Let \(\theta\) denote the angle at instant t
\(\frac{d\theta}{dt}=2\frac{d}{dt}(\sin \theta)\)
\(\frac{d\theta}{dt}=2\cos\theta.(\frac{d\theta}{dt})\)
\(1=2\cos\ \theta\)
\(2\cos\theta=1 cos\theta=\frac{1}{2}\)
\(\Rightarrow \theta=\cos^{-1}(\frac{1}{2})\)
Hence required angles is \(\frac{\pi}{3}\)

Let x=0.04
\(x+\Delta x=0.037\)
Then   \(\Delta x=0.037-0.040\)
\(\Rightarrow \Delta x=-0.003 \)
\(y=x^{1/2} \)
\(\Rightarrow (0.04)^{1/2}=0.2 \)
\(y=x^{1/2}\)
\(\Rightarrow \frac{dy}{dx} =\frac{1}{2\sqrt x}\)
\(\Rightarrow \Delta y=\frac{\Delta x}{2\sqrt x}\)
\(\Rightarrow \Delta y=\frac{-0.003}{2\sqrt {0.04}}=\frac{-0.003}{2\times 0.2}\)
\(=\frac{-3}{4\times 100}=\frac{-0.75}{100}\)
\(\Rightarrow \Delta y=-0.0075 \)
\(y+\Delta y=0.2-0.0075\)
\(=0.1925\)
\(\Rightarrow 0.037=0.1925\)

Given, \(y=x^2-7x+18\)
\(\frac{dy}{dx}=2x-7\)
\(\frac{dy}{dx} \ at\ (3,0)=2(3)-7 \)
\(=6-7=-1\)
\(\frac{dy}{dx}\ at\ (4,0)=2(4)-7\)
\(=8-7=1\)
\(m_1=-1, m_2=1\)
\(m_1\times m_2=-1\times 1=-1\)
Hence tangent are at right angles to each other,

\(y^2 =8x + 3\)
\(2y\frac{dy}{dt}=8\frac{dx}{dt}\)
\(\frac{dy}{dt}=8\frac{dx}{dt}\)
\(2y\cdot8\frac{dx}{dt}=8\frac{dx}{dt}\)
\(\Rightarrow y=\frac{8}{16}=\frac{1}{2}\)
\(y=\frac{1}{2},y^2=8x+3\)
\((\frac{1}{2})^2=8x+3\)
\(\frac{1}{4}-3=8x\)
\(\Rightarrow x=\frac{1-12}{4}\times \frac{1}{8}=-\frac{11}{32}\)
Points  \((-\frac{11}{32},\frac{1}{2})\)

Let x denote the edge of cube, v denote the volume and s denotes the surface area of cube at instant t.
\(\frac{dv}{dt}=9\ cm^2/sec.\ \ \ x=10cm\)
\(v=x^3\)
\(​​\frac{dv}{dt}=3x^2\frac{dx}{dt}\)
\(\frac{9}{3x^2}=\frac{dx}{dt}\)
Now, s=6x² 
\(\frac{ds}{dt}=12x\times\frac{dx}{dt}\)
\(\frac{ds}{dt}=12\times10 \times \frac{9}{3\times 10\times 10}\)
\(\frac{12\times 9}{3\times 10}=\frac{36}{10}\)
\(=3.6\ cm^2/sec\)

Maximum value = 3
Minimum value = nil

0.75

(i)\(y=\pm 5.\)  \(\therefore\) points are:\((0,\pm 5)\)
(ii)\(x=\pm2\)  \(\therefore\) points are:\((\pm2,0)\)

