Question

If $f\left(x\right)=\frac{\tan (\frac{\pi }{4}-x)}{\cot 2x},(x\ne \pi /4)$,find the value which can be assigned to $f\left(x\right)$ at $x=\pi /4$ that the function $f\left(x\right)$ becomes continuous everywhere in $[0,\pi /2]$.

• A. $1$
• B. $\frac{1}{2}$
• C. $\frac{1}{3}$
• D. $-1$

Answer 1

Answer: (B)

$\frac{1}{2}$

$f\left(x\right)=\frac{tan(\frac{\pi }{4}-x)}{cot2x},(x\ne \pi /4)$, is continuous at $x=\pi /4$. therefore,

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

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

Now, by applying L'Hospital's rule,
$\underset{x\to \frac{\pi }{4}}{lim}\frac{-se{c}^2\left(\frac{\pi }{4}x\right)}{-2cose{c}^2\left(2x\right)}=\frac{1}{2}$