Let AB represent the height of the street light from the ground.
At any time t s, let the man represented as ED of height 1.6 m be at a distance of x m from A B and the length of his shadow EC be y m.
From similarity of \(\triangle A B C\) and \(\triangle E D C\), we have
\( \frac{4}{1.6}=\frac{x+y}{y} \Rightarrow 4 y=1.6 x+1.6 y \Rightarrow 2.4 y=1.6 x \)
\(\Rightarrow  3 y=2 x\)
Now, differentiating both sides of Eq. (i) w.r.t. t, we get
\(3 \frac{d y}{d t} =2 \frac{d x}{d t} \Rightarrow \frac{d y}{d t}=\frac{2}{3} \times 0.3 \quad\left[\because \frac{d x}{d t}=0.3 \mathrm{~m} / \mathrm{s}\right] \)
\(\Rightarrow \frac{d y}{d t}  =0.2\)
At any time t s, the tip of his shadow is at a distance of (x+y) m from AB
\(\therefore\) The rate at which the tip of his shadow is moving
\(=\left(\frac{d x}{d t}+\frac{d y}{d t}\right)=(0.3+0.2)=0.5 \mathrm{~m} / \mathrm{s}\)
The rate at which his shadow is lengthening = dy/dx = 0.2 m/s

Given, \(y=\tan x+\sec x\)
\(\Rightarrow \quad y=\frac{\sin x}{\cos x}+\frac{1}{\cos x}=\frac{\sin x+1}{\cos x}\)
On differentiating both sides w.r.t. x, we get
\(\frac{d y}{d x} =\frac{\cos x \frac{d}{d x}(\sin x+1)-(\sin x+1) \frac{d}{d x}(\cos x)}{(\cos x)^2}\)
\( =\frac{\cos x(\cos x)-(\sin x+1)(-\sin x)}{\cos ^2 x} \)
\(=\frac{\cos ^2 x+\sin ^2 x+\sin x}{\cos ^2 x}=\frac{1+\sin x}{\cos ^2 x} \)
\(\Rightarrow \frac{d y}{d x} =\frac{1+\sin x}{1-\sin ^2 x}=\frac{1}{1-\sin x}\)
Again, differentiating w.r.t. x, we get
\(\frac{d^2 y}{d x^2}=\frac{(1-\sin x)(0)-1(0-\cos x)}{(1-\sin x)^2}=\frac{\cos x}{(1-\sin x)^2} \)
Hence proved.

Given, \((\cos x)^y=(\cos y)^x\)
On taking log both sides, we get
On differentiating both sides w,r,t, x, we get
\(\log (\cos x)^y=\log (\cos y)^x\)
\(\Rightarrow \quad y \log (\cos x)=x \log (\cos y) \)
\({\left[\because \log x^n=n \log x\right]}\)
\(​​​​​​​ y \cdot \frac{d}{d x} \log (\cos x)+\log \cos x \cdot \frac{d}{d x}(y)\)
\(=x \frac{d}{d x} \log (\cos y)+\log (\cos y) \frac{d}{d x}(x) \)
\(\Rightarrow \quad y \cdot \frac{1}{\cos x} \frac{d}{d x}(\cos x)+\log (\cos x) \frac{d y}{d x}\)
\( =x \cdot \frac{1}{\cos y} \frac{d}{d x}(\cos y)+\log \cos y \cdot 1\)
\(\Rightarrow y \cdot \frac{1}{\cos x}(-\sin x)+\log (\cos x) \cdot \frac{d y}{d x}\)
\(=x \frac{1}{\cos y}(-\sin y) \cdot \frac{d y}{d x}+\log (\cos y) \cdot 1\)
\(\Rightarrow-y \tan x+\log (\cos x) \frac{d y}{d x}=-x \tan y \frac{d y}{d x}+\log (\cos y)\)
\(​​​​​​​\Rightarrow[x \tan y+\log (\cos x)] \frac{d y}{d x}=\log (\cos y)+y \tan x\)
\(\therefore \quad \frac{d y}{d x}=\frac{\log (\cos y)+y \tan x}{x \tan y+\log (\cos x)}\)

