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

The Correct Option is B

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