Differentiating w.r.t. $$x,$$
$$2y\frac{{dy}}{{dx}} - 6{x^2} - 4\frac{{dy}}{{dx}} = 0\,\,\,\,\,\, \Rightarrow \frac{{dy}}{{dx}} = \frac{{3{x^2}}}{{y - 2}}$$
$$\therefore $$ the equation of the tangent at $$\left( {\alpha ,\,\beta } \right)$$  is $$y - \beta = \frac{{3{\alpha ^2}}}{{\beta - 2}}\left( {x - \alpha } \right)$$
It passes through (1, 2) if $$2 - \beta = \frac{{3{\alpha ^2}}}{{\beta - 2}}\left( {1 - \alpha } \right)$$
or $${\left( {\beta - 2} \right)^2} = 3{\alpha ^2}\left( {\alpha - 1} \right)$$
Also, $$\left( {\alpha ,\,\beta } \right)$$  satisfies the equation of the curve.
$$\eqalign{ & \therefore {\beta ^2} - 2{\alpha ^3} - 4\beta + 8 = 0\,\,\,{\text{or}}\,\,\,{\left( {\beta - 2} \right)^2} = 2{\alpha ^3} - 4 \cr & \therefore {\left( {\beta - 2} \right)^2} = 3{\alpha ^2}\left( {\alpha - 1} \right) = 2{\alpha ^3} - 4 \cr & \therefore {\alpha ^3} - 3{\alpha ^2} + 4 = 0 \cr & {\text{or, }}\left( {\alpha - 2} \right)\left( {{\alpha ^2} - \alpha - 2} \right) = 0 \cr & {\text{or, }}{\left( {\alpha - 2} \right)^2}\left( {\alpha + 1} \right) = 0 \cr & {\text{When }}\alpha = 2,\,{\left( {\beta - 2} \right)^2} = 12 \cr & {\text{or, }}\beta = 2 \pm 2\sqrt 3 \cr & {\text{When }}\alpha = - 1,\,{\left( {\beta - 2} \right)^2} = - 6 \cr & {\text{or, }}\beta = {\text{nonreal number}} \cr & \therefore \,\left( {\alpha ,\,\beta } \right)\,{\text{has two values}}{\text{.}} \cr} $$

$$\eqalign{ & f\left( x \right) = {x^3} + b{x^2} + cx + d,\,0 < {b^2} < c \cr & f'\left( x \right) = 3{x^2} + 2bx + c \cr & {\text{Discriminant}} = 4{b^2} - 12c = 4\left( {{b^2} - 3c} \right) < 0 \cr & \therefore \,\,f'\left( x \right) > 0\,\forall x \in R \cr & \Rightarrow \,\,f\left( x \right)\,{\text{is}}\,{\text{strictly}}\,{\text{increasing}}\,\forall x \in R \cr} $$

Let $$A$$ be the vertex and $$AO$$  the axis of the cone.
Let $$O'A = h$$   be the depth of water in the cone.
$$\eqalign{ & {\text{In}}\,\Delta AO'C, \cr & \tan \,{30^ \circ } = \frac{{O'C}}{h}{\text{ or }}O'C = \frac{h}{{\sqrt 3 }} = {\text{radius}} \cr & V = {\text{Volume of water in the cone}} \cr & \,\,\,\,\,\,\, = \frac{1}{3}\pi {\left( {O'C} \right)^2} \times AO' \cr & \,\,\,\,\,\,\, = \frac{1}{3}\pi \left( {\frac{{{h^2}}}{3}} \right) \times h \cr & \,\,\,\,\,\,\, = \frac{\pi }{9}{h^3} \cr & {\text{or }}\frac{{dV}}{{dt}} = \frac{\pi }{3}{h^2}\frac{{dh}}{{dt}}......\left( 1 \right) \cr} $$
But given that depth of water increases at the rate of $$1\,cm/s.$$
So, $$\frac{{dh}}{{dt}} = 1\,cm/s.......\left( 2 \right)$$
From $$\left( 1 \right){\text{ and}}\,\left( 2 \right),\,\frac{{dV}}{{dt}} = \frac{{\pi {h^2}}}{3}$$
When $$h = 24\,cm,$$   the rate of increase of volume is $$\frac{{dV}}{{dt}} = \frac{{\pi {{\left( {24} \right)}^2}}}{3} = 192\,c{m^3}/s.$$

