if f(x)=2\tan ^{-1}x+\sin ^{-1}\left ( \frac{2x}{1+x^{2}} \right ),x>1then 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)
D Divya Saini

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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