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

  • question_answer
    The value of f(0) so that the function \[f(x)=\frac{2x-{{\sin }^{-1}}x}{2x+{{\tan }^{-1}}x}\]is continuous at each point in its domain, is equal to

    A) 2                     

    B) 1/3

    C) 2/3                   

    D) -1/3

    Correct Answer: B

    Solution :

    [b] The function f is clearly continuous at each point in its domain except possibly at x=0. Given that f(x) is continuous at x=0. \[\therefore f(0)=\underset{x\to 0}{\mathop{\lim }}\,f(x)\] \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{2x-{{\sin }^{-1}}x}{2x+{{\tan }^{-1}}x}\] \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{2-(si{{n}^{-1}}x)/x}{2+({{\tan }^{-1}}x)x}=\frac{1}{3}\]


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