The given function is $$f(x) = \left\{ {_{{x^{\frac{2}{3}}},}^{{{(2 + x)}^3},}\,_{ - 1 < x < 2}^{ - 3 < x \leqslant - 1}} \right.$$
The graph of $$y = f\left( x \right)$$   is as shown in the figure. From graph, clearly, there is one local maximum $$\left( {{\text{at}}\,x = - 1} \right)$$   and one local minima $$\left( {{\text{at}}\,x = 0} \right)$$
∴ total number of local maxima or minima = 2.

Two curves cuts at right angle if product of their slopes is $$ –1.$$
Two given curves are
$$\eqalign{ & {x^3} - 3x{y^2} + 2 = 0......\left( {\text{i}} \right) \cr & {\text{and }}3{x^2}y - {y^3} - 2 = 0......\left( {{\text{ii}}} \right) \cr} $$
Differentiate equation $$\left( {\text{i}} \right),$$
$$\eqalign{ & 3{x^2} - 3\left[ {{y^2} + 2xy\frac{{dy}}{{dx}}} \right] = 0 \cr & \Rightarrow 3\left( {{x^2} - {y^2}} \right) = 6xy\frac{{dy}}{{dx}} \cr & \Rightarrow {m_1} = \frac{{dy}}{{dx}} = \frac{{3\left( {{x^2} - {y^2}} \right)}}{{6xy}} \cr} $$
Differentiate equation $$\left( {\text{ii}} \right),$$
$$\eqalign{ & 3{x^2}y - {y^3} - 2 = 0 \cr & \Rightarrow 3\left[ {{x^2}\frac{{dy}}{{dx}} + 2xy} \right] - 3{y^2}\frac{{dy}}{{dx}} = 0 \cr & \Rightarrow {x^2}\frac{{dy}}{{dx}} + 2xy - {y^2}\frac{{dy}}{{dx}} = 0 \cr & \Rightarrow \left( {{x^2} - {y^2}} \right)\frac{{dy}}{{dx}} = - 2xy \cr & \Rightarrow {m_2} = \frac{{dy}}{{dx}} = \frac{{ - 2xy}}{{\left( {{x^2} - {y^2}} \right)}} \cr & \therefore \,{m_1} \times {m_2} = \frac{{\left( {{x^2} - {y^2}} \right)}}{{2xy}} \times \frac{{ - 2xy}}{{\left( {{x^2} - {y^2}} \right)}} \cr & \Rightarrow {m_1} \times {m_2} = - 1 \cr} $$
i.e., curves cuts at right angle.

Given, velocity is $$v = {x^2}\log \left( {\frac{1}{x}} \right) = - {x^2}\log \,x$$       where $$x$$ is displacement.
For maximum velocity, $$\frac{{dv}}{{dx}} = 0$$
Now, $$\frac{{dv}}{{dx}} = - {x^2}\frac{1}{x} + \log \,x\left( { - 2x} \right) = - x - 2x\,\log \,x$$
$$\eqalign{ & \frac{{dv}}{{dx}} = 0 \cr & \Rightarrow - x - 2x\,\log \,x = 0 \cr & \Rightarrow x = - 2x\,\log \,x \cr & \Rightarrow \frac{{ - 1}}{2} = \log \,x \cr & \Rightarrow x = {e^{ - \frac{1}{2}}} \cr} $$
Hence, for maximum velocity $$x = {e^{ - \frac{1}{2}}}$$

From graph it is clear that both $$\sin x$$  and $$\cos x$$  in the interval $$\left( {\frac{\pi }{2},\pi } \right)$$  are decreasing function.


