Question

A function f(x) is defined by,
$f\left(x\right)=\{\begin{array}{c}\frac{\left[x^2\right]-1}{x^2-1},for{x}^2\ne 1\\ 0,for{x}^2=1\end{array}Where[.]denotesGIF$

• A. Continuous at X= -1
• B. Discontinuous at X = 1
• C. Differentiable at X = 1
• D. None of these

Answer 1

The correct option is B Discontinuous at X = 1

$f\left(x\right)=\{\begin{array}{c}\frac{\left[x^2\right]-1}{x^2-1},for{x}^2\ne 1\\ 0,for{x}^2=1\end{array}=\{\begin{array}{ccc}\frac{-1}{x^2-1},& for& 0<x^2<1\\ 0,& for& x^2=1\\ 0,& for& 1<x^2<2\end{array}$
$\therefore$ RHL at x = 1 is 0
Also LHL at x = 1 is $\infty$