$$\eqalign{ & f'\left( x \right) = 6\,{\sin ^2}x\,\cos \,x - 6\,\sin \,x\,\cos \,x + 12\,\cos \,x \cr & = 6\,\cos \,x\left\{ {{{\sin }^2}x - \sin \,x + 2} \right\} \cr & = 6\,\cos \,x\left\{ {{{\left( {\sin \,x - \frac{1}{2}} \right)}^2} + \frac{7}{4}} \right\} \cr & \therefore \,{\text{in}}\left[ {0,\,\frac{\pi }{2}} \right],\,f'\left( x \right) \geqslant 0. \cr & {\text{So, }}f\left( x \right){\text{ is increasing in }}\left[ {0,\,\frac{\pi }{2}} \right] \cr} $$

Let length and breadth of rectangular field be $$x$$ and $$y$$ respectively
$$\therefore \,2\left( {x + y} \right) = 200 \Rightarrow y = 100 - x$$
and area $$A = xy = x\left( {100 - x} \right)$$
$$\because \,\frac{{dA}}{{dx}} = 100 - 2x$$
Put $$\frac{{dA}}{{dx}} = 0$$   for maxima or minima
$$\eqalign{ & 100 - 2x = 0 \cr & \Rightarrow x = 50\,\,\, \Rightarrow y = 50 \cr} $$
Now, $$\frac{{{d^2}A}}{{d{x^2}}} = - 2 < 0,$$    which shows maximum, independent of values of $$x$$ and $$y,$$ but only when they are equal.
$$\therefore \,A$$  is maximum at $$x = 50$$
Hence, required area $$ = 50\left( {100 - 50} \right) = 50 \times 50 = 2500\,{m^2}$$

Here,
$$\eqalign{ & {f_1}\left( x \right) = {x^2} - x + 1\,\,{\text{and }}{f_2}\left( x \right) = {x^3} - {x^2} - 2x + 1 \cr & {\text{or }}{f_1}'\left( {{x_1}} \right) = 2{x_1} - 1\,\,{\text{and }}{f_2}'\left( {{x_2}} \right) = 3x_2^2 - 2{x_2} - 2 \cr} $$
Let the tangents drawn to the curves $$y = {f_1}\left( x \right)$$   and $$y = {f_2}\left( x \right)$$   at $$\left( {{x_1},\,{f_1}\left( {{x_1}} \right)} \right)$$   and $$\left( {{x_2},\,{f_2}\left( {{x_2}} \right)} \right)$$   be parallel. Then
$$\eqalign{ & 2{x_1} - 1 = 3x_2^2 - 2{x_2} - 2 \cr & {\text{or }}2{x_1} = \left( {3x_2^2 - 2{x_2} - 1} \right) \cr} $$
So, which is possible for infinite numbers of ordered pairs. So, there are infinite solutions.

The equation of the line is
$$\eqalign{ & y - 3 = \frac{{3 + 2}}{{0 - 5}}\left( {x - 0} \right),\,{\text{i}}{\text{.e}}{\text{., }}x + y - 3 = 0 \cr & y = \frac{c}{{x + 1}}{\text{ or }}\frac{{dy}}{{dx}} = \frac{{ - c}}{{{{\left( {x + 1} \right)}^2}}} \cr} $$
Let the line touch the curve at $$\left( {\alpha ,\,\beta } \right).$$
Then $$\alpha + \beta - 3 = 0\,{\left( {\frac{{dy}}{{dx}}} \right)_{\alpha ,\,\beta }} = \frac{{ - c}}{{{{\left( {\alpha + 1} \right)}^2}}} = - 1,$$          and $$\beta = \frac{c}{{\alpha + 1}}$$
$$\eqalign{ & \therefore \,\frac{c}{{\left( {\frac{c}{\beta }} \right)}} = 1 \cr & {\text{or, }}{\beta ^2} = c \cr & {\text{or, }}{\left( {3 - \alpha } \right)^2} = c = {\left( {\alpha + 1} \right)^2} \cr & {\text{or, }}{\left( {3 - \alpha } \right)^2} = {\left( {\alpha + 1} \right)^2} \cr & {\text{or, }}3 - \alpha = \pm \left( {\alpha + 1} \right) \cr & {\text{or, 3}} - \alpha = \alpha + 1 \cr & {\text{or, }}\alpha = 1 \cr & {\text{So, }}c = {\left( {1 + 1} \right)^2} = 4 \cr} $$

$$\eqalign{ & f'\left( x \right) = {x^2} + 2x + 2 = {\left( {x + 1} \right)^2} + 1 > 0\,{\text{for all}}\,x \cr & {\text{So,}}\,f\left( x \right)\,{\text{is m}}{\text{.i}}{\text{. in}}\,\left[ {2,\,4} \right] \cr & \therefore \,\min f\left( x \right) = f\left( 2 \right){\text{ and }}\max f\left( x \right) = f\left( 4 \right) \cr & \therefore \,\min f\left( x \right) = \int_0^2 {\left( {{t^2} + 2t + 2} \right)dt} = \left[ {\frac{{{t^3}}}{3} + {t^2} + 2t} \right]_0^2 = \frac{{32}}{3} \cr & \therefore \,\max f\left( x \right) = \int_0^4 {\left( {{t^2} + 2t + 2} \right)dt} = \left[ {\frac{{{t^3}}}{3} + {t^2} + 2t} \right]_0^4 = \frac{{136}}{3} \cr} $$

