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JEE Main & Advanced Mathematics Functions Question Bank Continuity

  • question_answer
    If \[f(x)=\left\{ \begin{matrix}    \frac{1-\sin x}{\pi -2x}, & x\ne \frac{\pi }{2}  \\    \,\,\,\,\,\,\,\,\,\,\,\,\,\lambda \,, & x=\frac{\pi }{2}  \\ \end{matrix} \right.\], be continuous at \[x=\pi /2,\] then value of \[\lambda \] is                                           [RPET 2002]

    A)            ?1

    B)            1           

    C)            0

    D)            2

    Correct Answer: C

    Solution :

               \[f(x)\] is continuous at \[x=\frac{\pi }{2}\], then            \[\underset{x\to \pi /2}{\mathop{\lim }}\,f(x)=f(0)\] or \[\lambda =\underset{x\to \pi /2}{\mathop{\lim }}\,\frac{1-\sin x}{\pi -2x}\],  \[\left( \frac{0}{0}\text{ form} \right)\]            Applying L-Hospital?s rule,            \[\lambda =\underset{x\to \pi /2}{\mathop{\lim }}\,\frac{-\cos x}{-2}\] Þ \[\lambda =\underset{x\to \pi /2}{\mathop{\lim }}\,\frac{\cos x}{2}=0.\]


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