therefore \quad \tan \theta & =\left|\frac{\left(\frac{d y}{d x}\right)_1-\left(\frac{d y}{d x}\right)_2}{1+\left(\frac{d y}{d x}\right)_1\left(\frac{d y}{d x}\right)_2}\right| \\ & =\left|\frac{a-(-a)}{1+(a)(-a)}\right|=\frac{2|a|}{\left|1-a^2\right|} \end{aligned} \end{aligned} $

Here, $x$ and $y$ are two positive integers, then

$ \begin{array}{rlrl} & & x+2 y & =10 \\ \Rightarrow & y & =\frac{10-x}{2} \end{array} $

Let $P=x^2 y^3$

$ \begin{aligned} \Rightarrow P & =x^2\left(\frac{10-x}{2}\right)^3=\frac{1}{8} x^2(10-x)^3 \\ \Rightarrow \frac{d P}{d x} & =\frac{1}{8}\left\{2 x(10-x)^3-3 x^2(10-x)^2\right\} \\ & =\frac{x(10-x)^2}{8}(2(10-x)-3 x) \end{aligned} $

$ \frac{d P}{d x}=\frac{x(10-x)^2(20-5 x)}{8} $

For $P$ to be maximum, $\frac{d P}{d x}=0$

$ \begin{aligned} & \Rightarrow \quad \frac{x(10-x)^2(20-5 x)}{8}=0 \\ & \Rightarrow x=4, x \neq 0,10 \\ & \begin{aligned} \frac{d^2 P}{d x^2}= & \frac{1}{8}\left[(10-x)^2(20-5 x)+x[ \right. \\ & \left.-2(10-x)(20-5 x)-(10-x)^2 \cdot 5\right] \end{aligned} \\ & \text { At } x=4, \frac{d^2 P}{d x^2}<0 \end{aligned} $

So, $x=4$ is point of maximum

$ y=\frac{10-x}{2}=\frac{10-4}{2}=3 $

So, $x^2+2 y^3=16+2 \times 27=70$

Given, equation of curve

$ \sin y=\sqrt{3} x \sin \left(\frac{\pi}{6}+y\right) $

When $y=0$,

$ \begin{array}{rlrl} \sin 0 & =\sqrt{3} x \sin \left(\frac{\pi}{6}+0\right) \\ 0 & =\sqrt{3} x \times \frac{\sqrt{3}}{2} \\ \Rightarrow \quad x & =0 \end{array} $

Now, differentiate equation of curve w.r.t. $y$,

we get, $\cos y=\sqrt{3} x \cos \left(\frac{\pi}{6}+y\right)$

$ \begin{array}{r} +\sqrt{3} \sin \left(\frac{\pi}{6}+y\right) \frac{d x}{d y} \\ \Rightarrow \quad-\frac{d x}{d y}=\frac{\sqrt{3} x \cos \left(\frac{\pi}{6}+y\right)-\cos y}{\sqrt{3} \sin \left(\frac{\pi}{6}+y\right)} \end{array} $

Slope of normal at $(0,0)$

$ =-\frac{\cos 0}{\sqrt{3} \sin \frac{\pi}{6}}=-\frac{2}{\sqrt{3}} $

Equation of normal is

$ \begin{aligned} y-0 & =-\frac{2}{\sqrt{3}}(x-0) \\ \Rightarrow \quad 2 x+\sqrt{3} y & =0 \end{aligned} $

Assertion : Given, two curve $y^2=4 x$ and $x^2=-2 y$. Solving the two equation to determine the intersecting points $=\left(-\frac{x^2}{2}\right)^2=4 x$

$ \Rightarrow \quad x^4-16 x=0 $

Either $x=0$

or $\quad x^3=16$

or $\quad x=2(2)^{1 / 3}$

If $x=0, y=0$

If $x=2(2)^{\frac{1}{3}}$, then

$ \begin{aligned} y & =-\frac{x^2}{2} \\ & =-\frac{4(2)^{2 / 3}}{2}=-2(2)^{2 / 3} \end{aligned} $

∴ The two curve does not intersect at the point

$(1,2)$.

Reason : The curve are said to intersect orthogonally if the product of the slopes of the tangents drawn to two curves at their point of intersection is -1 .

Hence, Assertion is false but Reason is true.