Given, \(f(x)= \begin{cases}a x+b, & x>-1 \\ b x^2-3, & x \leq-1\end{cases}\)
is differentiable at x = -1
\(\therefore\) Lf'(-1) = Rf'(-1)
here, Lf'(-1) = \(\lim _{h \rightarrow 0} \frac{f(-1-h)-f(-1)}{-h}\)
\(\begin{aligned} =\lim _{h \rightarrow 0} \frac{\left[b(-1-h)^2-3\right]-\left[b(-1)^2-3\right]}{-h} \end{aligned}\)
\(\begin{aligned} =\lim _{h \rightarrow 0} \frac{\left[b\left(1+h^2+2 h\right)-3\right]-[b-3]}{-h} \end{aligned}\)
\(\begin{aligned} =\lim _{h \rightarrow 0} \frac{b+b h^2+2 b h-3-b+3}{-h} \end{aligned}\)
\(\begin{aligned} =\lim _{h \rightarrow 0} \frac{h(b h+2 b)}{-h}=-\lim _{h \rightarrow 0}(2 b+b h) \end{aligned}\)
= -2b
and \(\begin{aligned} R f^{\prime}(-1) & =\lim _{h \rightarrow 0} \frac{f(-1+h)-f(-1)}{h} \end{aligned}\)
\(\begin{aligned} =\lim _{h \rightarrow 0} \frac{[a(-1+h)+6]-[a(-1)+6]}{h} \end{aligned}\)
\(\begin{aligned} =\lim _{h \rightarrow 0} \frac{[-a+a h+6+a-6]}{h} \end{aligned}\)
\(\begin{aligned} =\lim _{h \rightarrow 0} \frac{a h}{h}=a \end{aligned}\)
and f(-1) = b(-1)² - 3 = b - 3
Since, f(-1) = Lf'(-1) = Rf'(-1)
So, f(-1) = Lf'(-1)
\(\Rightarrow\)   b - 3 = -2b
\(\Rightarrow\)   b + 2b = 3
\(\Rightarrow\)        3b = 3
\(\Rightarrow\)          b = 1
and f(-1) = Rf'(-1)
\(\Rightarrow\)            b - 3 = a
\(\Rightarrow\)            1 - 3 =a                           [\(\because\)  b = 1]
\(\Rightarrow\)             -2 = a
\(\therefore\)   a = -2 and b = 1

\(Let\quad y={ cot }^{ -1 }\left[ \frac { \sqrt { 1+sin\quad x } +\sqrt { 1-sin\quad x } }{ \sqrt { 1+sin\quad x } +\sqrt { 1-sin\quad x } } \right]\)
\( Now\quad \sqrt { 1+sin\quad x } \)
\(=\sqrt { { sin }^{ 2 }\frac { x }{ 2 } { cos }^{ 2 }\frac { x }{ 2 } +2sin\frac { x }{ 2 } cos\frac { x }{ 2 } } \)
\(=\sqrt { { \left( cos\frac { x }{ 2 } +sin\frac { x }{ 2 } \right) }^{ 2 } } =cos\frac { x }{ 2 } +sin\frac { x }{ 2 } \)
\(Similarly\quad \sqrt { 1-sin\quad x } =cos\frac { x }{ 2 } -sin\frac { x }{ 2 } \)
\(y={ cot }^{ -1 }\left\{ \frac { \left( cos\frac { x }{ 2 } +sin\frac { x }{ 2 } \right) +\left( cos\frac { x }{ 2 } -sin\frac { x }{ 2 } \right) }{ \left( cos\frac { x }{ 2 } +sin\frac { x }{ 2 } \right) -\left( cos\frac { x }{ 2 } -sin\frac { x }{ 2 } \right) } \right\} \)
\(={ cot }^{ -1 }\left\{ \frac { 2cos\frac { x }{ 2 } }{ 2sin\frac { x }{ 2 } } \right\} \)
\(={ cot }^{ -1 }\left\{ cot\frac { x }{ 2 } \right\} =\frac { x }{ 2 } \)
\( \frac { dy }{ dx } =0\)

We have: \(f(x)=x^{ 2 }-4x+6\)
\(\therefore \) \(f'(x)=2x-4.\)
(a) Therefore, f ′(x) = 0 gives x = 2. Now the point x = 2 divides the real line into two disjoint intervals namely, (– ∞, 2) and (2, ∞). In the interval (– ∞, 2), f ′(x) = 2x – 4 < 0.
(b) Therefore, f is decreasing in this interval. Also, in the interval (2, ∞) , f'(x) > 0 and so the function f is increasing in this interval.

Let x be the length of a side, V be the volume, and s be the surface area of the cube.
Then, V = x³ and S = 6x² where x is a function of time t.
 It is given that dVdt=8 cm3/s . " id="MathJax-Element-558-Frame" role="presentation" style="position: relative;" tabindex="0"> It is given that  \( \frac{d V}{d t}=8 \mathrm{~cm}^{3} / \mathrm{s} \text { . }\)
 Then, by using the chain rule, we have: " id="MathJax-Element-559-Frame" role="presentation" style="position: relative;" tabindex="0">Then, by using the chain rule, we have: 
\(\therefore 8=\frac{d V}{d t}=\frac{d}{d t}\left(x^{3}\right)=\frac{d}{d x}\left(x^{3}\right) \cdot \frac{d x}{d t}=3 x^{2} \cdot \frac{d x}{d t}\)
\(\Rightarrow \frac{d x}{d t}=\frac{8}{3 x^{2}} \ \text { (1) } \quad \text { [By chain rule] } \)
\(\text { Now, } \frac{d \mathrm{~S}}{d t}=\frac{d}{d t}\left(6 x^{2}\right)=\frac{d}{d x}\left(6 x^{2}\right) \cdot \frac{d x}{d t} \ 0 .\)
\(=12 x \cdot \frac{d x}{d t}=12 x .\left(\frac{8}{3 x^{2}}\right)=\frac{32}{x}\)
\(\text {Thus, when } x=12 \mathrm{~cm}, \frac{d S}{d t}=\frac{32}{12} \mathrm{~cm}^{2} / \mathrm{s}=\frac{8}{3} \mathrm{~cm}^{2} / \mathrm{s}\)
Hence, if the length of the edge of the cube is 12 cm, then the surface area is increasing at the rate of \(\frac{8}{3}\) cm²/s

