# Solve this

Question:

Let $f(x)=\frac{\tan \left(\frac{\pi}{4}-x\right)}{\cot 2 x}, x \neq \frac{\pi}{4}$. The value which should be assigned to $f(x)$ at $x=\frac{\pi}{4}$, so that it is continuous everywhere is

(a) 1

(b) $1 / 2$

(c) 2

(d) none of these

Solution:

(b) $\frac{1}{2}$

If $f(x)$ is continuous at $x=\frac{\pi}{4}$, then

$\lim _{x \rightarrow \frac{\pi}{4}} f(x)=f\left(\frac{\pi}{4}\right)$

$\Rightarrow \lim _{x \rightarrow \frac{\pi}{4}} \frac{\tan \left(\frac{\pi}{4}-x\right)}{\cot 2 x}=f\left(\frac{\pi}{4}\right)$

If $\frac{\pi}{4}-x=y$, then $x \rightarrow \frac{\pi}{4}$ and $y \rightarrow 0$.

$\therefore \lim _{y \rightarrow 0}\left(\frac{\tan y}{\cot 2\left(\frac{\pi}{4}-y\right)}\right)=f\left(\frac{\pi}{4}\right)$

$\Rightarrow \lim _{y \rightarrow 0}\left(\frac{\tan y}{\cot \left(\frac{\pi}{2}-2 y\right)}\right)=f\left(\frac{\pi}{4}\right)$

$\Rightarrow \lim _{y \rightarrow 0}\left(\frac{\tan y}{\tan 2 y}\right)=f\left(\frac{\pi}{4}\right)$

$\Rightarrow \lim _{y \rightarrow 0}\left(\frac{\frac{\tan y}{y}}{\frac{\tan 2 y}{y}}\right)=f\left(\frac{\pi}{4}\right)$

$\Rightarrow \lim _{y \rightarrow 0}\left(\frac{\frac{\tan y}{y}}{\frac{2 \tan 2 y}{2 y}}\right)=f\left(\frac{\pi}{4}\right)$

$\Rightarrow \frac{1}{2} \lim _{y \rightarrow 0}\left(\frac{\frac{\tan y}{y}}{\frac{\tan 2 y}{2 y}}\right)=f\left(\frac{\pi}{4}\right)$

$\Rightarrow \frac{1}{2}\left(\frac{\lim _{y \rightarrow 0} \frac{\tan y}{y}}{\lim _{y \rightarrow 0} \frac{\tan 2 y}{2 y}}\right)=f\left(\frac{\pi}{4}\right)$

$\Rightarrow \frac{1}{2}\left(\frac{1}{1}\right)=f\left(\frac{\pi}{4}\right)$

$\Rightarrow f\left(\frac{\pi}{4}\right)=\frac{1}{2}$