$$\eqalign{ & f'\left( x \right) = \frac{{1 + x\tan \,x - x\left\{ {\tan \,x + x{{\sec }^2}x} \right\}}}{{{{\left( {1 + x\tan \,x} \right)}^2}}} = \frac{{1 - {x^2}{{\sec }^2}x}}{{{{\left( {1 + x\tan \,x} \right)}^2}}} \cr & f'\left( x \right) = 0\,\,\,\,\,\,\, \Rightarrow 1 - {x^2}{\sec ^2}x = 0 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow \cos \,x = \pm x \cr} $$
$$\therefore \,$$ in the interval $$\left( {0,\,\frac{\pi }{2}} \right),\,f'\left( x \right) = 0$$     can hold for only one value of $$x,$$ say $$\alpha, $$ where $$\cos \,\alpha = \alpha $$   (see the graph).

Verify $$f''\left( x \right) < 0{\text{ at }}x = \alpha $$

Given equation is $$s = 64t - 16{t^2}$$
$$\therefore $$  On differentiating w.r.t. $$t,$$ we get $$\frac{{ds}}{{dt}} = 64 - 32t$$
Put $$\frac{{ds}}{{dt}} = 0$$   for maximum height
$$\eqalign{ & \Rightarrow 64 - 32t = 0 \cr & \Rightarrow t = 2 \cr} $$
Now, $$\frac{{{d^2}s}}{{d{t^2}}} = - 32$$
At $$t = 2,\,\frac{{{d^2}s}}{{d{t^2}}} = - 32$$
Since, $${\left( {\frac{{{d^2}s}}{{d{t^2}}}} \right)_{t = 2}} < 0$$
$$\therefore $$  Required time $$= 2$$  seconds

$$\eqalign{ & {\text{Here, }}a + b = 8. \cr & {\text{Let }}y = \frac{1}{a} + \frac{1}{b} = \frac{1}{a} + \frac{1}{{8 - a}} = \frac{8}{{a\left( {8 - a} \right)}} \cr & \therefore \frac{{dy}}{{da}} = \frac{{ - 8}}{{{a^2}{{\left( {8 - a} \right)}^2}}}\left\{ {8 - 2a} \right\} \cr & \therefore \frac{{dy}}{{da}} = 0{\text{ if }}8 - 2a = 0,\,{\text{i}}{\text{.e}}{\text{., }}a = 4 \cr & {\text{Also}},{\left. {{\text{ }}\frac{{dy}}{{dx}}} \right)_{4 - \in }} = \frac{{ - 8}}{{{{\left( {4 - \in } \right)}^2}.{{\left( {4 + \in } \right)}^2}}}\left( {2 \in } \right) < 0 \cr & {\left. {{\text{ }}\frac{{dy}}{{dx}}} \right)_{4 + \in }} = \frac{{ - 8}}{{{{\left( {4 + \in } \right)}^2}.{{\left( {4 - \in } \right)}^2}}}\left( { - 2 \in } \right) > 0 \cr & \therefore a = 4,\,y{\text{ is minimum}}{\text{.}} \cr & {\text{So min }}y = \frac{1}{4} + \frac{1}{{8 - 4}} = \frac{1}{2} \cr} $$

Consider the function $$f\left( x \right) = a{x^3} + b{x^2} + cx$$     on $$\left[ {0, 1} \right]$$  then being a polynomial. It is continuous on $$\left[ {0, 1} \right],$$ differentiable on $$\left( {0, 1} \right)$$  and
$$f\left( 0 \right) = f\left( 1 \right) = 0$$     [as given $$a + b + c = 0$$   ]
$$\therefore $$ By Rolle's theorem $$\exists \,x \in \left( {0,1} \right)$$   such that
$$f'\left( x \right) = 0 \Rightarrow 3a{x^2} + 2bx + c = 0$$
Thus equation $$3a{x^2} + 2bx + c = 0$$     has at least one root in [0, 1].

$$\eqalign{ & f\left( x \right) = 2{x^3} - 15{x^2} + 36x + 1 \cr & f'\left( x \right) = 6{x^2} - 30x + 36 = 6\left( {x - 2} \right)\left( {x - 3} \right) \cr} $$
Thus, $$f\left( x \right)$$  is increasing in $$\left[ {0,\,2} \right]$$  and decreasing in $$\left[ {2,\,3} \right].$$  Therefore $$f\left( x \right)$$  is many-one
$$f\left( 0 \right) = 1\,;\,\,f\left( 2 \right) = 29\,;\,\,f\left( 3 \right) = 28$$
Range is $$\left[ {1,\,29} \right]$$
Hence, $$f\left( x \right)$$  is many-one-onto.