Let r be the radius, h be the height and V be the volume of the sand cone
Also given that, \(\frac{d V}{d t}=12 \mathrm{~cm}^{3} / \mathrm{s}, h=\frac{1}{6} r\) 
\(\Rightarrow r=6 h \text { and } h=4 \mathrm{~cm}\) 
Volume of sand cone,
\(V=\frac{1}{3} \pi r^{2} h\) 
\( \Rightarrow V=\frac{1}{3} \pi(6 h)^{2} h \)
\(\Rightarrow V=\frac{1}{3} \pi \times 36 h^{2} \times h=12 \pi h^{3} \) 
On differentiating both sides w.r.t. t, we get
\( \frac{d V}{d t}=12 \pi \times 3 h^{2} \frac{d h}{d t}=36 \pi h^{2} \frac{d h}{d t} \)
\(\Rightarrow 12=36 \pi(4)^{2} \frac{d h}{d t} \)
\(\Rightarrow \frac{d h}{d t}=\frac{12}{36 \pi \times 16}=\frac{1}{48 \pi} \mathrm{cm} / \mathrm{s} \) 
Hence, the height of the sand cone is increasing at the rate of \(\frac{1}{48 \pi} \mathrm{cm} / \mathrm{s}\). when the height is 4 cm,

Given equations are
\(x=\frac{\sin ^3 t}{\sqrt{\cos 2 t}}, y=\frac{\cos ^3 t}{\sqrt{\cos 2 t}}\)
Now, differentiate both w.r.t t
We get,
\(\frac{d x}{d t}=\frac{d\left(\frac{\sin ^3 t}{\sqrt{\cos 2 t}}\right)}{d t}=\frac{\sqrt{\cos 2 t} \cdot \frac{d\left(\sin ^3 t\right)}{d t}-\sin ^3 t \cdot \frac{d(\sqrt{\cos 2 t})}{d t}}{(\sqrt{\cos 2 t})^2}=\frac{3 \sin ^2 t \cos t \cdot \sqrt{\cos 2 t}-\sin ^3 t \cdot \frac{1}{2 \sqrt{\cos 2 t}} \cdot(-2 \sin 2 t)}{\cos 2 t}\)
\(=\frac{3 \sin ^2 t \cos t \cdot \cos 2 t+\sin ^3 t \sin 2 t}{\cos 2 t \sqrt{\cos 2 t}} \)
\(=\frac{\sin ^3 t \sin 2 t(3 \cot t \cot 2 t+1)}{\cos 2 t \sqrt{\cos 2 t}} \quad\left(\because \frac{\cos x}{\sin x}=\cot x\right)\)
Similarly,
\(\frac{d y}{d t}=\frac{d\left(\frac{\cos ^3 t}{\sqrt{\cos 2 t})}\right.}{d t}=\frac{\sqrt{\cos 2 t} \cdot \frac{d\left(\cos ^3 t\right)}{d t}-\cos ^3 t \cdot \frac{d(\sqrt{\cos 2 t})}{d t}}{(\sqrt{\cos 2 t})^2}=\frac{3 \cos ^2 t(-\sin t) \cdot \sqrt{\cos 2 t}-\cos ^3 t \cdot \frac{1}{2 \sqrt{\cos 2 t}} \cdot(-2 \sin 2 t)}{(\sqrt{\cos 2 t})^2}\)
\(=\frac{-3 \cos ^2 t \sin t \cos 2 t+\cos ^3 t \sin 2 t}{\cos 2 t \sqrt{\cos 2 t}} \)
\( =\frac{\sin 2 t \cos ^3 t(1-3 \tan t \cot 2 t)}{\cos 2 t \sqrt{\cos 2 t}} \)
\( \text { Now, } \frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}=\frac{\frac{\sin 2 t \cos ^3 t(1-3 \tan t \cot 2 t)}{\cos 2 t \sqrt{\cos 2 t}}}{\frac{\sin ^3 t \sin 2 t(3 \cot t \cot 2 t+1)}{\cos 2 t \sqrt{\cos 2 t}}}=\frac{\cot ^3 t(1-3 \tan t \cot 2 t)}{(3 \cot t \cot 2 t+1)} \)
\(=\frac{\cos ^3 t\left(1-3 \cdot \frac{\sin t}{\cos t} \cdot \frac{\cos 2 t}{\sin 2 t}\right)}{\sin ^3 t\left(3 \cdot \frac{\cos t}{\sin t} \cdot \frac{\cos 2 t}{\sin 2 t}+1\right)}=\frac{\cos ^2 t(\cos t \sin 2 t-3 \sin t \cos 2 t)}{\sin ^2 t(3 \cos t \cos 2 t+\sin t \sin 2 t)} \)
\( =\frac{\cos ^2 t\left(\cos t \cdot 2 \sin t \cos t-3 \sin t\left(2 \cos ^2 t-1\right)\right)}{\sin ^2 t\left(3 \cos t\left(1-2 \sin ^2 2 t\right)+\sin t \cdot 2 \sin t \cos t\right)} \)
\( \left(\because \sin 2 x=2 \sin x \cos x \text { and } \cos 2 x=2 \cos ^2 x-1 \text { and } \cos 2 x=1-2 \sin ^2 x\right) \)
\( =\frac{\cos ^2 t\left(2 \sin t \cos ^2 t-6 \sin t \cos ^2 t+3 \sin t\right)}{\sin ^2 t\left(3 \cos t-6 \cos t \sin ^2 t+2 \sin ^2 \cos t\right)} \)
\( =\frac{\sin t \cos t\left(-4 \cos ^3 t+3 \cos t\right)}{\sin t \cos t\left(3 \sin t-4 \sin ^3 t\right)} \)
\( \frac{d y}{d x}=\frac{-4 \cos ^3 t+3 \cos t}{3 \sin t-4 \sin ^3 t}=\frac{-\cos 3 t}{\sin 3 t}=-\cot 3 t \)
\( \left(\because \sin 3 t=3 \sin t-4 \sin ^3 t \text { and } \cos 3 t=4 \cos ^3 t-3 \cos t\right) \)
Therefore, the answer is \(\frac{d y}{d x}=-\cot 3 t\)

