A) 4
B) 2
C) 1
D) 1/4
Correct Answer: D
Solution :
\[f(x)=\left\{ \begin{matrix} \frac{1-\sqrt{2}\sin x}{\pi -4x}, & if\,x\ne \frac{\pi }{4} \\ a, & if\,x=\frac{\pi }{4} \\ \end{matrix} \right.\] \[\underset{x\to \frac{\pi }{4}}{\mathop{\lim }}\,f(x)=\underset{x\to \frac{\pi }{4}}{\mathop{\lim }}\,\frac{1-\sqrt{2}\sin x}{\pi -4x}\] \[=\underset{x\to \frac{\pi }{4}}{\mathop{\lim }}\,\frac{-\sqrt{2}\cos x}{-4}\] (by L Hospitals rule) \[=\frac{\sqrt{2}.\frac{1}{\sqrt{2}}}{4}=\frac{1}{4}\] Since,\[f(x)\] is continuous at\[x=\frac{\pi }{4}.\] \[\therefore \] \[\underset{x\to \frac{\pi }{4}}{\mathop{\lim }}\,f(x)=f\left( \frac{\pi }{4} \right)\] \[\Rightarrow \] \[\frac{1}{4}=a\]
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