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JAMIA MILLIA ISLAMIA Jamia Millia Islamia Solved Paper-2007

  • question_answer
        Let \[f(x)=\frac{1-\tan x}{4x-n},x\ne \frac{\pi }{4},x\in \left[ 0,\frac{\pi }{2} \right]\]. If  \[f(x)\]is continuous in\[\left[ 0,\frac{\pi }{2} \right],\]then\[f\left( \frac{\pi }{4} \right)\]is

    A)  1                                            

    B)  \[1/2\]

    C)  \[-1/2\]                              

    D)  \[-1\]

    Correct Answer: C

    Solution :

                     Since, \[f(x)=\frac{1-\tan x}{4x-\pi }\] \[\underset{x\to \pi /4}{\mathop{\lim }}\,f(x)=\underset{x\to \pi /4}{\mathop{\lim }}\,\left( \frac{1-\tan x}{4x-\pi } \right)\] By LHospitals rule \[=\underset{x\to \pi /4}{\mathop{\lim }}\,\left( \frac{-{{\sec }^{2}}x}{4} \right)=\frac{-{{\sec }^{2}}(\pi /4)}{4}=-\frac{2}{4}\] \[\Rightarrow \] \[\underset{x\to \pi /4}{\mathop{\lim }}\,f(x)=-\frac{1}{2}\] Also,\[f(x)\] is continuous in\[[0,\pi /2]\], so\[f(x)\]will be continuous at\[\pi /4\]. \[\therefore \]Value of function = Value of limit \[\Rightarrow \]               \[f\left( \frac{\pi }{4} \right)=-\frac{1}{2}\]


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