Let \(\theta\) be the semi-vertical angle of the cone.

It is clear that \(\theta \in\left(0, \frac{\pi}{2}\right)\).
Let r, h and l be the radius, height and the slant height of the cone, respectively.
Since, slant height of the cone is given, so consider it as constant.
Now, in \(\triangle A B C, r=l \sin \theta\) and \(h=l \cos \theta\)
Let V be the volume of the cone.
Then, \(V=\frac{\pi}{3} r^2 h \Rightarrow V=\frac{1}{3} \pi\left(l^2 \sin ^2 \theta\right)(l \cos \theta)\)
\(\Rightarrow \quad V=\frac{1}{3} \pi l^3 \sin ^2 \theta \cos \theta\)
On differentiating both sides w.r.t. \(\theta\) two times, we get
\(\frac{d V}{d \theta} =\frac{l^3 \pi}{3}\left[\sin ^2 \theta(-\sin \theta)+\cos \theta(2 \sin \theta \cos \theta)\right] \)
\(=\frac{l^3 \pi}{3}\left(-\sin ^3 \theta+2 \sin \theta \cos ^2 \theta\right)\)
\( \text { and } \frac{d^2 V}{d \theta^2}=\frac{l^3 \pi}{3}\left(-3 \sin ^2 \theta \cos \theta+2 \cos ^3 \theta\right. - 4 \sin^2 \theta \cos \theta)\)
\(\Rightarrow \frac{d^2 V}{d \theta^2}=\frac{l^3 \pi}{3}\left(2 \cos ^3 \theta-7 \sin ^2 \theta \cos \theta\right)\)
For maxima or minima, put\(\frac{d V}{d \theta}=0\)
\(\Rightarrow \quad \sin ^3 \theta=2 \sin \theta \cos ^2 \theta \Rightarrow \tan ^2 \theta=2 \)
\( \Rightarrow \quad \tan \theta=\sqrt{2} \Rightarrow \theta=\tan ^{-1} \sqrt{2}\)
Now, when \(\theta=\tan ^{-1} \sqrt{2}\), then \(\tan ^2 \theta=2\)
\(\Rightarrow \quad \sin ^2 \theta=2 \cos ^2 \theta\)
Now, we have \(\frac{d^2 V}{d \theta^2} =\frac{l^3 \pi}{3}\left(2 \cos ^3 \theta-14 \cos ^3 \theta\right)\)
\(=-4 \pi l^3 \cos ^3 \theta<0, \text { for } \theta \in\left(0, \frac{\pi}{2}\right)\)
\(\therefore V\) is maximum, when \(\theta=\tan ^{-1} \sqrt{2}\) or
\(\theta  =\cos ^{-1} \frac{1}{\sqrt{3}} \)
\({[\because \cos \theta}  \left.=\frac{1}{\sqrt{1+\tan ^2 \theta}}=\frac{1}{\sqrt{1+2}}=\frac{1}{\sqrt{3}}\right]\)
Hence, for given slant height, the semi-vertical angle of the cone of maximum volume is \(\cos ^{-1} \frac{1}{\sqrt{3}}\).

Let radius of cone be x and its height be h.
\(\therefore\) OD = (h - r)

