JEE Main & Advanced JEE Main Online Paper (Held On 08 April 2018)

  • question_answer
    Let \[S=\{t\in \mathbf{R}:f(x)=|x-\pi |\cdot ({{e}^{|x|}}-1)\sin |x|\] is not differentiable at\[\text{t }\!\!\}\!\!\text{ }\]. Then the set S is equal to:                         [JEE Main Online 08-04-2018]

    A)  \[\text{ }\!\!\{\!\!\text{ }\pi \text{ }\!\!\}\!\!\text{ }\]                 

    B)  \[\text{ }\!\!\{\!\!\text{ 0, }\pi \text{ }\!\!\}\!\!\text{ }\]

    C)  \[\phi \](an empty set)

    D)  \[\{0\}\]

    Correct Answer: C

    Solution :

      \[f(x)=|x-\pi |({{e}^{|x|}}-1)\sin |x|\] \[f(0)=0\] \[f(0)=\underset{h\to 0}{\mathop{\lim }}\,\frac{f(h)-f(0)}{h}=\underset{h\to 0}{\mathop{\lim }}\,\frac{(\pi -h)({{e}^{h}}-1)(\sinh )}{h}=0\]\[\Rightarrow \] Differentiable at \[x=0\] \[f(\pi )=0\] \[f(\pi )=\underset{h\to 0}{\mathop{\lim }}\,\frac{f(\pi +h)-f(0)}{h}=\underset{h\to 0}{\mathop{\lim }}\,\frac{h({{e}^{\pi +h}}-1)sin(\pi +h)}{h}=0\] \[\Rightarrow \text{differentiable at x=}\pi \] \[\Rightarrow S\text{ is an empty set}\text{.}\]


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