$ \begin{aligned} & f(x)=\left\{\begin{array}{cl} 1+6 x-3 x^2, & x \leq 1 \\ x+\log _2\left(b^2+7\right), & x>1 \end{array}\right. \\ & f^{\prime}(x)=\left\{\begin{array}{cl} 6-6 x, & x \leq 1 \\ 1, & x>1 \end{array}\right. \end{aligned} $

$f(1)$ is the maximum value of $f(x)$,

$ \begin{aligned} & & f^{\prime}(x) & \geq 0 \\ \Rightarrow & & 6-6 x & \geq 0 \Rightarrow 6 x \leq 6 \\ \Rightarrow & & x & \leq 1 \\ \therefore& & & x \in[-1,1] \end{aligned} $

$ \begin{aligned} &\text {Given, equation of curves, }\\ &\begin{aligned} & & x^2+y^2=4 \text { and } y^2 & =3 x \\ & & x^2+3 x-4 & =0 \\ &\Rightarrow & x^2+4 x-x-4 & =0 \\ & \Rightarrow & x(x+4)-1(x+4) & =0 \\ & \Rightarrow & (x+4)(x-1) & =0 \\ &\Rightarrow & x & =1\,\,\,\,[x \neq-4] \\ & & y & = \pm \sqrt{3} \end{aligned} \end{aligned} $

The intersecting points are

$ \begin{aligned} & (1, \sqrt{3}) \text { and }(1,-\sqrt{3}) \\ \therefore \quad & x^2+y^2=4 \end{aligned} $

On differentiating w.r.t. $x$, we get

$ \begin{aligned} & & 2 x+2 y \frac{d y}{d x} & =0 \\ & \Rightarrow & \frac{d y}{d x} & =\frac{-x}{y} \end{aligned} $

$\begin{aligned} \frac{d y}{d x} \text { at }(1, \sqrt{3}) & =\frac{-1}{\sqrt{3}} \quad\left(\text { let } m_1\right) \\ \frac{d y}{d x} \text { at }(1,-\sqrt{3}) & =\frac{1}{\sqrt{3}} \quad\left(\text { let } m_2\right) \\ y^2=3 x \Rightarrow \frac{d y}{d x} & =\frac{3}{2 y} \\ \frac{d y}{d x} \text { at } x=(1, \sqrt{3}) & =\frac{3}{2 \sqrt{3}}=\frac{\sqrt{3}}{2} \quad\left(\text { let } m_3\right) \\ \frac{d y}{d x} \text { at } x=(1,-\sqrt{3}) & =\frac{-3}{2 \sqrt{3}}=\frac{-\sqrt{3}}{2}\left(\text { let } m_4\right) \\ \tan \theta & =\left|\frac{m_1-m_3}{1+m_1 m_3}\right|\end{aligned}$

$ \begin{aligned} &=\left|\frac{-\frac{1}{2 \sqrt{3}}-\sqrt{3}}{1-\frac{1}{2}}\right|=\left|\frac{-5}{\sqrt{3}}\right|=\frac{5}{\sqrt{3}}\\ &\text { Also, }\\ &\tan \theta=\left|\frac{m_2-m_4}{1+m_2 m_4}\right|=\left|\frac{\frac{1}{2 \sqrt{3}}-\sqrt{3}}{1-\frac{1}{2}}\right|=\frac{5}{\sqrt{3}} \end{aligned} $

$ \text { In } \triangle A B C \text {, } $

$ \begin{aligned} \tan \pi / 3 & =\frac{\sqrt{3}}{H} \Rightarrow \sqrt{3}=\frac{\sqrt{3}}{H} \\ H & =1 \text { unit } \end{aligned} $

In $\triangle A D E$

$ \begin{aligned} \tan \pi / 3 & =r / h \Rightarrow \sqrt{3}=r / h \\ r & =\sqrt{3} h \Rightarrow r=\pi r^2(H-h) \\ r & =\pi 3 h^2(1-h)=3 \pi\left(h^2-h^3\right) \\ \frac{d v}{d h} & =3 \pi\left(2 h-3 h^2\right) \Rightarrow \frac{d v}{d h}=0 \\ \Rightarrow\,\,\,\,\,\,\,\,\,\,\, h & =2 / 3 \quad \quad[\because h \neq 0] \\ \frac{d^2 v}{d h^2} & =3 \pi(2-6 h] \\ \frac{d^2 v}{d h^2} \text { at } h & =\frac{2}{3}=3 \pi\left(2-6\left(\frac{2}{3}\right)\right) \\ & =3 \pi[2-4]<0 \end{aligned} $