Volume of cone (V)
\(=\frac { 1 }{ 3 } \pi { x }^{ 2 }h\) ...(i)
In \(\Delta OCD,\quad { x }^{ 2 }+({ h-r) }^{ 2 }={ r }^{ 2 }or\quad { x }^{ 2 }={ r }^{ 2 }-{ (h-r) }^{ 2 }\)
\(\therefore V=\frac { 1 }{ 3 } \pi h\{ { r }^{ 2 }-(h-r{ ) }^{ 2 }\} \)
\(=\frac { 1 }{ 3 } \pi (-{ h }^{ 3 }+{ 2h }^{ 2 }r)\)
\(\Rightarrow \frac { dV }{ dh } =\frac { \pi }{ 3 } (-3{ h }^{ 2 }+4hr)\)
\(\therefore \quad \frac { dV }{ dh } =0\Rightarrow h=\frac { 4r }{ 3 } \)
\(\frac { { d }^{ 2 }V }{ { dh }^{ 2 } } =\frac { \pi }{ 3 } (-6h+4r)\)
\(=\frac { \pi }{ 3 } \left( -6\left( \frac { 4r }{ 3 } \right) +4r \right) \)
\(=-\frac { 4\pi r }{ 3 } <0\)
\(\therefore \ at\quad h=\frac { 4r }{ 3 } \), Volume is maximum
Maximum volume
\(=\frac { 1 }{ 3 } \pi .\left\{ -{ \left( \frac { 4r }{ 3 } \right) }^{ 3 }+2{ \left( \frac { 4r }{ 3 } \right) }^{ 2 }r \right\} \)
\(=\frac { 8 }{ 27 } .\left( \frac { 4 }{ 3 } \pi { r }^{ 3 } \right) \)
\(=\frac { 8 }{ 27 } \) (volume of sphere)

Given: \(y=(x+\sqrt{x^2+1})^m\)
Differentiate with respect to 'x',
\(y'=m\{x+\sqrt{x^2+1}\}^{m-1}\{1+\frac{2x}{2\sqrt{x^2+1}}\}\)
\(=m\{x+\sqrt{x^2+1}\}^{m-1}\{1+\frac{\sqrt{x^2+1}+x}{\sqrt{x^2+1}}\}\)
\(=\{\frac{my}{\sqrt{x^2+1}}\}\)
\(\Rightarrow y'\sqrt{x^2+1}=my\)
Squaring both sides,
\(y'^2(x^2+1)=(m^2y^2)\)
Differentiate again with respect to 'x'
\(2y'y"(x^2+1)+2xy'^2=2m^2yy'\)
\(\Rightarrow y'[(x^2+1)y"+xy']=m^2yy'\)
\(\Rightarrow \ (x^2+1)\frac{d^2y}{dx^2}+x\frac{dy}{dx}-m^2y=0\)

Given \(asec^{ 3 }\theta \) 
\(\therefore \frac { dx }{ d\theta } =3asec^{ 3 }\theta tan\theta \)
\(y=atan^{ 3 }\theta \)
\(\therefore \frac { dx }{ d\theta } =tan^{ 2 }\theta sec^{ 2 }\theta \)
\(\frac { dy }{ dx } =\frac { dy/d\theta }{ dx/d\theta } \)
\(\Rightarrow \frac { dy }{ dx } =\frac { 3atan^{ 2 }\theta sec^{ 2 }\theta }{ 3asec^{ 3 }\theta tan\theta } =sin\theta \)
\(\frac { dy }{ dx } =sin\frac { \pi }{ 4 } \)
\( =\frac { 1 }{ \sqrt { 2 } } \)

\(f(x)=x^5-5x^4+5x^3-1\)
\(\Rightarrow f'(x)=5x^4-20x^3+15x^2\)
\(\Rightarrow f'(x)=5x^2(x-1)(x-3)\)
\(f'(x)=0\Rightarrow x=0,x=1,x=3\)
\(f''(x)=20x^3-60x^2+30x\)
\(=10x[2x^2-6x+3]\)
\(f'(0)=0,f'(1)=-ve,f'(3)=+ve\)
also , \(f'(<0)=-ve\)
\(f'(>0)=-ve\)
x = 0 is a point of inflexion
x = 1 is a point of local maxima
x = 3 is a point of local minima
∴ Local max. value = f(1) = 0
and Local min. value = f(3) = -28

\(\frac { dy }{ dx } =-2{ cos }^{ 3 }y\quad sin \ \ y\)

We have,\(f(x)=x^{3}-3 x, x \in[0, \sqrt{3}]\) 
For f(x), Rolle's theorem is satisfied
\( f^{\prime}(c) =0 \left[\because f^{\prime}(x)=3 x^{2}-3\right] \)
\(\Rightarrow 3 c^{2}-3 =0 \)
\( \Rightarrow c^{2}=\frac{3}{3}=1 \)
\( \Rightarrow c=\pm 1, \text { where } 1 \in(0, \sqrt{3}) \)
\(\therefore c=1 \)

\(y=\sin \left(\cos x^{2}\right) \) 
Therefore,\(\frac{d y}{d x}=\frac{d}{d x} \sin \left(\cos x^{2}\right)\) 
\( =\cos \left(\cos x^{2}\right) \frac{d}{d x}\left(\cos x^{2}\right) \\ =\cos \left(\cos x^{2}\right)\left(-\sin x^{2}\right) \frac{d}{d x}\left(x^{2}\right) \\ =-\sin x^{2} \cos \left(\cos x^{2}\right)(2 x) \\ =-2 x \sin x^{2} \cos \left(\cos x^{2}\right) \)

As y’ = 3x² – 3, for a point of absolute maximum or minimum y’=0 ⇒ x = ± 1.
y]_x=0 = 2,
y]_x=1 = 1 – 3 + 2 = 0,
y]_x=-1 = -1 +3+ 2 = 4,
y]_x=2 = 8 – 6 + 2 = 4

