Since, $$f$$ and $$g$$ both are continuous functions on [0, 1] and differentiable on (0, 1) then $$\exists \,c \in \left( {0,1} \right)$$   such that
$$\eqalign{ & f'\left( c \right) = \frac{{f\left( 1 \right) - f\left( 0 \right)}}{1} = \frac{{6 - 2}}{1} = 4 \cr & {\text{and }}g'\left( c \right) = \frac{{g\left( 1 \right) - g\left( 0 \right)}}{1} = \frac{{2 - 0}}{1} = 2 \cr & {\text{Thus, we get }}f'\left( c \right) = 2g'\left( c \right) \cr} $$

$$\eqalign{ & f\left( x \right) = {\tan ^{ - 1}}\left( {\sqrt {\frac{{1 + \sin x}}{{1 - \sin x}}} } \right) \cr & = {\tan ^{ - 1}}\left( {\sqrt {\frac{{{{\left( {\sin \frac{x}{2} + \cos \frac{x}{2}} \right)}^2}}}{{{{\left( {\sin \frac{x}{x} - \cos \frac{x}{2}} \right)}^2}}}} } \right) \cr & = {\tan ^{ - 1}}\left( {\frac{{1 + \tan \frac{x}{2}}}{{1 - \tan \frac{x}{2}}}} \right) \cr & = {\tan ^{ - 1}}\left( {\tan \left( {\frac{\pi }{4} + \frac{x}{2}} \right)} \right) \cr & \Rightarrow y = \frac{\pi }{4} + \frac{x}{2} \cr & \Rightarrow \frac{{dy}}{{dx}} = \frac{1}{2} \cr & {\text{Slope of normal }} = \frac{{ - 1}}{{\left( {\frac{{dy}}{{dx}}} \right)}} = - 2 \cr & {\text{At }}\left( {\frac{\pi }{6},\frac{\pi }{4} + \frac{\pi }{{12}}} \right) \cr & y - \left( {\frac{\pi }{4} + \frac{\pi }{{12}}} \right) = - 2\left( {x - \frac{\pi }{6}} \right) \cr & y - \frac{{4\pi }}{{12}} = - 2x + \frac{{2\pi }}{6} \cr & y - \frac{\pi }{3} = - 2x + \frac{\pi }{3} \cr & y = - 2x + \frac{{2\pi }}{3} \cr} $$
This equation is satisfied only by the point $$\left( {0,\frac{{2\pi }}{3}} \right)$$

$$\eqalign{ & \frac{{dx}}{{d\theta }} = a\left( { - \sin \theta + \sin \theta + \theta \cos \theta } \right) = a\theta \cos \theta \,......\left( 1 \right) \cr & \frac{{dy}}{{d\theta }} = a\left( {\cos \theta - \cos \theta + \theta \sin \theta } \right) = a\theta \sin \theta \,......\left( 2 \right) \cr} $$
Dividing (2) by (1), we get
$$\eqalign{ & \frac{{dy}}{{dx}} = \tan \theta \,\left( {{\text{slope of tangent}}} \right) \cr & \therefore {\text{Slope of normal }} = - \cot \theta \cr & \therefore {\text{Equation of normal is}} \cr & y - a\left( {\sin \theta - \theta \cos \theta } \right) = - \frac{{\cos \theta }}{{\sin \theta }}\left( {x - a\left( {\cos \theta + \theta \sin \theta } \right)} \right) \cr & \Rightarrow y\sin \theta - a{\sin ^2}\theta + a\sin \theta \cos \theta \cr & = - x\cos \theta + a{\cos ^2}\theta + a\theta \sin \theta \cos \theta \cr & \Rightarrow x\cos \theta + y\sin \theta = a \cr} $$
As $$\theta $$ varies inclination is not constant.
∴ (A) is not correct.
Clearly does not pass through (0, 0).
It's distance from origin $$ = \left| {\frac{a}{{\sqrt {{{\cos }^2}\theta + {{\sin }^2}\theta } }}} \right| = a$$
which is constant

