Given $$f\left( x \right) = k{x^3} - 9{x^2} + 9x + 3$$
On differentiating w.r.t. $$x$$, we get
$$f'\left( x \right) = 3k{x^2} - 18x + 9$$
For a function to be monotonically increasing.
$${b^2} - 4ac < 0$$
Here, $$a = 3k,\,\,b = - 18,\,\,c = 9$$
$$\eqalign{ & \therefore \,{b^2} - 4ac = {\left( { - 18} \right)^2} - 4\left( {3k} \right)\left( 9 \right) \cr & = \left( { - 18} \right)\left( { - 18} \right) - \left( {3k} \right)18 \times 2 \cr & \Rightarrow 36 - 12k < 0 \cr & \Rightarrow k > 3 \cr} $$

$$\eqalign{ & f\left( x \right) = {\left( {x + 1} \right)^3} + 1 \cr & \therefore \,f'\left( x \right) = 3{\left( {x + 1} \right)^2} \cr & f'\left( x \right) = 0\,\, \Rightarrow x = - 1 \cr & {\text{Now, }}f'\left( { - 1 - \in } \right) = 3{\left( { - \in } \right)^2} > 0,\,f'{\left( { - 1 + \in } \right)^2} = 3{ \in ^2} > 0 \cr} $$
$$\therefore \,f\left( x \right)$$   has neither a maximum nor a minimum at $$x = - 1.$$
$$\eqalign{ & {\text{Let }}f'\left( x \right) = \phi \left( x \right) = 3{\left( {x + 1} \right)^2} \cr & \therefore \,\phi '\left( x \right) = 6\left( {x + 1} \right) \cr & \phi '\left( x \right) = 0\,\, \Rightarrow x = - 1 \cr & \phi '\left( { - 1 - \in } \right) = 6\left( { - \in } \right) < 0,\,\,\phi '\left( { - 1 + \in } \right) = 6 \in > 0 \cr} $$
$$\therefore \,\phi \left( x \right)$$   has a minimum at $$x = - 1.$$

We have to find $$\frac{{d\left( {{y^2}} \right)}}{{d\left( {{x^2}} \right)}}.$$   Here, $$\frac{{dy}}{{dx}} = 1 - 2x$$
$$\eqalign{ & \therefore \frac{{d\left( {{y^2}} \right)}}{{d\left( {{x^2}} \right)}} \cr & = \frac{{\frac{{d\left( {{y^2}} \right)}}{{dx}}}}{{\frac{{d\left( {{x^2}} \right)}}{{dx}}}} \cr & = \frac{{2y\frac{{dy}}{{dx}}}}{{2x}} \cr & = \frac{y}{x}.\left( {1 - 2x} \right) \cr & = \frac{{\left( {x - {x^2}} \right)\left( {1 - 2x} \right)}}{x} \cr} $$

Let $$y = {x^2} - 4x + 3$$
Differentiate both sides w.r.t. $$'x’$$
$$\frac{{dy}}{{dx}} = 2x - 3$$
So, slope $$ = 2x - 3$$
Since, tangent is $$||$$ to $$x$$-axis
$$\therefore $$  slope $$ = 0$$
$$\eqalign{ & \Rightarrow \frac{{dy}}{{dx}} = 0 \cr & \Rightarrow 2x - 3 = 0 \cr & \Rightarrow x = \frac{3}{2} \cr & \Rightarrow {\text{one tangent}} \cr} $$

Let $$f\left( x \right) = 3\,\tan \,x + {x^3} - 2$$
Then $$f'\left( x \right) = 3\,{\sec ^2}\,x + 3{x^2} > 0.$$
Hence, $$f\left( x \right)$$  increases.
Also, $$f\left( 0 \right) = - 2$$   and $$f\left( {\frac{\pi }{4}} \right) > 0.$$
So, by intermediate value theorem, $$f\left( c \right) = 2$$   for some $$c\, \in \left( {0,\,\frac{\pi }{4}} \right)\,$$
Hence, $$f\left( x \right) = 0$$   has only one root.