As y' = 5Ae^5x - 5Be^-5x
and y'' = 25Ae^5x + 25Be^-5x
= 25y

As x^o = \(\frac { \pi }{ 180 } { x }^{ c }\)
\(\therefore \frac { d }{ dx } (cot{ x }^{ o })=\)\(\frac { d }{ dx } \left( cot\frac { \pi }{ 180 } x \right) \)
\(=-\frac { \pi }{ 180 } { cosec }^{ 2 }\frac { \pi }{ 180 } x\)
\(=-{ 1 }^{ o }{ cosec }^{ 2 }\frac { \pi }{ 180 } x\)
\(=-{ 1 }^{ o }{ cosec }^{ 2 }{ x }^{ 0 }\)

We have, \(f(x)=\tan x-4 x\)
\(\Rightarrow \quad f^{\prime}(x)=\sec ^2 x-4\)
(i) For f(x) to be strictly increasing
\( f^{\prime}(x)>0 \)
\( \Rightarrow \quad \sec ^2 x-4>0 \Rightarrow \frac{1-4 \cos ^2 x}{\cos ^2 x}>0\)
\( \Rightarrow \frac{(1-2 \cos x)(1+2 \cos x)}{\cos ^2 x}>0\)
\( \therefore \quad \cos x \in\left(-\frac{1}{2}, \frac{1}{2}\right) \)
\( \Rightarrow \quad x \in\left(\frac{\pi}{3}, \frac{\pi}{2}\right)\)
\(\therefore f(x)\) is strictly increasing on \(\left(\frac{\pi}{3}, \frac{\pi}{2}\right)\)
\(\text { (ii) For } f(x) \text { to be strictly decreasing } f^{\prime}(x)<0\)
\( \Rightarrow \quad \sec ^2 x-4<0\)
\( \Rightarrow \quad \frac{1-4 \cos ^2 x}{\cos ^2 x}<0\)
\( \Rightarrow \quad \frac{(1-2 \cos x)(1+2 \cos x)}{\cos ^2 x}<0\)
\( \Rightarrow \cos x \in\left(-\infty,-\frac{1}{2}\right) \cup\left(\frac{1}{2}, \infty\right)\)
\( \Rightarrow \quad x \in\left(0, \frac{\pi}{3}\right)\)  \( {\left[\because x \in\left(0, \frac{\pi}{2}\right)\right]}\)
\(\Rightarrow \quad \sec ^2 x-4<0\)
\(\Rightarrow \quad \frac{1-4 \cos ^2 x}{\cos ^2 x}<0\)
\(\Rightarrow \quad \frac{(1-2 \cos x)(1+2 \cos x)}{\cos ^2 x}<0\)
\(\Rightarrow \cos x \in\left(-\infty,-\frac{1}{2}\right) \cup\left(\frac{1}{2}, \infty\right)\)
\(\Rightarrow \quad x \in\left(0, \frac{\pi}{3}\right)\) \(\left[\because x \in\left(0, \frac{\pi}{2}\right)\right]\)
\(\therefore f(x)\) is strictly dcreasing on \(\left(\frac{\pi}{3}, 0\right)\)

Let a be the side of cube, s be the surface area and v be volume of cube.
Given, \(\frac{d s}{d t}=72 \mathrm{~cm}^2 / \mathrm{sec}\) and a = 3 cm
Surface area of cube = 6a²
\(\Rightarrow \quad \frac{d s}{d t}=12 a \cdot \frac{d a}{d t} \Rightarrow \frac{d a}{d t}=\frac{72}{12 a}=\frac{6}{a}\)
Now, volume of cube, v = a³
On differentiating w.r.t. t, we get
\(\frac{d v}{d t}=3 a^2 \cdot \frac{d a}{d t}=3 a^2 \cdot \frac{6}{a} \Rightarrow \frac{d v}{d t}=18 a\)
\(\Rightarrow \quad\left(\frac{d v}{d t}\right)_{a=3}=18 \times 3=54 \mathrm{~cm}^3 / \mathrm{sec}\)

We have, f(x) = |x - 5|
Test for differentiability at x =5
LHD = f'(5^-) \(=\lim _{h \rightarrow 0} \frac{f(5-h)-f(5)}{-h}\)
\(\begin{aligned} {\left[\because L f^{\prime}(a)=\lim _{h \rightarrow 0} \frac{f(a-h)-f(a)}{-h}\right]} \end{aligned}\)
\(\begin{aligned} =\lim _{h \rightarrow 0} \frac{|5-h-5|-|5-5|}{-h} \end{aligned}\)
\(\begin{aligned} =\lim _{h \rightarrow 0} \frac{|-h|}{-h}=\lim _{h \rightarrow 0} \frac{h}{-h}=-1 \end{aligned}\)
\([\because|-x|=x \text {, if } x>0]\)
RHD = f'(5⁺) \(=\lim _{h \rightarrow 0} \frac{f(5+h)-f(5)}{h}\)
\(\left[\because R f^{\prime}(a)=\lim _{h \rightarrow 0} \frac{f(a+h)-f(a)}{h}\right]\)
\(\begin{aligned} =\lim _{h \rightarrow 0} \frac{|5+h-5|-|5-5|}{h} \end{aligned}\)
\(\begin{aligned} =\lim _{h \rightarrow 0} \frac{|h|}{h}=\lim _{h \rightarrow 0} \frac{h}{h}=1 \quad[\because|x|=x \text {, if } x>0] \end{aligned}\)
Since, LHD \(≠\) RHD at x = 5
So, f is not differentiable.