$$\therefore $$ $$S$$ is correct.
To disprove $$R$$ let us consider the counter example :
$$f\left( x \right) = \sin x\,{\text{on}}\,\left( {0,\frac{\pi }{2}} \right)\,{\text{so}}\,{\text{that}}\,f'\left( x \right) = \cos x$$
Again from graph it is clear that $$f\left( x \right)$$  is increasing on $$\left( {0,\frac{\pi }{2}} \right)$$   but $$f'\left( x \right)$$  is decreasing on $$\left( {0,\frac{\pi }{2}} \right)$$
∴ $$R$$ is wrong.

$$\eqalign{ & {\text{We}}\,{\text{have}}\,f\left( x \right) = \frac{x}{{\sin x}},0 < x \leqslant 1 \cr & \Rightarrow f'\left( x \right) = \frac{{\sin x - x\cos x}}{{{{\sin }^2}x}} \cr & {\text{where}}\,{\sin ^2}x{\text{ is always }} + ve,{\text{ when }}0 < x \leqslant 1.{\text{ But to check Nr}}{\text{., we again let}} \cr & h\left( x \right) = \sin x - x\cos x \cr & \Rightarrow h'\left( x \right) = x\sin x > 0\,{\text{for}}\,0 < x \leqslant 1 \Rightarrow h\left( x \right)\,{\text{is increasing}} \cr & \Rightarrow h\left( 0 \right) < h\left( x \right),\,{\text{when}}\,0 < x \leqslant 1 \cr & \Rightarrow 0 < \sin x - x\cos x,\,{\text{when}}\,0 < x \leqslant 1 \cr & \Rightarrow \sin x - x\cos x > 0,\,{\text{when}}\,0 < x \leqslant 1 \cr & \Rightarrow f'\left( x \right) > 0,x \in \left( {0,1} \right] \cr & \Rightarrow f\left( x \right)\,{\text{is}}\,{\text{increasing}}\,{\text{on}}\,\left( {0,1} \right] \cr & {\text{Again}}\,g\left( x \right) = \frac{x}{{\tan x}} \cr & \Rightarrow g'\left( x \right) = \frac{{\tan x - x{{\sec }^2}x}}{{{{\tan }^2}x}},\,{\text{when}}\,0 < x \leqslant 1 \cr & {\text{Here}}\,{\tan ^2}x > 0\,{\text{But}}\,{\text{to check Nr}}{\text{. we consider}} \cr & p\left( x \right) = \tan x - x{\sec ^2}x \cr & p'\left( x \right) = {\sec ^2}x - {\sec ^2}x - x.2\sec x.\sec x\tan x \cr & \Rightarrow p'\left( x \right) = - 2x{\sec ^2}x\tan x < 0\,{\text{for}}\,0 < x \leqslant 1 \cr & \Rightarrow p\left( x \right)\,{\text{is decreasing, when}}\,0 < x \leqslant 1 \cr & \Rightarrow p\left( 0 \right) > p\left( x \right) \Rightarrow 0 > \tan x - x{\sec ^2}x \cr & \therefore g'\left( x \right) < 0 \cr & {\text{Hence }}g\left( x \right){\text{ is decreasing when}}\,0 < x \leqslant 1. \cr} $$

$$f'\left( x \right) < 0\,\,\,\,\, \Rightarrow 3{x^2} + 8x + \lambda < 0$$
So, $$\lambda $$ should be such that $$3{x^2} + 8x + \lambda = 0$$    has roots $$ = - 2,\, - \frac{2}{3}$$
Hence, $$\left( { - 2} \right)\,\left( { - \frac{2}{3}} \right) = \frac{\lambda }{3}\,\,\,\,\,\,\,\,\,\therefore \,\lambda = 4$$

$$\eqalign{ & f'\left( x \right) = x\left( {4 + {{4.2}^2}.{x^2} + ..... + {{20.2}^{10}}{x^{18}}} \right) \cr & \therefore \,f'\left( x \right) = 0\,\,\,\,\,\,\,\,\,\, \Rightarrow x = 0{\text{ only}} \cr & {\text{Also,}}\,\,f''\left( 0 \right) > 0 \cr} $$