Question

Let f(x)=\left\\{ \begin{matrix} \frac{\tan x-\cot x}{x-\frac{\pi }{4}}, & x\ne \frac{\pi }{4} \\\ a, & x=\frac{\pi }{4} \\\ \end{matrix} \right. the value of a so that $f(x)$ is continuous at $x=\frac{\pi }{4}$

• A. 2
• B. 4
• C. 3
• D. 1

Answer 1

Answer: (B)

4

Answer 2

A function is said to be a continuous function at $x=a$, if
$LHL = RHL =$ Value of function at $x=a .$
Since, $f(x)=\begin{cases}\frac{\tan x-\cot x}{x-\frac{\pi}{4}}, & x \neq \frac{\pi}{4} \\\ a, & x=\frac{\pi}{4}\end{cases}$
$LHL =\displaystyle\lim _{h \rightarrow 0} f\left(\frac{\pi}{4}-h\right)$
$=\displaystyle\lim _{h \rightarrow 0} \frac{\tan \left(\frac{\pi}{4}-h\right)-\cot \left(\frac{\pi}{4}-h\right)}{\frac{\pi}{4}-h-\frac{\pi}{4}}$
Applying L' Hospital's rule
$=\displaystyle\lim _{h \rightarrow 0} \frac{\sec ^{2}\left(\frac{\pi}{4}-h\right)+\text{cosec}^{2}\left(\frac{\pi}{4}-h\right)}{1}$
$=2+2=4 .$
$RHL =\displaystyle\lim _{h \rightarrow 0} f\left(\frac{\pi}{4}+h\right)$
$=\displaystyle\lim _{h \rightarrow 0} \frac{\tan \left(\frac{\pi}{4}+h\right)-\cot \left(\frac{\pi}{4}+h\right)}{\frac{\pi}{4}+h-\frac{\pi}{4}}$
$=\displaystyle\lim _{h \rightarrow 0} \frac{\sec ^{2}\left(\frac{\pi}{4}+h\right)+\text{cosec}^{2}\left(\frac{\pi}{4}+h\right)}{1}$
$=2+2=4$
$\because$ Function is continuous at $x=\frac{\pi}{4}$.
$\therefore f\left(\frac{\pi}{4}\right)= RHL = LHL$
$\therefore a=4$
Alternate Method : $\because f(x)$ is continuous function at $x=\frac{\pi}{4}$.
$\therefore \displaystyle\lim _{x \rightarrow \frac{\pi}{4}} \frac{\tan x-\cot x}{x-\frac{\pi}{4}}=a$
Applying L' Hospital's rule
$\Rightarrow \displaystyle\lim _{x \rightarrow \frac{\pi}{4}} \frac{\sec ^{2} x+\text{cosec}^{2} x}{1}=a$
$\Rightarrow a=2+2=4$

So, the correct option is (B): 4