A) 2
B) 4
C) 3
D) 1
Correct Answer: B
Solution :
Key Idea: A function is said to be a continuous function at \[x=a,\]if \[\text{LHL}\,\text{=}\,\text{RHL}\,\text{=}\,\text{Value}\,\text{of}\,\text{function}\,\text{at}\,x=a.\] Since, \[f(x)=\left\{ \begin{matrix} \frac{\tan x-\cot x}{x-\frac{\pi }{4}}, & x\ne \frac{\pi }{4} \\ a, & x=\frac{\pi }{4} \\ \end{matrix} \right.\] \[\text{LHL}=\underset{h\to 0}{\mathop{\lim }}\,f\left( \frac{\pi }{4}-h \right)\] \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{\tan \left( \frac{\pi }{4}-h \right)-\cot \left( \frac{\pi }{4}-h \right)}{\frac{\pi }{4}-h-\frac{\pi }{4}}\] Applying L Hospitals rule \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{{{\sec }^{2}}\left( \frac{\pi }{4}-h \right)+\cos e{{c}^{2}}\left( \frac{\pi }{4}-h \right)}{1}\] \[=2+2=4.\] \[RHL=\underset{h\to 0}{\mathop{\lim }}\,f\left( \frac{\pi }{4}+h \right)\] \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{\tan \left( \frac{\pi }{4}+h \right)-\cot \left( \frac{\pi }{4}+h \right)}{\frac{\pi }{4}+h-\frac{\pi }{4}}\] \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{{{\sec }^{2}}\left( \frac{\pi }{4}+h \right)+\cos e{{c}^{2}}\left( \frac{\pi }{4}+h \right)}{1}\] \[=2+2=4\] \[\because \]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 \[x=\frac{\pi }{4}.\] \[\therefore \] \[\underset{x\to \frac{\pi }{4}}{\mathop{\lim }}\,\frac{\tan x-\cot x}{x-\frac{\pi }{4}}=a\] Applying L Hospitals rule \[\Rightarrow \]\[\underset{x\to \frac{\pi }{4}}{\mathop{\lim }}\,\frac{{{\sec }^{2}}x+\cos e{{c}^{2}}x}{1}=a\] \[\Rightarrow \]\[a=2+2=4.\]
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