$$\eqalign{ & {\text{In }} - 1 \leqslant x < 0,\,f'\left( x \right) = 3{x^2} + 2x - 10 = 2{x^2} + {\left( {x + 1} \right)^2} - 11 < 0 \cr & \therefore f\left( x \right){\text{ is m}}{\text{.d}}{\text{.}} \cr & {\text{In }}0 < x < \frac{\pi }{2},\,f'\left( x \right) = - \sin \,x < 0 \cr & \therefore \,f\left( x \right){\text{ is m}}{\text{.d}}{\text{.}} \cr & {\text{In }}\frac{\pi }{2} < x \leqslant \pi ,\,f'\left( x \right) = \cos \,x < 0 \cr & \therefore \,f\left( x \right){\text{ is m}}{\text{.d}}{\text{.}} \cr} $$

Displaying the trend of values of the function in different intervals, we get the adjoining graph.
$$\eqalign{ & \therefore f\left( {\frac{\pi }{2} - h} \right) < 2 = f\left( {\frac{\pi }{2}} \right),\,\,\,\,\,f\left( {\frac{\pi }{2} + h} \right) < f\left( {\frac{\pi }{2}} \right) \cr & \therefore f\left( x \right){\text{ has a local maximum at }}x = \frac{\pi }{2} \cr} $$

$$\eqalign{ & {\text{Let}}\,f\left( x \right) = {e^x} - 1 - x\,{\text{then}}\,f'\left( x \right) = {e^x} - 1 > 0\,{\text{for}}\,x \in \left( {0,1} \right) \cr & \therefore f\left( x \right)\,{\text{is}}\,{\text{an increasing function}}{\text{.}} \cr & \therefore f\left( x \right) > f\left( 0 \right),\forall \,x\, \in \left( {0,1} \right) \cr & \Rightarrow {e^x} - 1 - x > 0 \Rightarrow {e^x} > 1 + x \cr & \therefore \left( {\text{a}} \right)\,{\text{does not hold}}. \cr & \left( {\text{b}} \right)\,{\text{Let}}\,g\left( x \right) = \log \left( {1 + x} \right) - x \cr & {\text{then}}\,g'\left( x \right) = \frac{1}{{1 + x}} - 1 = - \frac{x}{{1 + x}} < 0,\forall \,x \in \left( {0,1} \right) \cr & \therefore g\left( x \right)\,{\text{is decreasing on}}\,\left( {0,1} \right)\,\therefore x > 0 \cr & \Rightarrow g\left( x \right) < g\left( 0 \right) \cr & \Rightarrow \log \left( {1 + x} \right) - x < 0 \Rightarrow \log \left( {1 + x} \right) < x \cr & \therefore \left( {\text{b}} \right){\text{ holds}}{\text{. Similarly it can be shown that}}\,\left( {\text{c}} \right)\,{\text{and}}\,\left( {\text{d}} \right)\,{\text{do}}\,{\text{not hold}}{\text{.}} \cr} $$

Let $$y = x{e^x}$$
Differentiate both side w.r.t. $$‘x’.$$
$$\eqalign{ & \Rightarrow \frac{{dy}}{{dx}} = {e^x} + x{e^x} = {e^x}\left( {1 + x} \right) \cr & {\text{Put }}\frac{{dy}}{{dx}} = 0 \cr & \Rightarrow {e^x}\left( {1 + x} \right) = 0 \Rightarrow x = - 1 \cr & {\text{Now, }}\frac{{{d^2}y}}{{d{x^2}}} = {e^x} + {e^x}\left( {1 + x} \right) = {e^x}\left( {x + 2} \right) \cr & {\left( {\frac{{{d^2}y}}{{d{x^2}}}} \right)_{\left( {x = - 1} \right)}} = \frac{1}{e} + 0 > 0 \cr} $$
Hence, $$y = x{e^x}$$   is minimum function and $${y_{\min }} = - \frac{1}{e}.$$

Let one side of quadrilateral be $$x$$ and another side be $$y$$
So, $$2\left( {x + y} \right) = 34\,\,{\text{or }}\left( {x + y} \right) = 17......\left( {\text{i}} \right)$$
We know from the basic principle that for a given perimeter square has the maximum area, so, $$x = y$$  and putting this value in equation $$\left( {\text{i}} \right)$$
$$x = y = \frac{{17}}{2}$$
Area $$ = x.y = \frac{{17}}{2} \times \frac{{17}}{2} = \frac{{289}}{4} = 72.25$$

