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JEE Main & Advanced Mathematics Limits, Continuity and Differentiability Question Bank Self Evaluation Test - Continuity and Differentiability

  • question_answer
    Let \[f(x)=\left\{ \begin{align}   & \frac{1-{{\sin }^{3}}x}{3{{\cos }^{2}}x},x<\frac{\pi }{2} \\  & p,x=\frac{\pi }{2} \\  & \frac{q(1-\sin x)}{{{(\pi -2x)}^{2}}},x>\frac{\pi }{2} \\ \end{align} \right.\] If f(x) is continuous at \[x=\frac{\pi }{2},(p,q)=\]

    A) \[(1,4)\]

    B) \[\left( \frac{1}{2},2 \right)\]

    C) \[\left( \frac{1}{2},4 \right)\]

    D) None of these

    Correct Answer: C

    Solution :

    [c] \[f[{{(\pi /2)}^{-}}]=\underset{h\to 0}{\mathop{\lim }}\,\frac{1-{{\sin }^{3}}\left[ (\pi /2)-h \right]}{3{{\cos }^{2}}\left[ (\pi /2)-h \right]}\] \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{1-{{\cos }^{3}}h}{3{{\sin }^{2}}h}=\frac{1}{2}\] \[f[{{(\pi /2)}^{+}}]=\underset{h\to 0}{\mathop{\lim }}\,\frac{q[1-sin\{(\pi /2)+h\}]}{{{[\pi -2\{(\pi /2)+h\}]}^{2}}}\] \[=\,\,\,\underset{h\to 0}{\mathop{\lim }}\,\frac{q(1-cos\,h)}{4{{h}^{2}}}=\frac{q}{8}\] \[\therefore p=\frac{1}{2}=\frac{q}{8}\Rightarrow p=\frac{1}{2},q=4.\]


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