Question

Let $f\left(x\right)=\frac{tan(\frac{\pi }{4}-x)}{cot2x}$. The value which should be assigned to $f$ at $x=\pi /4$. So that it is continuous every where, is

• A. 1/2
• B. 1
• C. 2
• D. none of these

Answer 1

The correct option is A 1/2

Given $f\left(x\right)=\frac{tan(\frac{\pi }{4}-x)}{cot2x}$

Given function is continuous everywhere and we have to find $f\left(x\right)$ at $x=\frac{\pi }{4}$, we can write,

$f\left(\frac{\pi }{4}\right)=\underset{x\to \frac{\pi }{4}}{lim}f\left(x\right)$

$\therefore f\left(\frac{\pi }{4}\right)=\underset{x\to \frac{\pi }{4}}{lim}\frac{tan(\frac{\pi }{4}-x)}{cot2x}$

We have to make limit tending to zero.

$\therefore$ put $t=x-\frac{\pi }{4}$

$\therefore \frac{\pi }{4}-x=-t$

Similarly, $x=t+\frac{\pi }{4}$

$\therefore 2x=2t+\frac{\pi }{2}$

$\therefore f\left(\frac{\pi }{4}\right)=\underset{t\to 0}{lim}\frac{tan(-t)}{cot(2t+\frac{\pi }{2})}$

Now, $tan(-\theta )=-tan\theta$ and $cot(\frac{\pi }{2}+\theta )=-tan\theta$

$\therefore f\left(\frac{\pi }{4}\right)=\underset{t\to 0}{lim}\frac{-tan\left(t\right)}{-tan\left(2t\right)}$

$\therefore f\left(\frac{\pi }{4}\right)=\underset{t\to 0}{lim}\frac{tan\left(t\right)}{\left(\frac{2tant}{1-{tan}^{2}t}\right)}$

$\therefore f\left(\frac{\pi }{4}\right)=\underset{t\to 0}{lim}\frac{tan\left(t\right)\times (1-{tan}^{2}t)}{2tan\left(t\right)}$

$\therefore f\left(\frac{\pi }{4}\right)=\underset{t\to 0}{lim}\frac{(1-{tan}^{2}t)}{2}$

$\therefore f\left(\frac{\pi }{4}\right)=\frac{(1-{tan}^2\left(0\right))}{2}$

$\therefore f\left(\frac{\pi }{4}\right)=\frac{(1-0)}{2}$

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