$$\eqalign{ & {\text{Slope}}\,{\text{of}}\,{\text{tangent}}\,{\text{at}}\,\left( {x,f\left( x \right)} \right)\,{\text{is}}\,2x + 1 \cr & \Rightarrow f'\left( x \right) = 2x + 1 \Rightarrow f\left( x \right) = {x^2} + x + c \cr & {\text{Also}}\,{\text{the}}\,{\text{curve}}\,{\text{passes}}\,{\text{through}}\,\left( {1,2} \right) \cr & \therefore f\left( 1 \right) = 2 \cr & \Rightarrow 2 = 1 + 1 + c \Rightarrow c = 0,\,\therefore f\left( x \right) = {x^2} + x \cr & \therefore {\text{Required}}\,{\text{area}} = \int_0^1 {\left( {{x^2} + x} \right)} dx \cr & = \left( {\frac{{{x^3}}}{3} + \frac{{{x^2}}}{2}} \right)_0^1 = \frac{1}{3} + \frac{1}{2} = \frac{5}{6} \cr} $$

Area of rectangle $$ABCD = 2a\cos \theta $$
$$\left( {2b\sin \theta } \right) = 2ab\sin 2\theta $$

$$ \Rightarrow \,$$ Area of greatest rectangle is equal to $$2ab$$
When $$\sin 2\theta = 1$$

$$\eqalign{ & {\text{We}}\,{\text{have}}\,\,f\left( x \right) = \frac{{\ln \left( {\pi + x} \right)}}{{\ln \left( {e + x} \right)}} \cr & \therefore \quad {f^\prime }(x) = \frac{{\left( {\frac{1}{{\pi + x}}} \right)\ln (e + x) - \frac{1}{{(e + x)}}\ln (\pi + x)}}{{{{[\ln (e + x)]}^2}}} \cr & = \frac{{(e + x)\ln (e + x) - (\pi + x)\ln (\pi + x)}}{{(e + x)(\pi + x){{(\ln (e + x))}^2}}} \cr & < 0{\text{ on }}(0,\infty ){\text{ since }}1 < e < \pi \cr & \therefore f\left( x \right)\,{\text{decraeses}}\,{\text{on}}\,(0,\infty ). \cr} $$

$$\eqalign{ & f'\left( x \right) = \frac{{{p^2} - 1}}{{{p^2} + 1}}.3{x^2} - 3 < 0 \cr & \Rightarrow \frac{{{p^2} - 1}}{{{p^2} + 1}}{x^2} < 1 \cr & \Rightarrow \left( {{p^2} - 1} \right){x^2} - \left( {{p^2} + 1} \right) < 0 \cr} $$
The inequation is satisfied for all $$x\, \in \,R$$   if $${p^2} - 1 \leqslant 0,{\text{ i}}{\text{.e}}{\text{., }} - 1 \leqslant p \leqslant 1$$

$$\eqalign{ & {\text{Let }}f\left( \theta \right) = \tan \,\theta - \theta \cr & {\text{Then }}f'\left( \theta \right) = {\sec ^2}\theta - 1 = {\tan ^2}\theta > 0 \cr} $$
$$\therefore \,f\left( \theta \right)$$   is increasing. Therefore, $$f\left( \theta \right) > f\left( 0 \right)\,\,\,\,\, \Rightarrow \tan \,\theta - \theta > 0$$
Let $$\phi \left( \theta \right) = \theta - \sin \,\theta $$
Then $$\phi '\left( \theta \right) = 1 - \cos \,\theta = 2{\sin ^2}\frac{\theta }{2} > 0$$
$$\therefore \,\phi \left( \theta \right)$$   is increasing. Therefore, $$\phi \left( \theta \right) > \phi \left( 0 \right)\,\, \Rightarrow \sin \,\theta - \,\theta > 0$$

