$$\eqalign{ & f\left( x \right) = 2{x^3} - 9a{x^2} + 12{a^2}x + 1 \cr & f'\left( x \right) = 6{x^2} - 18ax + 12{a^2};f''\left( x \right) = 12x - 18a \cr & {\text{For max}}{\text{. or min}}{\text{.}} \cr & 6{x^2} - 18ax + 12{a^2} = 0 \Rightarrow {x^2} - 3ax + 2{a^2} = 0 \cr & \Rightarrow x = a\,{\text{or}}\,x = 2a.\,{\text{At}}\,x = a\max .\,{\text{and}}\,{\text{at}}\,x = 2a\min \cr & \therefore p = a\,{\text{and}}\,q = 2a \cr & {\text{As}}\,{\text{per}}\,{\text{question}}\,{p^2} = q \cr & \therefore {a^2} = 2a \Rightarrow a = 2\,{\text{or}}\,a = 0 \cr & {\text{but}}\,a > 0,\,{\text{therefore}},\,a = 2. \cr} $$

$$\eqalign{ & 3\sin x - 4{\sin ^3}x = \sin 3x\,{\text{which}}\,{\text{increase}}\,{\text{for}}\, \cr & 3x \in \left( { - \frac{\pi }{2},\frac{\pi }{2}} \right) \Rightarrow x \in \left( { - \frac{\pi }{6},\frac{\pi }{6}} \right)\,{\text{whose}}\,{\text{length}}\,{\text{is}}\,\frac{\pi }{3}. \cr} $$

$$\eqalign{ & {\text{Let volume}} = V = \frac{4}{3}\pi {r^3}......\left( 1 \right) \cr & {\text{and surface area}} = S = 4\pi {r^2}......\left( 2 \right) \cr & {\text{Now,}} \cr & \left( 1 \right) \Rightarrow \frac{{dv}}{{dt}} = \frac{4}{3} \times 3\pi {r^2} \times \frac{{dr}}{{dt}} \cr & = 4\pi {r^2}\frac{{dr}}{{dt}}......\left( 3 \right) \cr & \left( 2 \right) \Rightarrow \frac{{ds}}{{dt}} = 4\pi \times 2 \times r\frac{{dr}}{{dt}} \cr & = \frac{{8\pi {r^2}}}{r}\frac{{dr}}{{dt}} \cr & = \frac{2}{r}\left[ {4\pi {r^2}\frac{{dr}}{{dt}}} \right]\, \cr & = \frac{2}{r}\frac{{dv}}{{dt}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{\text{from 3}}} \right) \cr} $$

$$\eqalign{ & f\left( x \right) = \frac{x}{2} + \frac{2}{x} \Rightarrow f'\left( x \right) = \frac{1}{2} - \frac{2}{{{x^2}}} = 0 \cr & \Rightarrow {x^2} = 4{\text{ or }}x = 2, - 2;\,f''\left( x \right) = \frac{4}{{{x^3}}} \cr & {\left. {f''\left( x \right)} \right]_{x = 2}} = + ve \Rightarrow f\left( x \right){\text{ has local min at }}x = 2. \cr} $$

$$\frac{{dy}}{{dx}} = \frac{{\frac{{dy}}{{d\theta }}}}{{\frac{{dx}}{{d\theta }}}} = \frac{{a\,\sin \,\theta }}{{a\left( {1 + \cos \,\theta } \right)}} = \tan \,\frac{\theta }{2}$$
$$\therefore $$ the rate of change of the slope, i.e., $$\frac{{dy}}{{dx}}$$
$$ = \frac{{d\left( {\frac{{dy}}{{dx}}} \right)}}{{dt}} = \frac{1}{2}{\sec ^2}\frac{\theta }{2}.\frac{{d\theta }}{{dt}} = \frac{k}{2}{\sec ^2}\frac{\theta }{2}$$
$$\therefore $$ the required rate $$ = \frac{k}{2}.{\sec ^2}\frac{\pi }{6} = \frac{k}{2}.{\left( {\frac{2}{{\sqrt 3 }}} \right)^2}$$