Let 'r' be the radius and 'h' the length of the cylinder
By the question \(V=\pi r^{ 2 }h\) ..(1) [Given]
Now \(S=2\pi rh+\pi r^{ 2 }\)
\(=2\pi r\frac { V }{ \pi r^{ 2 } } +\pi r^{ 2 }\)  [Using (1)]
=\(2\frac { V }{ r } +\pi r^{ 2 }\)
\(\therefore \)  \(\frac { ds }{ dr } =-\frac { 2V }{ r^{ 2 } } =2\pi r\)
and   \(\frac { d^{ 2 }S }{ dr^{ 2 } } =\frac { 4V }{ r^{ 3 } } +2r>0\)
S is minimum when  \(\frac { dS }{ dr } =0and\frac { d^{ 2 }S }{ dr^{ 2 } } >0\)
Now \(\frac { dS }{ dr } =0\Rightarrow -\frac { 2V }{ r^{ 2 } } +2\pi r=0\)
\(\Rightarrow \)  \(V=\pi r^{ 3 }\)
Putting (1), \(\pi r^{ 3 }=\pi r^{ 2 }h\Rightarrow h=r\)
Hence S in maximum when the height is equal to the radius of the base.   

Let 'r' and 'h' be the radius and height respectively of the cylinder.
\(\therefore \) S, the surface area = \(\pi r^{ 2 }+2\pi rh\)
\(\Rightarrow \)  \(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)....(1)
And V,the volume = \(\pi r^{ 2 }h\)
i.e. \(V=\pi r^{ 2 }\left( \frac { S-\pi r^{ 2 } }{ 2\pi r } \right) \)  [Using (1)]
\(\Rightarrow \) \(V=\frac { r }{ 2 } (S-\pi r^{ 2 })\)
\(\Rightarrow \) \(V=\frac { r }{ 2 } (Sr-\pi r^{ 2 })\)
\(\therefore \)  \(\frac { dV }{ dr } =\frac { 1 }{ 2 } (S-3\pi r^{ 2 })\)...(2)
and   \(\frac { d^{ 2 }V }{ dr } =-\frac { 3\pi }{ 2 } (2r)=-3\pi r\)   (3)
For greatest volume \(\frac { dV }{ dr } =0\Rightarrow -\frac { 1 }{ 2 } (S-3\pi r^{ 2 })=0\)
\(\Rightarrow \) \(S=3\pi r^{ 2 }\Rightarrow r=\sqrt { \frac { S }{ 3\pi } } \)
Putting in (3),
\(\frac { d^{ 2 }V }{ dr^{ 2 } } =-3\pi \sqrt { \frac { S }{ 3\pi } } =-\sqrt { 3\pi S } \) which is -ve.
Thus V is the greatest when \(r=\sqrt { \frac { s }{ 3\pi } } \)
And from (1),
\(h=\frac { S-\pi \left( \frac { S }{ 3\pi } \right) }{ 2\pi \sqrt { \frac { S }{ 3\pi } } } =\frac { \frac { 2S }{ 3 } }{ \frac { 2 }{ \sqrt { 3 } } \sqrt { \pi S } } =\sqrt { \frac { s }{ 3\pi } } \)
Hence, height = Radius of the base.                   

Let \(x_{ 1 }\quad and\quad x_{ 2 }\in R\)
Now  \(x_{ 1 }<{ x }_{ 2 }\)
\(\Rightarrow 7x_{ 1 }<7x_{ 2 }\)
\( \Rightarrow 7x_{ 1 }-3<7x_{ 2 }-3\)
\(\Rightarrow f(x_{ 1 })\)
Hence,\(f\) is strictly increasing on \(R\)

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.
Here \(\frac { dV }{ dt } =9cm3/s \)
\(9=\frac{d \mathrm{~V}}{d t}=\frac{d}{d t}\left(x^3\right)=\frac{d}{d x}\left(x^3\right) \cdot \frac{d x}{d t}\)(By Chain Rule)
\( =3 x^2 \cdot \frac{d x}{d t} \\ \frac{d x}{d t} =\frac{3}{x^2} \ \ ...(1) \)
\(\frac{d S}{d t}=\frac{d}{d t}\left(6 x^2\right)=\frac{d}{d x}\left(6 x^2\right) \cdot \frac{d x}{d t}\)(By Chain Rule)
\(=12 x \cdot\left(\frac{3}{x^2}\right)=\frac{36}{x}\)(Using (1))
Hence when,  \(x=10 \mathrm{~cm}, \frac{d S}{d t}=3.6 \mathrm{~cm}^2 / \mathrm{s}\)

