Question

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

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

Answer 1

Answer: (B)

$\pi$

Answer 2

The given function is $f(x) = 2tan^{-1}x + sin^{-1}(\frac{2x}{1 + x^2})$ for $x > 1$.

We use the identity for $sin^{-1}(\frac{2x}{1 + x^2})$ in terms of $tan^{-1}x$. The general identity is:

$sin^{-1}(\frac{2x}{1 + x^2}) = \begin{cases} 2tan^{-1}x & \text{if } |x| \le 1 \\ \pi - 2tan^{-1}x & \text{if } x > 1 \\ -\pi - 2tan^{-1}x & \text{if } x < -1 \end{cases}$

Given that $x > 1$, we use the second case of the identity: $sin^{-1}(\frac{2x}{1 + x^2}) = \pi - 2tan^{-1}x$ for $x > 1$.

Substitute this into the expression for $f(x)$:

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

For $x > 1$, this becomes:

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

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

$f(x) = \pi$

So, for all $x > 1$, the function $f(x)$ has a constant value of $\pi$.

We need to find $f(5)$. Since $5 > 1$, the value of $f(5)$ is $\pi$. Therefore, $f(5) = \pi$.