$$V = {x^3}$$   and the per cent error in measuring $$x = \frac{{dx}}{x} \times 100 = k$$
The per cent error in measuring volume $$ = \frac{{dV}}{V} \times 100$$
$$\eqalign{ & {\text{Now, }}\frac{{dV}}{{dx}} = 3{x^2}\,\,\,\,\,\,\,\therefore dV = 3{x^2}dx \cr & \therefore \frac{{dV}}{V} = \frac{{3{x^2}dx}}{{{x^3}}} = 3\frac{{dx}}{x} \cr & \therefore \frac{{dV}}{V} \times 100 = 3\frac{{dx}}{x} \times 100 = 3k \cr} $$

$${\text{Let}}\,f\left( x \right) = {a_n}{x^n} + {a_{n - 1}}{x^{n - 1}} + ........ + {a_1}x = 0$$
The other given equation,
$$\eqalign{ & n{a_n}{x^{n - 1}} + \left( {n - 1} \right){a_{n - 1}}{x^{n - 2}} + ....... + {a_1} = 0 = f'\left( x \right) \cr & {\text{Given}}\,{a_1} \ne 0 \Rightarrow f\left( 0 \right) = 0 \cr & {\text{Again}}\,f\left( x \right)\,{\text{has}}\,{\text{root}}\,\alpha , \Rightarrow f\left( \alpha \right) = 0 \cr & \therefore f\left( 0 \right) = f\left( \alpha \right) \cr} $$
$$\therefore $$ By Roll’s theorem $$f'\left( x \right) = 0$$   has root between $$\left( {0,\alpha } \right)$$
Hence $$f'\left( x \right)$$  has a positive root smaller than $${\alpha .}$$

$$\frac{{dy}}{{dx}} = \frac{1}{{1 + {x^3}}}\,\,\,\,\,\,\,\therefore {\left. {\frac{{dy}}{{dx}}} \right)_{x = 1}} = \frac{1}{2}$$

$$\eqalign{ & f''\left( x \right) > 0,\forall x \in R \cr & f\left( {\frac{1}{2}} \right) = \frac{1}{2},f\left( 1 \right) = 1 \cr} $$
∴ $$f'$$ is an increasing function on $$R.$$
By Lagrange's Mean Value theorem.
$$\eqalign{ & f'\left( \alpha \right) = \frac{{f\left( 1 \right) - f\left( {\frac{1}{2}} \right)}}{{1 - \frac{1}{2}}},\alpha \in \left( {\frac{1}{2},1} \right) \cr & \Rightarrow f'\left( \alpha \right) = 1\,{\text{for}}\,{\text{some}}\,\alpha \in \left( {\frac{1}{2},1} \right) \cr & \Rightarrow f'\left( 1 \right) > 1 \cr} $$

$$\eqalign{ & f\left( x \right) = 4 - {\cos ^2}x + \cos \frac{x}{2} + \sin \frac{x}{2} \cr & \therefore f'\left( x \right) = \sin \,2x - \frac{1}{2}\left( {\sin \frac{x}{2} - \cos \frac{x}{2}} \right) \cr & = 2\sin \,x.\left( {{{\cos }^2}\frac{x}{2} - {{\sin }^2}\frac{x}{2}} \right) + \frac{1}{2}\left( {\cos \frac{x}{2} - \sin \frac{x}{2}} \right) \cr & = \left( {\cos \frac{x}{2} - \sin \frac{x}{2}} \right)\left\{ {2\sin \,x\left( {\cos \frac{x}{2} + \sin \frac{x}{2}} \right) + \frac{1}{2}} \right\} \cr & = \left( {\cos \frac{x}{2} - \sin \frac{x}{2}} \right)\left\{ {2\sqrt 2 \sin \,x.\sin \left( {\frac{x}{2} + \frac{\pi }{4}} \right) + \frac{1}{2}} \right\} \cr & \therefore \,\,f'\left( x \right) = 0\,\,\,\, \Rightarrow \cos \frac{x}{2} - \sin \frac{x}{2} = 0 \cr & \Rightarrow x = \frac{\pi }{2} \cr & f''\left( x \right) = 2\cos \,2x - \frac{1}{2}.\frac{1}{2}\left( {\cos \frac{x}{2} + \sin \frac{x}{2}} \right) \cr & \therefore f''\left( {\frac{\pi }{2}} \right) = - 2 - \frac{1}{4}\left( {\frac{1}{{\sqrt 2 }} + \frac{1}{{\sqrt 2 }}} \right) < 0 \cr & \therefore \max \,f\left( x \right) = f\left( {\frac{\pi }{2}} \right) = 4 - 0 + \frac{1}{{\sqrt 2 }} + \frac{1}{{\sqrt 2 }} = 4 + \sqrt 2 \cr} $$

