A) \[\frac{1}{2}\]
B) \[1\]
C) \[\frac{1}{\sqrt{2}}\]
D) \[2\]
Correct Answer: A
Solution :
\[\therefore \]function should be continuous at\[x=\frac{\pi }{4}\] \[\therefore \]\[\underset{x\to \frac{\pi }{4}}{\mathop{\lim }}\,f(x)=f\left( \frac{\pi }{4} \right)\]\[\Rightarrow \underset{x\to \frac{\pi }{4}}{\mathop{\lim }}\,\frac{\sqrt{2}\cos sx-1}{\cot x-1}=k\] \[\Rightarrow \underset{x\to \frac{\pi }{4}}{\mathop{\lim }}\,\frac{-\sqrt{2}\operatorname{sinx}}{\cot \,e{{c}^{2}}x}=k\](Using L'H\[\hat{o}\]pital rule) \[\underset{x\to \frac{\pi }{4}}{\mathop{\lim }}\,\sqrt{2}{{\sin }^{3}}x=k\] \[\Rightarrow k=\sqrt{2}{{\left( \frac{1}{\sqrt{2}} \right)}^{3}}=\frac{1}{2}\]
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