Given $$f\left( x \right) = {\tan ^{ - 1}}\left( {\sin x + \cos x} \right)$$

$$f'\left( x \right) = {1 \over {1 + {{\left( {\sin x + \cos x} \right)}^2}}}.\left( {\cos x - \sin x} \right)$$

$$ = {{\sqrt 2 .\left( {{1 \over {\sqrt 2 }}\cos x - {1 \over {\sqrt 2 }}\sin x} \right)} \over {1 + {{\left( {\sin x + \cos x} \right)}^2}}}$$

$$ = {{\left( {\cos {\pi \over 4}.\cos x - \sin {\pi \over 4}.\sin x} \right)} \over {1 + {{\left( {\sin x + \cos x} \right)}^2}}}$$

$$\therefore$$ $$f'\left( x \right) = {{\sqrt 2 \cos \left( {x + {\pi \over 4}} \right)} \over {1 + {{\left( {\sin x + \cos x} \right)}^2}}}$$

if $$f'\left( x \right) > O$$ then $$f\left( x \right)$$ is increasing function.

Hence $$f(x)$$ is increasing, if $$ - {\pi \over 2} < x + {\pi \over 4} < {\pi \over 2}$$

$$ \Rightarrow - {{3\pi } \over 4} < x < {\pi \over 4}$$

Hence, $$f(x)$$ is increasing when $$n \in \left( { - {\pi \over 2},{\pi \over 4}} \right)$$

Area $$ = {1 \over 2}{x^2}\,\sin \,\theta $$



Maximum value of $$\sin \theta $$ is $$1$$ at $$\theta = {\pi \over 2}$$

$${A_{\max }} = {1 \over 2}{x^2}$$

$$f\left( x \right) = {x \over 2} + {2 \over x} \Rightarrow f'\left( x \right) = {1 \over 2} - {2 \over {{x^2}}} = 0$$

$$ \Rightarrow {x^2} = 4$$ or $$x=2,-2;$$ $$\,\,\,\,\,f''\left( x \right) = {4 \over {{x^3}}}$$

$$f''{\left. {\left( x \right)} \right]_{x = 2}} = + ve \Rightarrow f\left( x \right)$$

has local min at $$x=2.$$

Let $$f\left( x \right) = {a_n}{x^n} + {a_{n - 1}}{x^{n - 1}} + ........... + {a_1}x = 0$$

The other given equation,

$$na{}_n{x^{n - 1}} + \left( {n - 1} \right){a_{n - 1}}{x^{n - 2}} + .... + {a_1} = 0 = f'\left( x \right)$$

Given $${a_1} \ne 0 \Rightarrow f\left( 0 \right) = 0$$

Again $$f(x)$$ has root $$\alpha ,\,\,\, \Rightarrow f\left( \alpha \right) = 0$$

$$\therefore$$ $$\,\,\,\,\,f\left( 0 \right) = f\left( \alpha \right)$$

$$\therefore$$ By Roll's theorem $$f'(x)=0$$ has root be-

tween $$\left( {0,\alpha } \right)$$

Hence $$f'(x)$$ has a positive root smaller than $$\alpha .$$