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JEE Main & Advanced JEE Main Paper (Held On 11 April 2015)

  • question_answer
    Let k be a non-zero real number. If\[\begin{align}   & f(x)=\left\{ \frac{{{({{e}^{x}}-1)}^{2}}}{\sin \left( \frac{x}{k} \right)\log \left( 1+\frac{x}{4} \right)},x\ne 0 \right. \\  & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,12\,\,\,\,\,\,\,\,\,\,\,\,\,,\,\,\,\,\,\,\,\,x=0 \\ \end{align}\] is a continuous function, then the value of k is : [JEE Main Online Paper (Held On 11 April 2015)]  

    A)  1

    B)  2

    C)  3

    D)  4

    Correct Answer: C

    Solution :

      \[\underset{x\to 0}{\mathop{\lim }}\,f\left( x \right)=f(0)\] \[\Rightarrow \underset{x\to 0}{\mathop{\lim }}\,\frac{\frac{{{\left( {{e}^{x}}-1 \right)}^{2}}}{{{x}^{2}}}}{\frac{\sin \left( {}^{x}/{}_{k} \right)}{k\,\,\,\,\,\,{}^{x}/{}_{k}}\frac{\log \left( 1+{}^{x}/{}_{4} \right)}{4\,\,\,\,\,\,\,\,{}^{x}/{}_{4}}}=12\] \[\Rightarrow 4k=12\] \[k=3\]


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