JEE Main & Advanced JEE Main Paper (Held on 09-4-2019 Morning)

  • question_answer If the function ? defined on\[\left( \frac{\pi }{6},\frac{\pi }{3} \right)\]by\[f(x)=\left\{ \begin{matrix}    \frac{\sqrt{2}\cos x-1}{\cot x-1}, & x\ne \frac{\pi }{4}  \\    k, & x=\frac{\pi }{4}  \\ \end{matrix} \right.\] is continuous, then k is equal to [JEE Main 9-4-2019 Morning]

    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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