Given, \(f(x)=\left\{\begin{array}{cc} x \sin \frac{1}{x}, & \text { if } x \neq 0 \\ 0, & \text { if } x=0 \end{array}\right.\) and point is x = 0.
At x = 0, f(0) = 0
LHL = \(\lim _{x \rightarrow 0^{-}} f(x)=\lim _{h \rightarrow 0} f(0-h)\)
\(\begin{aligned} =\lim _{h \rightarrow 0} f(-h) \end{aligned}\)
\(\begin{aligned} =\lim _{h \rightarrow 0}(-h) \sin \left(\frac{1}{-h}\right) \end{aligned}\)
\(\begin{aligned} =\lim _{h \rightarrow 0} h \sin \frac{1}{h} \quad[\because \sin (-\theta)=-\sin \theta] \end{aligned}\)
= 0 \(\times\) (An oscillating value lies between - 1 and 1)
= 0
RHL = \(\lim _{x \rightarrow 0^{+}} f(x)=\lim _{h \rightarrow 0} f(0+h)\)
\(=\lim _{h \rightarrow 0} f(h)=\lim _{h \rightarrow 0} h \sin \left(\frac{1}{h}\right)\)
= 0 \(\times\) (An oscillating value lies between -1 and 1)
= 0
Here, LHL = f(0) = RHL = 0
Hence, f(x) is continuous at x = 0.

Volume and surface area of sphere  \(\mathrm{V}=\frac{4}{3} \pi \mathrm{r}^{3}\) 
\(\mathrm{S}=4 \pi \mathrm{r}^{2}\) 
Rate of volume and surface area is \(\frac{d V}{d t}=\frac{4}{3} \pi \times 3 r^{2} \times d r / d t\) 
\( \frac{\mathrm{dS}}{\mathrm{dt}}=4 \pi \times 2 \mathrm{r} \times \mathrm{dr} / \mathrm{dt} \)
\(\frac{\mathrm{d} \mathrm{V}}{\mathrm{dt}}=8 \)
\(\Rightarrow \frac{4}{3} \pi \times 3 \mathrm{r}^{2} \times \frac{\mathrm{dr}}{\mathrm{dt}}=8 \)
\(\Rightarrow \frac{\mathrm{dr}}{\mathrm{dt}}=\frac{8 \times 7}{22 \times 4 \times 12^{2}}=0.00441 \)
\(\text { Now } \frac{\mathrm{dS}}{\mathrm{dt}}=4 \pi \times 2 \mathrm{r} \times \frac{\mathrm{dr}}{\mathrm{dt}}=\frac{4 \times 22 \times 2 \times 12}{7} \times 0.00441=65.33 \mathrm{~cm}^{2} / \mathrm{s}\)

The area of a circle (A) with radius (r) is given by.
Therefore, the rate of change of area (A) with respect to time (t) is given by,
[By chain rule]
It is given that.
Thus, when r = 8 cm,

Hence, when the radius of the circular wave is 8 cm, the enclosed area is increasing at the rate of 80 π cm²/s.

If continuous at x = 0
\(\underset{x=0}{\mathrm{LHL}}=\underset{x=0}{\mathrm{RHL}}=f(0) \)
\(\Rightarrow \operatorname{Lim}_{x \rightarrow 0} \frac{k x}{|x|}=\operatorname{Lim}_{x \rightarrow 0}(3)=3 \)
\(\Rightarrow \frac{-k x}{x}=3=3 \Rightarrow k=-3\)

(d) Given, f(x) = 2x³ - 3x² - 36x + 7
\(\Rightarrow\) Differentiating w.r.t. x,
f'(x) = \(\frac{d}{dx}\)(2x³ - 3x² - 36x + 7)
= 2.3 x² - 3. 2x - 36.1 + 0
= 6x² - 6x - 36 = 6(x² - x - 6)
\(\Rightarrow\) f'(x) = 6(x - 3)(x + 2)
On putting f'(x) = 0, we get
6(x - 3)(x+2) = 0 \(\Rightarrow\) x = 3 and -2
which divides real line into three intervals namely
(-\(\infty\), -2), (-2, 3) and (3, \(\infty\)).

Thus, the given function f is strictly increasing in intervals (-\(\infty\),- 2) and(3, \(\infty\)), while function f is strictly decreasing in the interval (- 2, 3).
Hence, Assertion is incorrect but Reason is correct.