$$\eqalign{ & {\text{Let }}y = \frac{1}{{40}}\left( {3{x^4} + 8{x^3} - 18{x^2} + 60} \right) \cr & \Rightarrow \frac{{dy}}{{dx}} = \frac{1}{{40}}\left( {12{x^3} + 24{x^2} - 36x} \right) \cr & {\text{and }}\frac{{{d^2}y}}{{d{x^2}}} = \frac{1}{{40}}\left( {36{x^2} + 48x - 36} \right) \cr & {\text{Now, }}\frac{{dy}}{{dx}} = 0 \Rightarrow {x^3} + 2{x^2} - 3x = 0 \cr & {\text{or }}x\left( {x - 1} \right)\left( {x + 3} \right) = 0 \cr & {\text{or}}\,\,x = 0,\,1,\, - 3 \cr & {\text{At }}x = 0,\,\frac{{{d^2}y}}{{d{x^2}}} = - 36 < 0 \cr & \therefore y{\text{ is maximum at }}x = 0 \cr & \Rightarrow {\text{The given function i}}{\text{.e}}{\text{., }}\frac{1}{y}{\text{ is minimum at }}x = 0 \cr & \therefore {\text{minimum value of the function}} = \frac{{40}}{{60}} = \frac{2}{3} \cr} $$

$$\eqalign{ & y = \cos \left( {x + y} \right)......\left( 1 \right) \cr & \therefore \,\frac{{dy}}{{dx}} = - \sin \left( {x + y} \right)\left\{ {1 + \frac{{dy}}{{dx}}} \right\} \cr & = - \frac{{\sin \left( {x + y} \right)}}{{1 + \sin \left( {x + y} \right)}} \cr & = - \frac{1}{2} \cr & \Rightarrow \sin \left( {x + y} \right) = 1,{\text{ so}}\,\cos \left( {x + y} \right) = 0 \cr & \therefore \,{\text{from equation}}\left( 1 \right)y = 0{\text{ and }}\left( {x + y} \right) = 2n\pi + \frac{\pi }{2} \cr & {\text{Tangent at }}\left( {\frac{\pi }{2},\,0} \right){\text{ is }}x + 2y = \frac{\pi }{2} \cr} $$

$$\eqalign{ & {\text{Here, }}f\left( x \right) = - \left( {2x + 1} \right) - 2\left\{ { - \left( {x - 1} \right)} \right\} = - 3,\,x < - \frac{1}{2} \cr & 2x + 1 - 2\left\{ { - \left( {x - 1} \right)} \right\} = 4x - 1,\, - \frac{1}{2} \leqslant x < 1 \cr & 2x + 1 - 2\left( {x - 1} \right) = 3,\,x \geqslant 1 \cr} $$
$$f\left( x \right)$$  is the constant $$-3$$  in $$\left( { - \infty ,\, - \frac{1}{2}} \right)$$   and the constant $$3$$ in $$\left[ {1,\, + \infty } \right)$$
In $$ - \frac{1}{2} \leqslant x < 1,\,f'\left( x \right) = 4 > 0$$       and so, $$f\left( x \right)$$  is m.i. in $$\left[ { - \frac{1}{2},\,1} \right)$$
In this interval, $$f\left( x \right)$$  increases from $$-3$$  to $$3 - \in $$
As the function is continuous, range $$ = \left[ { - 3,\,3} \right]$$

We have $$g'\left( x \right) = f'\left( {\frac{x}{2}} \right) - f'\left( {2 - x} \right)$$
Given $$f''\left( x \right) < 0\,\forall \,x\, \in \left( {0,\,2} \right)$$
So, $$f'\left( x \right)$$  is decreasing on $$\left( {0,\,2} \right)$$
Let $$\frac{x}{2} > 2 - x{\text{ or }}f'\left( {\frac{x}{2}} \right) < f'\left( {2 - x} \right)$$
Thus, $$\forall \,x > \frac{4}{3},\,g'\left( x \right) < 0$$
Therefore, $$g\left( x \right)$$  decreasing in $$\left( {\frac{4}{3},\,2} \right)$$  and increasing in $$\left( {0,\,\frac{4}{3}} \right).$$

For $$f\left( x \right) = {x^6}$$   and $$f\left( x \right) = {x^7},$$
$$f'\left( 0 \right) = f''\left( 0 \right) = f'''\left( 0 \right) = {f^{iv}}\left( 0 \right) = {f^v}\left( 0 \right) = 0$$
$$x = 0$$  is point of minima for $$f\left( x \right) = {x^6}$$
But $$x = 0$$  is not point of maxima/minima for $$f\left( x \right) = {x^7}$$
Hence, it is difficult to say anything.

$${\text{Area = }}\frac{1}{2}{x^2}\sin \theta $$


Maximum value of $$\sin \theta $$  is 1 $$\theta = \frac{\pi }{2}$$
$${A_{\max }} = \frac{1}{2}{x^2}$$

$$\eqalign{ & {\text{Let }}f\left( x \right) = \left| x \right| \cr & f'\left( 0 \right) \ne 0{\text{ but }}f\left( x \right){\text{ has a minimum at }}x = 0 \cr} $$