From the question, $$\left| {\frac{{dx}}{{dy}}} \right| > 1.$$   Differentiating $${x^3} = 12y$$   w.r.t. $$y$$
$$\eqalign{ & 3{x^2}\frac{{dx}}{{dy}} = 12\,\,\,\,\,\,\,\,\,\therefore \frac{{dx}}{{dy}} = \frac{4}{{{x^2}}} \cr & \therefore \frac{4}{{{x^2}}} > 1\,\,\,\,\, \Rightarrow {x^2} - 4 < 0\,\,\,\,\, \Rightarrow - 2 < x < 2 \cr} $$

$$x = {e^{\sin \,y}} \Rightarrow \ln \,x = \sin \,y \Rightarrow \frac{1}{x}\frac{{dx}}{{dy}} = \cos \,y$$
$$\therefore $$  slope of normal at $$\left( {1,\,0} \right)$$
$$ = - \frac{{dx}}{{dy}} = - 1 \times \cos \,0 = - 1$$
$$\therefore $$  Equation of normal is $$y - 0 = - 1\left( {x - 1} \right)\, \Rightarrow x + y = 1$$
Which forms triangle of area $$\frac{1}{2}$$ with axes.

Given, $$f'\left( x \right) < 0$$   and $$g'\left( x \right) > 0$$   therefore $$g\left( x \right)$$  is an increasing function and $$f'\left( x \right)$$  is a decreasing function
$$\eqalign{ & \therefore \,x + 1 < x + 5 \cr & \Rightarrow g\left( {x + 1} \right) < g\left( {x + 5} \right) \cr & \Rightarrow f\left( {g\left( {x + 1} \right)} \right) > f\left( {g\left( {x + 5} \right)} \right) \cr & {\text{Again }}x < x + 1 \Rightarrow g\left( {x + 1} \right) \Rightarrow f\left( {g\left( x \right)} \right) > f\left( {g\left( {x + 1} \right)} \right) \cr & x < x + 2 \Rightarrow f\left( x \right) > f\left( {x + 2} \right) \Rightarrow g\left( {f\left( x \right)} \right) > g\left( {f\left( {x + 1} \right)} \right) \cr & x > x - 2 \Rightarrow f\left( x \right) < f\left( {x - 2} \right) \Rightarrow g\left( {f\left( x \right)} \right) < g\left( {f\left( {x - 2} \right)} \right) \cr} $$

$$\eqalign{ & {\text{Let}}\,f\left( x \right) = \frac{{a{x^3}}}{3} + \frac{{b{x^2}}}{2} + cx \Rightarrow f\left( 0 \right) = 0\,{\text{and}}\,f\left( 1 \right) \cr & = \frac{a}{3} + \frac{b}{2} + c = \frac{{2a + 3b + 6c}}{6} = 0 \cr} $$
Also $$f(x)$$  is continuous and differentiable in [0, 1] and [0, 1[. So by Rolle’s theorem, $$f'\left( x \right) = 0.$$
i.e $$a{x^2} + bx + c = 0$$    has at least one root in [0, 1].

$$\eqalign{ & {\text{We have}} \cr & {\text{Total length}} = r + r + r\theta = 20 \cr & \Rightarrow 2r + r\theta = 20 \cr & \Rightarrow \theta = \frac{{20 - 2r}}{r}\,......\left( 1 \right) \cr & A = {\text{Area}} = \frac{\theta }{{2\pi }} \times \pi {r^2} = \frac{1}{2}{r^2}\theta = \frac{1}{2}{r^2}\left( {\frac{{20 - 2r}}{r}} \right) \cr & A = 10r - {r^2} \cr & {\text{For}}\,A\,{\text{to}}\,{\text{be}}\,{\text{maximum}}\,\frac{{dA}}{{dr}} = 0 \Rightarrow 10 - 2r = 0 \Rightarrow r = 5 \cr & \frac{{{d^2}A}}{{d{r^2}}} = - 2 < 0 \cr} $$

$$\eqalign{ & \therefore \,\,{\text{For}}\,r = 5\,A\,{\text{is}}\,{\text{maximum}} \cr & {\text{From}}\,\left( 1 \right) \cr & \theta = \frac{{20 - 2\left( 5 \right)}}{5} = \frac{{10}}{5} = 2 \cr & A = \frac{2}{{2\pi }} \times \pi {\left( 5 \right)^2} = 25\,sq.m. \cr} $$