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

  • question_answer
    Let \[f(x)=\frac{{{({{e}^{x}}-1)}^{2}}}{\sin \left( \frac{x}{a} \right)\log \left( 1+\frac{x}{4} \right)}\]  for \[x\ne 0,\] and \[f(0)=12\]. If f is continuous at \[x=0\], then the value of a is equal to

    A) 1

    B) -1

    C) 2

    D) 3

    Correct Answer: D

    Solution :

    [d] \[\underset{x\to 0}{\mathop{Lt}}\,\frac{{{({{e}^{x}}-1)}^{2}}}{\sin \left( \frac{x}{a} \right)\log \left( 1+\frac{x}{4} \right)}\] \[=\underset{x\to 0}{\mathop{Lt}}\,\frac{\frac{{{({{e}^{x}}-1)}^{2}}}{x}.{{x}^{2}}}{\frac{x}{a}.\frac{\sin \left( \frac{x}{a} \right)}{\left( \frac{x}{a} \right)}.\frac{\log \left( 1+\frac{x}{4} \right)}{\frac{x}{4}}.\frac{x}{4}}\Rightarrow 4a=12\] \[\Rightarrow a=3\]


You need to login to perform this action.
You will be redirected in 3 sec spinner