Using Lagrange's Mean Value Theorem Let $$f\left( x \right)$$  be a function defined on $$[a,b]$$
$$\eqalign{ & {\text{then}}\,f'\left( c \right) = \frac{{f\left( b \right) - f\left( a \right)}}{{b - a}}\,......\left( {\text{i}} \right) \cr & c \in \left[ {a,b} \right] \cr & \therefore {\text{Given }}f\left( x \right) = {\log _e}x\therefore f'\left( x \right) = \frac{1}{x} \cr & \therefore {\text{equation}}\,\left( {\text{i}} \right)\,{\text{become }}\frac{1}{c} = \frac{{f\left( 3 \right) - f\left( 1 \right)}}{{3 - 1}} \cr & \Rightarrow \frac{1}{c} = \frac{{{{\log }_e}3 - {{\log }_e}1}}{2} = \frac{{{{\log }_e}3}}{2} \cr & \Rightarrow c = \frac{2}{{{{\log }_e}3}} \Rightarrow c = 2{\log _3}e \cr} $$

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

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

$$\eqalign{ & x = a\left( {\cos \theta + \theta \sin \theta } \right) \cr & \Rightarrow \frac{{dx}}{{d\theta }} = a\left( { - \sin \theta + \sin \theta + \theta \cos \theta } \right) \cr & \Rightarrow \frac{{dx}}{{d\theta }} = a\theta \cos \theta \,......\left( 1 \right) \cr & y = a\left( {\sin \theta - \theta \cos \theta } \right) \cr & \frac{{dy}}{{d\theta }} = a\left[ {\cos \theta - \cos \theta + \theta \sin \theta } \right] \cr & \Rightarrow \frac{{dy}}{{d\theta }} = a\theta \sin \theta \,......\left( 2 \right) \cr & {\text{From equations}}\,\left( {\text{1}} \right)\,{\text{and}}\,\left( {\text{2}} \right){\text{, we get}} \cr & \frac{{dy}}{{dx}} = \tan \theta \Rightarrow {\text{Slope of normal}} = - \cot \theta \cr & {\text{Equation}}\,{\text{of}}\,{\text{normal}}\,{\text{at}}\,'\theta '\,{\text{is}}\,y - a\left( {\sin \theta - \theta \cos \theta } \right) \cr & = - \cot \theta \left( {x - a} \right.\left( {\cos \theta + \theta \sin \theta } \right) \cr & \Rightarrow y\sin \theta - a{\sin ^2}\theta + a\theta \cos \theta \sin \theta \cr & = - x\cos \theta + a{\cos ^2}\theta + a\theta \sin \theta \cos \theta \cr & \Rightarrow x\cos \theta + y\sin \theta = a \cr} $$
Clearly this is an equation of straight line which is at a constant distance $$'a'$$ from origin.

Let $$y = \cos \left( {x + y} \right)$$
$$ \Rightarrow \frac{{dy}}{{dx}} = - \sin \left( {x + y} \right)\left( {1 + \frac{{dy}}{{dx}}} \right)......\left( 1 \right)$$
Now, given equation of tangent is $$x + 2y = k$$
$$\eqalign{ & \Rightarrow {\text{Slope}} = \frac{{ - 1}}{2} \cr & {\text{So, }}\frac{{dy}}{{dx}} = \frac{{ - 1}}{2}\,{\text{put this value in equation}}\left( 1 \right),{\text{ we get}} \cr & \frac{{ - 1}}{2} = - \sin \left( {x + y} \right)\left( {1 - \frac{1}{2}} \right) \cr & \Rightarrow \sin \left( {x + y} \right) = 1 \cr & \Rightarrow x + y = \frac{\pi }{2} \cr & \Rightarrow y = \frac{\pi }{2} - x \cr & {\text{Now, }}\frac{\pi }{2} - x = \cos \left( {x + y} \right) \cr & \Rightarrow x = \frac{\pi }{2}{\text{ and }}y = 0 \cr & {\text{Thus }}x + 2y = k \Rightarrow \frac{\pi }{2} = k \cr} $$