2. Questions Bank
  3. JEE Main & Advanced
  4. Mathematics
  5. Limits, Continuity and Differentiability

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

  • question_answer
    If the function\[f(x)=\frac{x(x-2)}{{{x}^{2}}-4},x\ne \pm 2\]is continuous at\[x=2\], then what is \[f(2)\] equal to?

    A) 0

    B) \[\frac{1}{2}\]

    C) 1

    D) 2

    Correct Answer: B

    Solution :

    [b] Let\[f(x)=\frac{x(x-2)}{{{x}^{2}}-4}=\frac{x(x-2)}{(x-2)(x+2)}=\frac{x}{x+2}\] Since f(x) is continuous at x = 2 \[\therefore \underset{x\to 2}{\mathop{\lim }}\,f(x)=f(2)\] \[\Rightarrow \underset{x\to 2}{\mathop{\lim }}\,\frac{x}{x+2}=f(2)\Rightarrow f(2)=\frac{2}{4}=\frac{1}{2}\]


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