Volume is maximum at $h=2 / 3$

Height of cylinder $=H-h=1-\frac{2}{3}=\frac{1}{3}$

$ \begin{aligned} &\\ &\begin{aligned} & \text { } A=4 \pi r^2 \\ & \Rightarrow \quad d A=8 \pi r d r \Rightarrow 0.02=(8 \pi)(10)(d r) \\ & \Rightarrow \quad d r=\frac{0.02}{80 \pi} \Rightarrow V=\frac{4}{3} \pi r^3 \\ & \Rightarrow d V=4 \pi r^2 d r=4 \pi \times 100 \times \frac{0.02}{80 \pi} \\ & \quad=0.1 \mathrm{~cm}^3 \end{aligned} \end{aligned} $

$ \begin{aligned} & y^2=4 x, x^2+y^2=5 \\ & \frac{d y}{d x}=\frac{2}{y}, \frac{d y}{d x}=\frac{-x}{y} \end{aligned} $

For point of intersection,

$ \begin{aligned} & x^2+4 x-5=0 \\ & \Rightarrow \quad x=-5 \text { or } x=1 \end{aligned} $

∴ Points of intersection are $(1,2)$, (1, -2)

$\therefore$ Slopes of tangents to two curves at points of intersection are $1, \frac{-1}{2}$ or $-1, \frac{1}{2}$.

$ \therefore \quad|\tan \theta|=\left|\frac{1+1 / 2}{1-1 / 2}\right|=3 $

$ \begin{aligned} &\\ &\begin{aligned} & \text { } f(x)=-(x-2)^3(x+2)^2 \\ & f^{\prime}(x)=-\left[(x-2)^3 \cdot 2(x+2)\right. \left.+(x+2)^2 \cdot 3(x-2)^2\right] \\ & =-(x-2)^2(x+2)[2(x-2)+3(x+2)] \\ & =-(x-2)^2(x+2)[5 x+2] \\ & \text { on putting } f^{\prime}(x)=0 \\ & \Rightarrow \quad x=-2, \frac{-2}{5}, 2 \end{aligned} \end{aligned} $

$f(x)$ has local maxima, at $x=\frac{-2}{5}$

∴ Local maximum value

$ \begin{aligned} & =-\left(-\frac{2}{5}-2\right)^3\left(\frac{-2}{5}+2\right)^2 \\ & =\left(\frac{12}{5}\right)^3 \cdot\left(\frac{8}{5}\right)^2=\frac{(12)^3 8^2}{5^5} \end{aligned} $

$ \begin{aligned} &\text { Given curves, } y^2=4 x\\ &2 y \frac{d y}{d x}=4 \Rightarrow \frac{d y}{d x}=\frac{2}{y} \end{aligned} $

$ \text { At }(1,2) \frac{d y}{d x}=\frac{2}{2}=1 $

Equation of tangent

$ y-2=1(x-1) \Rightarrow y=x+1 \Rightarrow T(0,1) $

Slope of normal $=-1$

Equation of normal

$ y-2=-1(x-1) \quad x+y=3 \Rightarrow N(0,3) $

$ \begin{aligned}Area \,\,of \,\, \Delta T P N & =\frac{1}{2}\left|\begin{array}{lll} 0 & 1 & 1 \\ 0 & 3 & 1 \\ 1 & 2 & 1 \end{array}\right| \\ & =\frac{1}{2}|1-3|=1 \end{aligned} $

Given,

$ C_1: y^2=4 a x, C_2: x^2+\frac{y^2}{2}=c^2 $

Let point of intersection of two curves is

$ \begin{aligned} & \left(x_1, y_1\right) . C_1: y^2=4 a x \\ & \Rightarrow \quad 2 y\left(\frac{d y}{d x}\right)_1=4 a \Rightarrow\left(\frac{d y}{d x}\right)_1=\frac{2 a}{y_1} \end{aligned} $

Let angle between two curves be $\theta$.

