JEE Main 2016 (9)

Let $\begin{cases} \text{arccot} x = \alpha & \Rightarrow p = \sin \alpha & \Rightarrow \cot \alpha = x \\ \arcsin x = \beta & \Rightarrow q = \cot \beta & \Rightarrow \sin \beta = x \end{cases}$

From $\cot \alpha = x$ :

$\begin{aligned} \Rightarrow \frac{\cos^2 \alpha}{\sin^2 \alpha} & = x^2 \\ \frac{1-\sin^2 \alpha}{\sin^2 \alpha} & = x^2 \\ \sin^2 \alpha & = \frac{1}{1+x^2} \\ \Rightarrow p^2 & = \frac{1}{1+x^2} \end{aligned}$

From $q = \cot \beta$ :

$\begin{aligned} \Rightarrow q & = \frac{\cos \beta}{\sin \beta} \\ q^2 & = \frac{\cos^2 \beta}{\sin^2 \beta} = \frac{1-\sin^2 \beta}{\sin^2 \beta} = \frac{1-x^2}{x^2} \\ \Rightarrow x^2 & = \frac{1}{1+q^2} \end{aligned}$

Therefore, we have:

$\begin{aligned} p^2 & = \frac{1}{1+x^2} = \frac{1}{1+\frac{1}{1+q^2}} = \frac{1+q^2}{1+q^2+1} = \frac{f(q)}{1+f(q)} \\ \Rightarrow f(q) & = 1+q^2 \\ f(5) & = 1 + 5^2 = \boxed{26} \end{aligned}$