\(f^{\prime}(x) =1-\sin x \)
\(=\cos ^{2} \frac{x}{2}+\sin ^{2} \frac{x}{2}-2 \sin \frac{x}{2} \cos \frac{x}{2} \)
\(=\left(\cos \frac{x}{2}-\sin \frac{x}{2}\right)^{2}>0 \)
for [0, 2\(\pi\)] as square is always positive, Always increasing in [0, 2\(\pi\)]

Increasing:\((-\infty ,0);\)Decreasing:\((0,\infty)\)

The area A ofa circle with radius r is given by \(A=\pi r^{2}\)
\(\frac{d A}{d t}=2 \pi r \frac{d r}{d t} \Rightarrow 100=2 \pi r \times 5 \Rightarrow r=\frac{10}{\pi} \mathrm{cm} . \)
\({\left[\frac{d r}{d t}=5 \mathrm{~cm} / \mathrm{s}, \frac{d A}{d t}=100 \mathrm{~cm}^{2} / \mathrm{s}\right]}\) 
Hence, radius of the circle is \(r={10\over\pi}{cm}\)

\(V=\frac{4}{3} 4 \pi r^{3} \Rightarrow \frac{d V}{d t}=4 \pi r^{2} \frac{d r}{d t} \)
\(\left.\Rightarrow \frac{d r}{d t}=\frac{900}{4 \pi r^{2}} \Rightarrow \frac{d r}{d t}\right]_{r=30}=\frac{1}{4 \pi} \mathrm{cm} / \mathrm{s} ; \)
\(S=4 \pi r^{2} \Rightarrow \frac{d S}{d t}=8 \pi r \frac{d r}{d t}=\frac{8 \pi r \times 900}{4 \pi r^{2}} \)
\(\left.\Rightarrow \frac{d S}{d t}\right]_{r=30}=\frac{1800}{30}=60 \mathrm{~cm}^{2} / \mathrm{s} \)

\(x^{x^2-3}+(x-3)^{x^2}, \text { for } x>3\)
Let \(y=x^{x^2-3}+(x-3)^{x^2}\)
And let \(u=x^{x^2-3}, v=(x-3)^{x^2}\)
\(\boldsymbol{y}=\boldsymbol{u}+\boldsymbol{v}\)
Differentiating both sides w.r.t. x
\(\frac{d y}{d x}=\frac{d(u+v)}{d x} \)
\( \frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}\)
Differentiating both sides with respect to x, we obtain
\( \frac{1}{u} \frac{d u}{d x}=\log x \cdot \frac{d}{d x}\left(x^2-3\right)+\left(x^2-3\right) \cdot \frac{d}{d x}(\log x) \)
\( \Rightarrow \frac{1}{u} \frac{d u}{d x}=\log x \cdot 2 x+\left(x^2-3\right) \cdot \frac{1}{x} \)
\( \Rightarrow \frac{d u}{d x}=x^{x^2-1}\left[\frac{x^2-3}{x}+2 x \log x\right]\)
Now, v=(x-3)^x
Taking logarithm on both the sides, we obtain
\(\log v =\log (x-3)^{x^2} =x^2 \log (x-3)\)
Differentiating both sides with respect to x, we obtain
\(\frac{1}{v} \frac{d v}{d x}=\log (x-3) \cdot \frac{d}{d x}\left(x^2\right)+\left(x^2\right) \cdot \frac{d}{d x}[\log (x-3)] \)
\(\Rightarrow \frac{1}{v} \frac{d v}{d x}=\log (x-3) \cdot 2 x+x^2 \cdot \frac{1}{x-3} \cdot \frac{d}{d x}(x-3) \)
\( \Rightarrow \frac{d v}{d x}=v\left[2 x \log (x-3)+\frac{x^2}{x-3} \cdot 1\right] \)
\( \Rightarrow \frac{d v}{d x}=(x-3)^{x^2}\left[\frac{x^2}{x-3}+2 x \log (x-3)\right]\)
From (1), (2), and (3), we obtain