$ \begin{aligned} & C_2: x^2+\frac{y^2}{2}=c^2 \\ \Rightarrow & 2 x+\frac{2 y}{2}\left(\frac{d y}{d x}\right)_2=0 \\ \Rightarrow & \quad\left(\frac{d y}{d x}\right)_2=-\frac{2 x_1}{y_1} \\ \because & \tan \theta=\left|\frac{\left(\frac{d y}{d x}\right)_1-\left(\frac{d y}{d x}\right)_2}{1+\left(\frac{d y}{d x}\right)_1\left(\frac{d y}{d x}\right)_2}\right| \end{aligned} $

$ \begin{aligned} & =\left|\frac{\frac{2 a}{y_1}+\frac{2 x_1}{y_1}}{1-\frac{4 a x_1}{y_1^2}}\right|=\left|\frac{\frac{2 a+2 x_1}{y_1}}{1-\frac{y_1^2}{y_1^2}}\right| \\ \tan \theta & =\left|\frac{\frac{2 a+2 x_1}{y_1}}{0}\right|=\infty \\ \theta & =\frac{\pi}{2} \end{aligned} $

If $c \in(a, b)$

$ f(c-\delta) < f(c) \text { and } f(c+\delta) < f(c) $

So, $f(c)$ is point of local maxima Also $[f(\alpha-\delta)-f(\alpha)] f(\alpha+\delta)<0$

$\Rightarrow \quad f(\alpha-\delta)-f(\alpha)<0$ and $f(\alpha+\delta)>0$ or $\quad f(\alpha-\delta)-f(\alpha)>0$ and $f(\alpha+\delta)<0$

i.e. No maxima and minima at $\alpha, \alpha \in(a, b)$

,$3 f(\cos x)+2 f(\sin x)=5 x\quad \text{... (i)}$

$3 f(\sin x)+2 f(\cos x)=5\left(\frac{\pi}{2}-x\right)\quad \text{... (ii)}$

Solving Eqs. (i) and (ii) simultaneously, we get

$\begin{aligned} & f(\cos x)=5 x-\pi \\ \Rightarrow & -\sin x f^{\prime}(\cos x)=5 \\ \Rightarrow & f^{\prime}(\cos x)=\frac{-5}{\sin x} \end{aligned}$

Similarly,

$\begin{aligned} f^{\prime}(\sin x) & =\frac{-5}{\cos x} \\ & =f^{\prime}(\cos x)+f^{\prime}(\sin x) \\ & =\frac{-5}{\sin x}-\frac{5}{\cos x} \end{aligned}$

Given curve $3 y=6 x-5 x^3$

$\Rightarrow \quad \frac{d y}{d x}=2-5 x^2$

Slope of normal $=\frac{-1}{2-5 x^2}$

Equation of normal with slope $\frac{-1}{2-5 x^2}$ passing through origin $(0,0)$.

$\frac{\left(y-y_1\right)}{x-x_1}=m$

$\Rightarrow \frac{y-0}{x-0}=\frac{-1}{2-5 x^2} \Rightarrow \frac{y}{x}=\frac{-1}{2-5 x^2}$

Since, $(h, k)$ lies on normal

$\begin{aligned} & \frac{k}{h}=\frac{-1}{2-5 h^2} \\ \Rightarrow & \frac{k}{h}=\frac{1}{5 h^2-2} \quad \text{... (i)} \end{aligned}$

Since, $(h, k)$ lies on curve

$\begin{aligned} 3 k & =6 h-5 h^3 \\ \Rightarrow \quad 3 k & =h\left(6-5 h^2\right) \\ \Rightarrow \quad \frac{k}{h} & =\frac{6-5 h^2}{3}\quad \ \text{... (ii)} \end{aligned}$

On solving Eqs. (i) and (ii),

$\begin{array}{ll} & \frac{1}{5 h^2-2}=\frac{6-5 h^2}{3} \\ \Rightarrow & 3=\left(6-5 h^2\right)\left(5 h^2-2\right) \\ \Rightarrow & 3=30 h^2-12-25 h^4+10 h^2 \\ \Rightarrow & 15=-25 h^4+40 h^2 \\ \Rightarrow & 5 h^4-8 h^2+3=0 \end{array}$

For simplicity, let $h^2=z$

The equation becomes $5 z^2-8 z+3=0$

$(5 z-3)(z-1)=0$

Now, $z=\frac{3}{5}, 1$

Putting $h^2=z$

$\Rightarrow \quad h^2=\frac{3}{5} \Rightarrow h= \pm \sqrt{\frac{3}{5}}$

It is not in option.

or $(z-1)=0$

$z=1$

Putting $h^2=z$

$\Rightarrow \quad \begin{aligned} h^2 & =1 \\ h & = \pm 1 \end{aligned}$

But $h=-1$ is not in option.

