Question

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

• A. $\frac{\pi}{2}$
• B. $\pi$
• C. $4\, \tan^{-1} \left(5\right)$
• D. $ \tan^{-1} \left(\frac{65}{156}\right)$

Answer 1

The Correct Option is B

 The given function is:

\(f(x) = 2 \tan^{-1} x + \sin^{-1} \left(\frac{2x}{1+x^{2}}\right)\)

We are asked to find \(f(t)\) for some value \(t\).

Let's use the identity for the inverse sine function:

\(\sin^{-1} \left(\frac{2x}{1+x^2}\right) = 2 \tan^{-1} x\) for \(x > 1\)

Referencing this identity allows us to rewrite the given expression:

\(f(x) = 2 \tan^{-1} x + 2 \tan^{-1} x = 4 \tan^{-1} x\)

Since \(\tan^{-1} x\) asymptotically approaches \(\frac{\pi}{2}\) as \(x\) goes towards infinity, it results in:

\(4 \times \frac{\pi}{4} = \pi\)

Thus for \(t \to \infty\) or for sufficiently large \(t\) in \(x > 1\), we have:

\(f(t) = \pi\)

Therefore, the correct answer is:

\(\pi\)