JEE Main & Advanced Sample Paper JEE Main - Mock Test - 10

  • question_answer
    The function \[f:R-\{0\}\to R,\]\[f(x)=\frac{1}{x}-\frac{2}{{{e}^{2x}}-1}\]can be made continuous  at \[x=0\] by defining f(0) as

    A) 2                     

    B) -1   

    C) 0         

    D) 1

    Correct Answer: D

    Solution :

    [d]: \[\underset{x\to 0}{\mathop{\lim }} \left( \frac{1}{x}-\frac{2}{{{e}^{2x}}-1} \right)=\underset{x\to 0}{\mathop{\lim }}\,\frac{{{e}^{2x}}-1-2x}{x({{e}^{2x}}-1)}\] \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{\left\{ \left( 1+2x+\frac{4{{x}^{2}}}{2!}+\frac{8{{x}^{2}}}{3!}+... \right)-1-2x \right\}}{x\left\{ \left( 1+2x+\frac{4{{x}^{2}}}{2!}+\frac{8{{x}^{2}}}{3!}+... \right)-1 \right\}}=1=f(0)\]


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