$\therefore h=1$ is correct

Slope of line joining the points $(0,3)$ and $(5,-2) \text { is } \frac{3+2}{0-5}=-1$

This is equal to the slope of tangent on the curve and that is given by

$\begin{aligned} & \frac{d y}{d x}=\frac{-c}{(x+1)^2} \\ \Rightarrow & \frac{d y}{d x}=-1 \\ \Rightarrow & c=(x+1)^2\quad \text{... (i)} \end{aligned}$

Equation of line joining the points $(0,3)$ and $(5,2)$ is given by $(y-3)=-1(x-0)$.

Solving equation of tangent and the curve for point of intersection

$\frac{c}{x+1}+x=3\quad \text{... (ii)}$

Solving Eqs. (i) and (ii),

$x=1$

On putting this in Eq. (ii), we get $c=4$

Given, $y=\frac{b^2}{a-x}+\frac{a^2}{x}$

$\begin{aligned} & \therefore \quad \frac{d y}{d x}=\frac{-b^2(-1)}{(a-x)^2}+\left(\frac{-a^2}{x^2}\right) \Rightarrow \frac{d y}{d x}=\frac{b^2}{(a-x)^2}-\frac{a^2}{x^2} \\ & \because \quad \frac{d y}{d x}=0, x=\frac{a^2}{a \pm b} \\ & \therefore \frac{d^2 y}{d x^2}=\frac{2 a^2}{x^3}+\frac{2 b^2}{(a-x)^3} \end{aligned}$

So, $\left.\frac{d^2 y}{d x^2}\right|_{x=\frac{a^2}{a+b}}>0$

Therefore, $y$ is minimum at $x=\frac{a^2}{a+b}$

$\begin{aligned} y_{\min } & =\frac{b^2}{a-\left(\frac{a^2}{a+b}\right)}+\frac{a^2}{(a)^2}(a+b) \\ & =\frac{b^2(a+b)}{a^2+a b-a^2}+(a+b)=(a+b)\left[\frac{b}{a}+1\right] \\ & =\frac{(a+b)^2}{a} \end{aligned}$

Let $(x, y)$ be on the parabola $y=x^2+4 x+3$ which is closet to the line $y=3 x+2$

Perpendicular distance between a point $\left(x_1, y_1\right)$ and a line $(a x+b y+c=0)$

$D=\left|\frac{a x_1+b y_1+c}{\sqrt{a^2+b^2}}\right|$

$\left[\therefore\right.$ line $3 x-y+2=0$ at point $\left.\left(x_1, y_1\right)\right]$

$\begin{aligned} D & =\left|\frac{3 x-y+2}{\sqrt{(3)^2+(-1)^2}}\right|=\frac{\left|3 x-\left(x^2+4 x+3\right)+2\right|}{\sqrt{9+1}} \\ \Rightarrow \quad D & =\frac{\left|-x^2-x-1\right|}{\sqrt{10}} \\ \Rightarrow \quad D & =\frac{x^2+x+1}{\sqrt{10}}=\frac{\left(x+\frac{1}{2}\right)^2+\frac{3}{4}}{\sqrt{10}} \end{aligned}$

On differentiating w.r.t. $x$,

$\begin{aligned} & \frac{d D}{d x}=\frac{\left(x+\frac{1}{2}\right)^2}{\sqrt{10}}+\frac{\frac{3}{4}}{\sqrt{10}} \\ \Rightarrow & \frac{d D}{d x}=0, x=\frac{-1}{2} \Rightarrow y=x^2+4 x+3 \\ \Rightarrow & y=\left(\frac{-1}{2}\right)^2+(4)\left(\frac{-1}{2}\right)+3 \Rightarrow y=\frac{1}{4}+3-2=0 \\ \Rightarrow & y=\frac{1+12-8}{4} \Rightarrow y=\frac{5}{4} \\ & \frac{d D}{d x}=\frac{2\left(x+\frac{1}{2}\right)}{\sqrt{10}} \Rightarrow \frac{d^2 D}{d x^2}>0 \end{aligned}$

Hence, minimum at $\left(\frac{-1}{2}, \frac{5}{4}\right)$.

Given, equation of cuve $y^2-2 x^3-4 y+8=0$

$\Rightarrow 2 y y^{\prime}-6 x^2-4 y^{\prime}=0 \Rightarrow y^{\prime}=\frac{3 x^2}{y-2}$

Since, $p(x, y)$ is a point on this curve.

