STD 12 - 6. Application of derivatives — JEE STD 11 Science — Question

==> $f'(x) = \frac{{1 - 2{x^2}}}{{\sqrt {1 - {x^2}} }} = 0 \Rightarrow x = \pm \frac{1}{{\sqrt 2 }}$

But as $x > 0$, we have $x = \frac{1}{{\sqrt 2 }}$

Now, again $f''(x) = \frac{{\sqrt {1 - {x^2}} ( - 4x) - (1 - 2{x^2})\frac{{ - x}}{{\sqrt {1 - {x^2}} }}}}{{(1 - {x^2})}}$

$ = \frac{{2{x^3} - 3x}}{{{{(1 - {x^2})}^{3/2}}}}$

$ \Rightarrow f''\left( {\frac{1}{{\sqrt 2 }}} \right) = - ve$.

Then $f(x)$ is maximum at $x = \frac{1}{{\sqrt 2 }}$.