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JEE Main & Advanced Sample Paper JEE Main Sample Paper-35

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

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

    B)  \[\frac{1}{2}\]

    C)  1                                            

    D)  \[-1\]

    Correct Answer: D

    Solution :

    \[\underset{x\to \pi }{\mathop{Lim}}\,\,\,\frac{(1+\cos x)-\sin x}{(1+\cos x)+\sin x}\,=\underset{x\to \pi }{\mathop{Lim}}\,\,\frac{2{{\cos }^{2}}\,\frac{x}{2}-2\sin \,\frac{x}{2}.\cos \frac{x}{2}}{2{{\cos }^{2}}\,\frac{x}{2}\,+2\sin \,\frac{x}{2}\,.\cos \,\frac{x}{2}}\]\[=\underset{x\to \pi }{\mathop{Lim}}\,\tan \left( \frac{\pi }{4}-\frac{\pi }{2} \right)\,\,=-1\] Since, \[f(x)\] is continuous at \[x=\pi ,\] So, \[f(\pi )\,=\underset{x\to \pi }{\mathop{Lim}}\,\,f(x)\,=-1\]


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