Mathematics

Mathematics

Problem

Which of the following is true about f(x)={(x−2)∣x−2∣(x2−1x2+1),x≠235,x=2f(x) = \begin{cases} \frac{(x-2)}{|x-2|}\left(\frac{x^2-1}{x^2+1}\right), & x \neq 2 \\ \frac{3}{5}, & x = 2 \end{cases} (a) f(x)f(x) is continuous at x=2x = 2. (b) f(x)f(x) has removable discontinuity at x=2x = 2. (c) f(x)f(x) has non-removable discontinuity at x=2x = 2. (d) Discontinuity at x=2x = 2 can be removed by redefining the function at x=2x = 2.

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