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

  • question_answer
    The function \[f(x)=\frac{1-\sin x+\cos x}{1+\sin x+\cos x}\] is not defined at \[x=\pi \]. The value of \[f(\pi )\], so that \[f(x)\] is continuous at \[x=\pi \], is

    A) \[-\frac{1}{2}\]

    B) \[\frac{1}{2}\]

    C) \[-1\]

    D) 1

    Correct Answer: C

    Solution :

    [c] \[f(x)=\frac{2{{\cos }^{2}}\frac{x}{2}-2\sin \frac{x}{2}\cos \frac{x}{2}}{2{{\cos }^{2}}\frac{x}{2}+2\sin \frac{x}{2}\cos \frac{x}{2}}=\frac{\cos \frac{x}{2}-\sin \frac{x}{2}}{\cos \frac{x}{2}+\sin \frac{x}{2}}\] \[=\tan \left( \frac{x}{4}-\frac{x}{2} \right)\] at \[x=\pi .f(\pi )=-tan\frac{\pi }{4}=-1\].


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