$$\eqalign{ & {\text{Let the polynominal be }}P\left( x \right) = a{x^2} + bx + c \cr & {\text{Given}}\,P\left( 0 \right) = 0\,{\text{and}}\,P\left( 1 \right) = 1 \Rightarrow c = 0\,\,{\text{and}}\,a + b = 1 \Rightarrow a = 1 - b \cr & \therefore P\left( x \right) = \left( {1 - b} \right){x^2} + bx \cr & \Rightarrow P'\left( x \right) = 2\left( {1 - b} \right)x + b \cr & {\text{Given }}P'\left( x \right) > 0,\forall x \in \left[ {0,1} \right] \cr & \Rightarrow 2\left( {1 - b} \right)x + b > 0 \cr & \Rightarrow {\text{When }}x = 0,b > 0{\text{ and when }}x = 1,b < 2 \cr & \Rightarrow 0 < b < 2 \cr & \therefore S = \left\{ {\left( {1 - a} \right){x^2} + ax,a \in \left( {0,2} \right)} \right\} \cr} $$

$$f\left( x \right) = 2.\cos \frac{{\sqrt 2 + 1}}{2}x.\cos \frac{{\sqrt 2 - 1}}{2}x \leqslant 2,$$         and it is 2 only when $$\cos \frac{{\sqrt 2 + 1}}{2}x$$   and $$\cos \frac{{\sqrt 2 - 1}}{2}x$$   are both equal to 1 for a value of $$x.$$ This is possible only when $$x=0.$$

We have equation of tangent to any curve $$f\left( x \right)$$  at $$\left( {{x_1},\,{y_1}} \right){\text{ is }}\left( {y - {y_1}} \right) = {\left. {\frac{{dy}}{{dx}}} \right|_{\left( {{x_1},\,{y_1}} \right)}}\left( {x - {x_1}} \right)$$
Given curve is $$y = {e^{ - \left| x \right|}}$$
Point of intersection is $$\left( {1,\,\frac{1}{e}} \right){\text{ at }}x = 1,\,\left| x \right| = x$$
So, $$y = {e^{ - x}} \Rightarrow \frac{{dy}}{{dx}} = - {e^{ - x}}$$
$$\therefore \,{\left( {\frac{{dy}}{{dx}}} \right)_{x = 1}} = - {e^{ - 1}}$$
Therefore, equation of tangent is $$y - \frac{1}{e} = \frac{{ - 1}}{e}\left( {x - 1} \right) \Rightarrow x + ey = 2$$

$$\eqalign{ & f'\left( x \right) = \sqrt x \sin x,f'\left( x \right) = 0 \cr & \Rightarrow x = 0\,{\text{or}}\,\sin x = 0 \cr & \Rightarrow x = 2\pi ,\pi \in \left( {0,\frac{{5\pi }}{2}} \right) \cr & f''\left( x \right) = \sqrt x \cos x + \frac{1}{{2\sqrt x }}\sin x \cr & = \frac{1}{{2\sqrt x }}\left( {2x\cos x + \sin x} \right) \cr & {\text{At}}\,\,x = \pi ,f''\left( x \right) < 0 \cr & {\text{Hence, local maxima at}}\,x = \pi \cr & {\text{At}}\,\,x = 2\pi ,f''\left( x \right) > 0 \cr & {\text{Hence local minima at}}\,x = 2\pi \cr} $$

$$\eqalign{ & P\left( x \right) = - 3500 + \left( {400 - x} \right)x \cr & = - 3500 + 400x - {x^2} \cr} $$
On differentiating w.r.t. $$x,$$ we get
$$P'\left( x \right) = 400 - 2x$$
Put $$P'\left( x \right) = 0$$   for maxima or minima
$$\eqalign{ & \Rightarrow 400 - 2x = 0 \cr & \Rightarrow x = 200 \cr & {\text{Now }}P''\left( x \right) = - 2x \cr & \Rightarrow P''\left( {200} \right) = - 400 < 0 \cr} $$
$$\therefore \,P\left( x \right)$$   is maximum at $$x = 200$$
Hence, $$200$$  items should the firm sell so that the firm has maximum profit.

$$\eqalign{ & \frac{{dx}}{{dt}} = 2t + 3,\,\frac{{dy}}{{dt}} = 4t - 2 \cr & \frac{{dy}}{{dx}} = \frac{{\frac{{dy}}{{dt}}}}{{\frac{{dx}}{{dt}}}} = \frac{{4t - 2}}{{2t + 3}} \cr & {\text{slope at }}t = 2 = \frac{{\frac{{dy}}{{dx}}}}{t} = 2 = \frac{{4\left( 2 \right) - 2}}{{2\left( 2 \right) + 3}} = \frac{6}{7} \cr} $$