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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 33

  • question_answer
    In the function \[f(x)=\frac{2x-{{\sin }^{-1}}x}{2x+{{\tan }^{-1}}x},\,\,(x\ne 0)\], is continuous at each point of its domain, then the value of f(0) is

    A) 2                     

    B)        1/3

    C) 2/3                   

    D)        -1/3

    Correct Answer: B

    Solution :

    \[\underset{x\to 0}{\mathop{lim}}\,\,\,f(x)=\underset{x\to 0}{\mathop{lim}}\,\left( \frac{2x-{{\sin }^{-1}}x}{2x+{{\tan }^{-1}}x} \right)=f(0),\,\,\left( \frac{0}{0}\,\,form \right)\] Applying L-Hospital?s rule \[f(0)=\underset{x\to 0}{\mathop{\lim }}\,\frac{\left( 2-\frac{1}{\sqrt{1-{{x}^{2}}}} \right)}{\left( 2+\frac{1}{1+{{x}^{2}}} \right)}\,\,=\frac{2-1}{2+1}=\frac{1}{3}\]


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