$\therefore \quad \frac{y-2}{x-1}=\frac{3 x^2}{y-2}$

(for tangent lines passing through $(1,2)$)

$\begin{aligned} & \Rightarrow \quad(y-2)^2=3 x^2(x-1) \\ & \therefore \quad y^2-2 x^3-4 y+8=0 \\ & \Rightarrow \quad(y-2)^2-2 x^3+4=0 \\ & \Rightarrow \quad 3 x^2(x-1)-2 x^3+4=0 \\ & \Rightarrow \quad 3 x^3-3 x^2-2 x^3+4=0 \\ & \Rightarrow \quad x^3-3 x^2+4=0 \\ & \Rightarrow \quad(x+1)(x-2)=0 \\ & \therefore \quad x=-1,2 \end{aligned}$

For $x=-1$, there is no real value of $y$.

For $x=2, y=2 \pm 2 \sqrt{3}$

$\therefore \quad y^{\prime}=\frac{3 x^2}{y-2}= \pm 2 \sqrt{3}$

Thus, tangent lines drawn from the point $(1,2)$ is $y=2 \pm 2 \sqrt{3}(x-1)$.

$\therefore$ Two tangents are possible.

The given straight line is $x \cos \alpha+y \sin \alpha=p$ and given curve is $\left(\frac{x}{a}\right)^n+\left(\frac{y}{b}\right)^n=2$

On differentiating, $\frac{n x^{n-1}}{a^n}+\frac{n y^{n-1}}{b^n} y^{\prime}=0$

At $(a, b)$, then $y^1=-\frac{b}{a}$

Thus, equation of tangent at $(a, b)$ is

$y-b=-\frac{b}{a}(x-a)$

$\begin{aligned} & \Rightarrow \quad a y-a b+b x=a b \quad \text{... (i)}\\ & \Rightarrow \quad b x+a y=2 a b \quad \text{... (ii)} \end{aligned}$

Comparing Eqs. (i) and (ii) on both sides, we get

$\begin{aligned} & \quad \frac{b}{\cos \alpha}=\frac{a}{\sin \alpha}=\frac{2 a b}{p} \\ & \Rightarrow \quad \cos \alpha=\frac{p}{2 a} \text { and } \sin \alpha=\frac{p}{2 b} \\ & \therefore \text { Thus, } \sin ^2 \alpha+\cos ^2 \alpha=1 \\ & \Rightarrow \quad \frac{p^2}{4 a^2}+\frac{p^2}{4 b^2}=1 \Rightarrow \frac{1}{a^2}+\frac{1}{b^2}=\frac{4}{p^2}=\frac{k}{p^2} \end{aligned}$

Hence, the value of $k$ is 4.

Given, $y^2=4 a x$ and $x y=c^2$ cut orthogonally.

Let they intersect at $\left(x_1, y_1\right)$.

$\begin{aligned} \therefore \quad & y^2=4 a x \Rightarrow 2 y \frac{d y}{d x}=4 a \\ \Rightarrow \quad & \frac{d y}{d x}=\frac{2 a}{y} \end{aligned}$

$\therefore \quad \frac{d y}{d x}\left(x_1, y_1\right)=\frac{2 a}{y_1} \quad \text{... (i)}$

$\begin{aligned} \text { and } & & x y & =c^2 \\ \Rightarrow & & y+x \frac{d y}{d x} & =0 \\ \Rightarrow & & \frac{d y}{d x}\left(x_1, y_1\right) & =-\frac{y_1}{x_1} \quad \text{... (ii)} \end{aligned}$

From Eqs (i) and (ii), we get

$\begin{aligned} \frac{2 a}{y_1} \times-\frac{y_1}{x_1} & =-1 \\ \Rightarrow \quad x_1 & =2 a \end{aligned}$

From $y^2=4 a x$

$\begin{aligned} \Rightarrow \quad & y_1 =\sqrt{4 a \cdot 2 a}=2 \sqrt{2} a \Rightarrow x_1 y_1=c^2 \\ \Rightarrow \quad & 2 a(2 \sqrt{2} a) =c^2 \Rightarrow \frac{c^2}{a^2}=4 \sqrt{2} \\ \Rightarrow \quad & c^4 =32 a^4 \end{aligned}$

Let $h$ be the height and $r$ be the radius. The volume of cylinder is $\pi r^2 h$.

