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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] \[\underset{x\to \pi }{\mathop{\lim }}\,\frac{1-\sin x+\cos x}{1+\sin x+\cos x}\] Using L?hospital?s rule \[\Rightarrow \underset{x\to \pi }{\mathop{\lim }}\,\frac{-\cos x-sinx}{\cos x-\sin x}=\frac{-\cos \pi -\sin \pi }{\cos \pi -\sin \pi }\] \[=\frac{-(-1)-0}{-1-0}=-1\]


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