(c) Assertion is correct, Reason is incorrect

(i) Given, function is \(f(x)=-0.1 x^2+m x+98.6\)
\(\because f(x)\) is a polynomial function.
\(\therefore\) It is differentiable everywhere and hence, differentiable in (0,12).
(ii) We have, \(f(x)=-0.1 x^2+m x+98.6\)
\(\therefore \quad f^{\prime}(x)=-0.2 x+m\)
because 6 is the critical point of the function.
\(\therefore  f^{\prime}(6)=0 \)
\(\Rightarrow  -0.2(6)+m=0 \Rightarrow m=1.2\)
(iii) We have, \(f(x)=-0.1 x^2+1.2 x+98.6\)
\(\therefore\)\(f(x)=-0.2 x+1.2 = -0.2(x - 6)\)
 \(\begin{array}{c|c} \hline \text { Interval } & \text { Sign of } f^{\prime}(x) \\ \hline(0,6) & + \text { ve } \\ \hline(6,12) & - \text { ve } \\ \hline \end{array}\)
Hence, f(x) is strictly increasing in the interval (0, 6) and strictly decreasing in the interval (6, 12).
Or We have, \(f(x)=-0.1 x^2+1.2 x+98.6\)
\(\therefore  f^{\prime}(x)=-0.2 x+1.2=-0.2(x-6) \)
\(\therefore  f^{\prime}(x)=0 \Rightarrow x=6\)
Now, \(f^{\prime \prime}=-0.2\)
and \(f^{\prime \prime}(6)=-0.2<0\)
Hence, x=6 is the point of local maximum and the local maximum value.
\(\Rightarrow f(6)=-0.1 \times(6)^2+(1.2 \times 6)+98.6=102.2\)
Now, \(f(0)=98.6, f(6)=102.2\) and \(f(12)=98.6\)
Hence, x=6 is the point of absolute maximum and the absolute maximum value of the function at x=6 is 102.2 and x=0, x=12 both are the points of absolute minimum and the absolute minimum value of the function is 98.6.

Given, $x$ is the side of the square base of the tank (in m ) and $h$ is the depth of the tank (in m ).
$\therefore$ Area of the base of the tank $=x^2 \mathrm{~m}^2$
$\therefore$ Cost of land $=₹ 5000 x^2$
Cost of digging the tank $=₹ 40000 h^2$, where $h$ is the depth of the tank (in m).
$\therefore$ Total cost of the tank $=₹\left(5000 x^2+40000 h^2\right)$
Given, capacity of tank $=250 \mathrm{~m}^3$
Volume of tank, $x^2 h=250$

$$
\Rightarrow \quad h=\frac{250}{x^2}
$$

(i) The total cost $C$ of digging the tank

$$
\begin{aligned}
C(x) & =5000 x^2+40000\left(\frac{250}{x^2}\right)^2 \\
& =5000\left(x^2+\frac{8 \times 625 \times 100}{x^4}\right) \\
& =5000\left(x^2+\frac{5 \times 10^5}{x^4}\right) \\
\Rightarrow C(x) & =₹ 5000\left(\frac{x^6+5 \times 10^5}{x^4}\right)
\end{aligned}
$$

Given, x is the side of the square base of the tank (in m ) and h is the depth of the tank (in m ).
Therefore Area of the base of the tank \(=x^2 \mathrm{~m}^2\)
Therefore Cost of land =₹ 5000 x²
Cost of digging the tank =₹ 40000 h², where h is the depth of the tank (in m).
Therefore Total cost of the tank = (5000 x²+40000 h²)
Given, capacity of tank = 250 m³
Volume of tank, x² h=250
\(\Rightarrow \quad h=\frac{250}{x^2}\)
(i) The total cost C of digging the tank
\(C(x)  =5000 x^2+40000\left(\frac{250}{x^2}\right)^2\)
\(=5000\left(x^2+\frac{8 \times 625 \times 100}{x^4}\right)\)
\(=5000\left(x^2+\frac{5 \times 10^5}{x^4}\right)\)
\(\Rightarrow C(x)  =₹ 5000\left(\frac{x^6+5 \times 10^5}{x^4}\right)\)
​​(ii) \(\frac{d C(x)}{d x}\)
\(=5000\left[\frac{x^4\left(6 x^5+0\right)-\left(x^6+5 \times 10^5\right) \times 4 x^3}{x^8}\right] \)
\(=5000\left(\frac{2 x^9-20 \times 10^5 x^3}{x^8}\right) \)
\(=5000\left(\frac{2 x^6-20 \times 10^5}{x^5}\right)\)

\(\text { Given, } h(t)=-\frac{7}{2} t^2+\frac{13}{2} t+1\)
(i) Yes,
Since, h(t) is a polynomial function.
So, it is everywhere continuous.
(ii) \( h^{\prime}(t)=-\frac{7}{2}(2 t)+\frac{13}{2} \)
\(\Rightarrow \quad h^{\prime}(t)=-7 t+\frac{13}{2}\)
Also, \(h^{\prime \prime}(t)=-7\)
For maxima or minima, put \(h^{\prime}(t)=0\)
\(\Rightarrow  -7 t+\frac{13}{2}=0 \)
\(\Rightarrow  -7 t=-\frac{13}{2}\)
\(\Rightarrow  t=\frac{13}{14}\)
At \(t=\frac{13}{14}, h^{\prime \prime}(t)<0\)
So, \(t=\frac{13}{14}\) is point of maxima.
Thus, at \(t=\frac{13}{14}\) the height of the ball is maximum.