The surface area is $2 \pi r^2+2 \pi r h$

$\begin{aligned} & =2 \pi\left(r^2+\frac{V}{r \pi}\right) \\ & \Rightarrow \frac{d}{d r}(\text { surface })=2 \pi\left(2 r-\frac{V}{\pi r^2}\right) \end{aligned}$

$\text { Also } \frac{d}{d r}(\text { surface })=0$

$ \begin{aligned} & \Rightarrow \quad 2 \pi\left(2 r-\frac{V}{\pi r^2}\right)=0 \\ & \Rightarrow \quad r^3=\frac{V}{2 \pi} \Rightarrow r=\left(\frac{V}{2 \pi}\right)^{1 / 3} \\ & \therefore \quad h=(2)^{2 / 3} \cdot\left(\frac{V}{\pi}\right)^{1 / 3} \Rightarrow h: r=2: 1 \end{aligned}$

Given that, two particles $P$ and $Q$ located at the points with coordinate $P\left(t, t^3-16 t-3\right), Q\left(t+1, t^3-6 t-6\right)$ are moving in a plane. Thus,

$\begin{aligned} P Q & =\sqrt{(t+1-t)^2+\left(t^3-6 t-6-t^3+16 t+3\right)^2} \\ & =\sqrt{1+(10 t-3)^2} \end{aligned}$

$\therefore(P Q)^2=1+(10 t-3)^2 \geq 1$

Thus, minimum value of $P Q$ is 1.

$x^3-2 x^2 y^2+5 x+y-5=0$

On differentiating w.r.t. $x$, we get $3 x^2-4 x y^2-4 x^2 y \cdot y^{\prime}+5+y^{\prime}-0=0\quad \text{.... (i)}$

On putting $x=1, y=1$,

$\begin{aligned} & 3-4-4 y^{\prime}+5+y^{\prime}=0 \\ & \Rightarrow \quad y^{\prime}=\frac{4}{3} \\ & \text { or } \quad y^{\prime}(1)=\frac{4}{3} \\ & \end{aligned}$

On differentiating Eq. (i) w.r.t. $x$,

$6 x-4 y^2-8 x y y^{\prime}-8 x y y^{\prime}-4 x^2 y^{\prime} \cdot y^{\prime}-4 x^2 y y^{\prime \prime}+y^{\prime \prime}=0$

On putting $x=1, y=1$ and $y^{\prime}=\frac{4}{3}$ to obtain $y^{\prime \prime}(1)$.

$\begin{array}{ll} \Rightarrow & 6-4-8 \times \frac{4}{3}-8 \times \frac{4}{3}-4 \times \frac{4}{3} \times \frac{4}{3}-4 \times y^{\prime \prime}+y^{\prime \prime}=0 \\ \Rightarrow & 2-\frac{64}{3}-\frac{64}{9}-3 y^{\prime \prime}=0 \\ \Rightarrow & \frac{-238}{9}=3 y^{\prime \prime} \\ \Rightarrow & y^{\prime \prime}=-\frac{238}{27} \\ \text { or } & y^{\prime \prime}(1)=-\frac{238}{27} \end{array}$

$\begin{aligned} y & =x^3-3 x^2-8 x-4 \\ y & =3 x^2+7 x+4 \end{aligned}$

$\because$ Curves touch each other.

$\begin{aligned} & \therefore \quad x^3-3 x^2-8 x-4=3 x^2+7 x+4 \\ & \Rightarrow \quad x^3-6 x^2-15 x-8=0 \end{aligned}$

On putting $x=-1$

$\begin{aligned} \mathrm{LHS} & =-1-6+15-8 \\ & =-15+15=0=\text { RHS } \end{aligned}$

On putting $x=-1$ in any curve equation,

$y=-1-3+8-4=0$

$\therefore(-1,0)$ is the intersection point.

Equation of tangent,

$\begin{aligned} & y-y_1=m\left(x-x_1\right) \quad \text{... (i)}\\ & y=x^3-3 x^2-8 x-4 \\ & y^{\prime}=3 x^2-6 x-8 \end{aligned}$

$y^{\prime} |_{(-1,0)}=3+6-8=1 \Rightarrow m=1$

Put the values into Eq. (i),

$\begin{array}{rlrl} & y-0=1(x+1) \\ \Rightarrow & y=x+1 \\ \Rightarrow & x-y+1=0 \end{array}$

$\begin{aligned} & f(x)=\frac{x}{1+4 x+x^2} \\ & f^{\prime}(x)=\frac{\left(1+4 x+x^2\right)(1)-x(4+2 x)}{\left(1+4 x+x^2\right)^2}=\frac{1-x^2}{\left(1+4 x+x^2\right)^2} \\ & \text { Put } f^{\prime}(x)=0 \\ & \Rightarrow \quad x= \pm 1 \end{aligned}$

At $x=-1$, it gives minimum value.

