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RAJASTHAN ­ PET Rajasthan PET Solved Paper-2002

  • question_answer
    If the function\[f(x)=\left\{ \begin{matrix} \frac{1-\sin x}{\pi -2x}, & x\ne \frac{\pi }{2}  \\ \lambda , & x=\frac{\pi }{2}  \\ \end{matrix} \right.\] is continuous at\[x=\frac{\pi }{2},\]then\[\lambda \]is equal to

    A)  1              

    B)  \[-1\]

    C)  \[1/2\]             

    D)  0

    Correct Answer: D

    Solution :

     \[f(x)=\left\{ \begin{matrix}    \frac{1-\sin x}{\pi -2x}, & x\ne \frac{\pi }{2}  \\    \lambda , & x=\frac{\pi }{2}  \\ \end{matrix} \right.\] \[\because \]\[f(x)\]is continuous at\[x=\frac{\pi }{2},\] \[\therefore \] \[\underset{x\to \frac{\pi }{2}}{\mathop{\lim }}\,f(x)=f\left( \frac{\pi }{2} \right)\] \[\Rightarrow \] \[\underset{x\to \frac{\pi }{2}}{\mathop{\lim }}\,\frac{1-\sin x}{\pi -2x}=\lambda \] \[\Rightarrow \] \[\underset{x\to \frac{\pi }{2}}{\mathop{\lim }}\,\frac{-\cos x}{-2}=\lambda \] [using L-Hospital rule] \[\frac{\cos \frac{\pi }{2}}{2}=\lambda \] \[\Rightarrow \] \[\lambda =0\]


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