MCQ
${\tan ^{ - 1}}\frac{{1 - {x^2}}}{{2x}} + {\cos ^{ - 1}}\frac{{1 - {x^2}}}{{1 + {x^2}}} = $
- A$\frac{\pi }{4}$
- ✓$\frac{\pi }{2}$
- C$\pi $
${\tan ^{ - 1}}\frac{{1 - {x^2}}}{{2x}} + {\cos ^{ - 1}}\frac{{1 - {x^2}}}{{1 + {x^2}}}$
$ = {\tan ^{ - 1}}\left( {\frac{{1 - {{\tan }^2}\theta }}{{2\tan \theta }}} \right) + {\cos ^{ - 1}}\left( {\frac{{1 - {{\tan }^2}\theta }}{{1 + {{\tan }^2}\theta }}} \right)$
$ = {\tan ^{ - 1}}(\cot 2\theta ) + {\cos ^{ - 1}}(\cos 2\theta )$
$ = \frac{\pi }{2} - 2\theta + 2\theta = \frac{\pi }{2}$.
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($1$) $P(X>Y)$ is
($A$) $\frac{1}{4}$ ($B$) $\frac{5}{12}$ ($C$) $\frac{1}{2}$ ($D$) $\frac{7}{12}$
($2$) $P(X=Y)$ is
($A$). $\frac{11}{36}$ ($B$) $\frac{1}{3}$ ($C$) $\frac{13}{36}$ ($D$) $\frac{1}{2}$
Given the answer quetion ($1$) and ($2$)
$8 \sqrt{x}(\sqrt{9+\sqrt{x}}) d y=(\sqrt{4+\sqrt{9+\sqrt{x}}})^{-1} d x, \quad x>0$
and $y(0)=\sqrt{7}$, then $y(256)=$