At $x=1$, it gives maximum value.

Put $x=1$ in $f(x)$, we get

$f(1)=\frac{1}{1+4+1}=\frac{1}{6}$

$\begin{aligned} f(x) & =x+\frac{4}{x+2} \\ f^{\prime}(x) & =1+4\left(-\frac{1}{(x+2)^2}\right) \\ f^{\prime}(x) & =\frac{(x+2)^2-4}{(x+2)^2} \end{aligned}$

$\begin{array}{lr} \text { Put } & f^{\prime}(x)=0 \\ \Rightarrow & (x+2)^2-4=0 \\ \Rightarrow & (x+2)^2=4 \\ \Rightarrow & x+2= \pm 2 \end{array}$

For positive value, $x+2=2 \Rightarrow x=0$

For negative value, $x+2=-2 \Rightarrow x=-4$

$f(0)=0+\frac{4}{0+2}=\frac{4}{2}=2$

and $f(-4)=-4+\frac{4}{-4+2}=-4-2=-6$

As per the option given,

Minimum value $=2$

$\begin{aligned} & f(x)=a x^3+b x^2+c x+d \\ & f^{\prime}(x)=3 a x^2+2 b x+c \end{aligned}$

For no extrema, the roots of $f^{\prime}(x)$ will not be real i.e. $f^{\prime}(x)$ has imaginary roots.

$\begin{aligned} & \Rightarrow\quad(2 b)^2-4(3 a)(c)<0 \\ & \Rightarrow \quad 4 b^2<4(3 a c) \\ & \Rightarrow \quad b^2<3 a c \end{aligned} $

Length of subnormal $=y m$

$\begin{aligned} & \Rightarrow \quad y m=x-1 \\ & \Rightarrow \quad y \frac{d y}{d x}=x-1 \Rightarrow y d y=(x-1) d x \\ & \Rightarrow \int y d y=\int(x-1) d x \\ & \Rightarrow \quad \frac{y^2}{2}=\frac{(x-1)^2}{2}+c \quad \text{... (i)} \end{aligned}$

$\because$ Curve passes through $(1,2)$.

$\therefore$ On putting $x=1$ and $y=2$ into Eq. (i), we get

$\frac{4}{2}=\frac{0}{2}+c \Rightarrow c=2$

From Eq. (i), $y^2=(x-1)^2+4$

$\begin{array}{ll} \Rightarrow & \frac{y^2}{1}-\frac{(x-1)^2}{1}=4 \\ \Rightarrow & \frac{y^2}{2^2}-\frac{(x-1)^2}{2^2}=1 \\ \Rightarrow & \frac{(y-k)^2}{b^2}-\frac{(x-h)^2}{a^2}=1 \end{array}$

where, $b=a$

Vertices are $(0, \pm b) \equiv(0, \pm 2)$.

Perimeter = 440 ft

$\Rightarrow$ 2x + $\pi$r + $\pi$r = 440

2x + 2$\pi$r = 440

A = Area of the rectangular portion = x2r

A = $x\frac{(440-2x)}{\pi}=\frac{1}{\pi}(440x-2x^2)$

Let $\frac{dA}{dx}=\frac{1}{\pi}(440-4x)=0$

$\Rightarrow$ x = 110 for which $\frac{d^2A}{dx^2} < 0$

$\Rightarrow$ A is maximum when x = 110

$\Rightarrow$ $2r=\frac{440-2x}{\pi}=\frac{440-220}{22/7}=70$

$\therefore$ r = 35 ft and x = 110 ft

V = $\frac{4}{3}\pi r^3\Rightarrow4500\pi=\frac{4\pi r^3}{3}$

$\Rightarrow$ r = 15 m

After 49 min = (4500 $-$ 49.72) $\pi$

= 972 $\pi$m³

$\Rightarrow$ 972$\pi=\frac{4}{3}\pi r^3$

$\Rightarrow$ r³ = 3 $\times$ 243 = 3 $\times$ 3⁵

r = 9

$\frac{dv}{dt}=4 \pi r^2\left(\frac{dr}{dt}\right)$

72$\pi$ = 4$\pi$ $\times$ 9 $\times$ $\left(\frac{dr}{dt}\right)$

$\frac{dr}{dt}=\left(\frac{2}{9}\right)$