JEE Mains · Maths · STD 12 - 2. inverse trigonometric function
If \(f(x)\, = \,2\,{\tan ^{ - 1}}\,x\, + \,{\sin ^{ - 1}}\,\left( {\frac{{2x}}{{1 + {x^2}}}} \right),x > 1\,\) then \(f\,(5)\) is equal to
- A \(\,{\tan ^{ - 1}}\left( {\frac{{65}}{{156}}} \right)\)
- B \(\frac {\pi }{2}\)
- C \(\pi \)
- D \(\,4\,\,{\tan ^{ - 1}}(5)\)
Answer & Solution
Correct Answer
(C) \(\pi \)
Step-by-step Solution
Detailed explanation
\(f\left( x \right) = 2{\tan ^{ - 1}}x + {\sin ^{ - 1}}\left( {\frac{{2x}}{{1 + {x^2}}}} \right)\) \( \Rightarrow f\left( x \right) = 2{\tan ^{ - 1}}x + \pi - 2{\tan ^{ - 1}}x\) \( \Rightarrow f\left( x \right) = \pi \) \( \Rightarrow f\left( 5 \right) = \pi \)
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