We have $$AC = \sec \,\theta ,\,AG = 8$$
$$\therefore \,CG = 8 - \sec \,\theta $$    ($$C$$ being the peg)

$$\eqalign{ & {\text{But }}u = CG\,\sin \,\theta = \left( {8 - \sec \,\theta } \right)\sin \,\theta \cr & u = 8\,\sin \,\theta - \tan \,\theta \cr & \frac{{du}}{{d\theta }} = 8\,\cos \,\theta - {\sec ^2}\theta , \cr & \frac{{{d^2}u}}{{d{\theta ^2}}} = - 8\,\sin \,\theta - 2\,{\sec ^2}\theta \,\tan \,\theta \cr & \frac{{du}}{{d\theta }} = 0,\,{\text{when }}{\cos ^3}\theta = \frac{1}{8},\,\cos \,\theta = \frac{1}{2}, \cr & \frac{{{d^2}u}}{{d{\theta ^2}}} > 0\left( {{\text{at }}\theta = {{60}^ \circ }} \right), \cr & \therefore \,\theta = {60^ \circ } \cr} $$

At time $$t,$$ the distance $$z$$ between the cyclists is given by $${z^2} = {\left( {3vt} \right)^2} + {\left( {4vt} \right)^2}$$
$$\therefore \,\,z = 5vt\,\,\,\,\,\, \Rightarrow \frac{{dz}}{{dt}} = 5v$$

$$\eqalign{ & {x^{\frac{3}{2}}} + {y^{\frac{3}{2}}} = {a^{\frac{3}{2}}} \cr & \Rightarrow \frac{3}{2}.{x^{\frac{1}{2}}} + \frac{3}{2}.{y^{\frac{1}{2}}}\frac{{dy}}{{dx}} = 0\,\,\,\,\,\, \Rightarrow \frac{{dy}}{{dx}} = - \frac{{{x^{\frac{1}{2}}}}}{{{y^{\frac{1}{2}}}}} \cr & \therefore {\left. {\,\frac{{dy}}{{dx}}} \right)_{\alpha ,\,\beta }} = 1\,\,\,\, \Rightarrow {\alpha ^{\frac{1}{2}}} + {\beta ^{\frac{1}{2}}} = 0 \cr & {\text{Also }}{\alpha ^{\frac{3}{2}}} + {\beta ^{\frac{3}{2}}} = {a^{\frac{3}{2}}}\,\,\,\,\,\,\left\{ {\because \left( {\alpha ,\,\beta } \right)\,{\text{is on the curve}}} \right\} \cr & {\text{These give no values of }}\alpha ,\,\beta \cr & {\left. {\,{\text{Now, }}\frac{{dy}}{{dx}}} \right)_{\alpha ,\,\beta }} = - 1\,\,\,\,\,\,\,\,\,\, \Rightarrow \,{\alpha ^{\frac{1}{2}}} = {\beta ^{\frac{1}{2}}} \cr & {\text{Also }}{\alpha ^{\frac{3}{2}}} + {\beta ^{\frac{3}{2}}} = {a^{\frac{3}{2}}}\,\,\therefore \,\alpha = \beta = \frac{a}{{{2^{\frac{2}{3}}}}} \cr & \therefore \,\,{\text{there is only one point}}{\text{.}} \cr} $$

$$\eqalign{ & f'\left( x \right) = 4{x^3} - 2x - 2 \cr & \therefore \,f'\left( x \right) = 0 \cr & \Rightarrow 2{x^3} - x - 1 = 0 \cr & \Rightarrow \left( {x - 1} \right)\left( {2{x^2} + 2x + 1} \right) = 0 \cr & \therefore \,x = 1 \cr & {\text{Now, }}f''\left( x \right) = 12{x^2} - 2\,\,\,\,\,\,\,\,\,\therefore f''\left( 1 \right) = 12 - 2 > 0 \cr & {\text{Hence, global min}}f\left( x \right) = 1 - 1 - 2 + 6 = 4 \cr} $$

$$\eqalign{ & f\left( x \right) = \frac{1}{{{e^x} + 2{e^{ - x}}}} = \frac{{{e^x}}}{{{e^{2x}} + 2}} \cr & f'\left( x \right) = \frac{{\left( {{e^{2x}} + 2} \right){e^x} - 2{e^{2x}}.{e^x}}}{{{{\left( {{e^{2x}} + 2} \right)}^2}}} \cr & f'\left( x \right) = 0 \Rightarrow {e^{2x}} + 2 = 2{e^{2x}} \cr & {e^{2x}} = 2 \Rightarrow {e^x} = \sqrt 2 \cr & \operatorname{maximum} f\left( x \right) = \frac{{\sqrt 2 }}{4} = \frac{1}{{2\sqrt 2 }} \cr & 0 < f\left( x \right) \leqslant \frac{1}{{2\sqrt 2 }}\,\,\,\forall x \in R \cr & {\text{since }}0 < \frac{1}{3} < \frac{1}{{2\sqrt 2 }} \Rightarrow {\text{ for some }}c \in R \cr & f\left( c \right) = \frac{1}{3} \cr} $$