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if $f(x)=2\tan ^{-1}x+\sin ^{-1}\left ( \frac{2x}{1+x^{2}} \right ),x>1$ then $f(5)$ is equal to :

Option 1)
$\frac{\pi }{2}$

Option 2)
$\pi$

Option 3)
$4\tan ^{-1}(5)$

Option 4)
$\tan ^{-1}\left ( \frac{65}{156} \right )$

Answers (1)

best_answer

As we learnt in

Inverse Trigonometric Functions -

The functions $\sin ^{-1}x, \cos ^{-1}x, \tan ^{-1}x, \cot ^{-1}x, \csc ^{-1}x$ and $\sec ^{-1}x$ are the inverse trigonometric functions.

- wherein

If $\sin x = \frac{1}{2}$

then, $x = \sin ^{-1}\frac{1}{2}$

$f(5)=2\tan^{-1}5+\sin^{-1}\frac{2\times 5}{1+25}$ $=2\tan^{-1}5+\sin^{-1}\frac{5}{13}$

$=2\tan^{-1}5+\tan^{-1}\frac{5}{12}$ $\left [ \sin^{-1}\frac{5}{13}=\tan^{-1}\frac{5}{12} \right ]$

$=\tan^{-1}\frac{2\times 5}{1-25}+\tan^{-1}\frac{5}{12}$

$=\tan^{-1}\left( -\frac{5}{12}\right)=\tan^{-1}\frac{5}{12}$

now,

$\tan [f(5)]=\tan\: \left[\tan^{-1} \frac{5}{12}+\tan^{-1}\left ( -\frac{5}{12} \right )\right]=0$

$\Rightarrow [f(5)]=\tan^{-1}0=0\:\:or\:\pi$ .......................... ( 1 )

Now,

$x>0,\:\tan^{-1} 5>0,\:\sin\frac{5}{13}>0$ ............................( 2 )

$\Rightarrow f(5)>0$

$\Rightarrow f(5)=\pi$ [From ( 1 ) and ( 2 )]

Option 1)

$\frac{\pi }{2}$

This option is incorrect.

Option 2)

$\pi$

This option is correct.

Option 3)

$4\tan ^{-1}(5)$

This option is incorrect.

Option 4)

$\tan ^{-1}\left ( \frac{65}{156} \right )$

This option is incorrect.

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divya.saini

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