Consider a line
$$\eqalign{ & x\,\cos \,\theta + y\,\sin \,\theta = 2 \cr & \Rightarrow y\,\sin \,\theta = - x\,\cos \,\theta + 2 \cr & \Rightarrow y = - x\frac{{\cos \,\theta }}{{\sin \,\theta }} + \frac{2}{{\sin \,\theta }} \cr & \Rightarrow y = - x\,\cot \,\theta + 2\,{\text{cosec}}\,\theta \cr} $$
On comparing this equation with $$y = mx + c$$   we get
slope of line $$x\,\cos \,\theta + y\,\sin \,\theta = 2{\text{ is }} - \cot \,\theta $$
Also, we have a line $$x - y = 3\, \Rightarrow y = x - 3$$
slope of line $$x - y = 3$$  is $$1$$
Since, both the lines are perpendicular to each other.
$$\therefore $$  Product of their slopes $$ = - 1$$
$$\eqalign{ & \Rightarrow \left( { - \cot \,\theta } \right)\left( 1 \right) = - 1 \cr & \Rightarrow \cot \,\theta = 1 = \cot \frac{\pi }{4} \cr & \Rightarrow \theta = \frac{\pi }{4} \cr} $$

$$\eqalign{ & f\left( x \right) = \log \,x \cr & {\text{Clearly }}f\left( x \right){\text{ is increasing on}}\,\left( {0,\,\infty } \right) \cr & f\left( x \right) = {e^x} - x\,\log \,x \cr & f'\left( x \right) = {e^x} - \left( {\log \,x + 1} \right) \cr} $$

From the figure it is clear that $$f'\left( x \right) > 0$$   on $$\left( {1,\,\infty } \right).$$
So both statements (1) & (2) are correct.

$$\eqalign{ & f'\left( x \right) = - 12 + 18x - 6{x^2} \cr & = - 6\left( {{x^2} - 3x + 2} \right) \cr & = - 6\left( {x - 1} \right)\left( {x - 2} \right) \cr & \therefore \,f'\left( x \right) > 0\,{\text{if }}1 < x < 2\,{\text{and }}f'\left( x \right) < 0\,{\text{if }}2 < x \leqslant 4 \cr & \therefore \,f\left( x \right){\text{is}}\,\,{\text{m}}{\text{.i}}{\text{.}}\,\,{\text{in}}\,1 < x < 2\,{\text{and}}\,{\text{m}}{\text{.d}}{\text{.}}\,{\text{in}}\,\,2 < x \leqslant 4 \cr & \therefore {\text{absolute}}\,{\text{maximum}} = {\text{the}}\,{\text{greatest}}\,{\text{among}}\,\left\{ {f\left( 1 \right),\,f\left( 2 \right)} \right\} \cr & = {\text{the}}\,{\text{greatest}}\,{\text{among}}\,\left\{ {1,\,2} \right\} = 2 \cr} $$

$$\eqalign{ & {\text{If }} - \frac{\pi }{2} < x < 0{\text{ then,}} \cr & {\text{ }}y = {\cos ^{ - 1}}\left( {\cos \,x} \right) = {\cos ^{ - 1}}\left\{ {\cos \left( { - x} \right)} \right\} = - x \cr & \therefore \,\frac{{dy}}{{dx}} = - 1 \cr & \therefore {\left. {\frac{{dy}}{{dx}}} \right)_{x = - \frac{\pi }{4}}} = - 1 \cr} $$

$$\eqalign{ & f\left( x \right) = 2\,\log \left( {x - 2} \right) - {x^2} + 4x + 1 \cr & \Rightarrow f'\left( x \right) = \frac{2}{{x - 2}} - 2x + 4 \cr & \Rightarrow f'\left( x \right) = 2\left[ {\frac{{1 - {{\left( {x - 2} \right)}^2}}}{{x - 2}}} \right] \cr & \Rightarrow f'\left( x \right) = - 2\frac{{\left( {x - 1} \right)\left( {x - 3} \right)}}{{x - 2}} \cr & \Rightarrow f'\left( x \right) = - \frac{{2\left( {x - 1} \right)\left( {x - 3} \right)\left( {x - 2} \right)}}{{{{\left( {x - 2} \right)}^2}}} \cr & \therefore \,f'\left( x \right) > 0 \cr & \Rightarrow - 2\left( {x - 1} \right)\left( {x - 3} \right)\left( {x - 2} \right) > 0 \cr & \Rightarrow \left( {x - 1} \right)\left( {x - 2} \right)\left( {x - 3} \right) < 0 \cr & \Rightarrow x\, \in \left( { - \infty ,\,1} \right) \cup \left( {2,\,3} \right) \cr} $$
Thus, $$f\left( x \right)$$  is increasing on $$\left( { - \infty ,\,1} \right) \cup \left( {2,\,3} \right).$$