Question

Let $f$ be a composite function of $x$ defined by
$f \left(u\right)=\frac{1}{u^{2}+u-2}, u \left(x\right)=\frac{1}{x-1}$.
Then the number of points $x$ where $f$ is discontinuous is :

• A. $4$
• B. $3$
• C. $2$
• D. $1$

Answer 1

Answer: (B)

$\mu\left(x\right)=\frac{1}{x-1},$ which is discontinous at $x=1$
$f \left(u\right)=\frac{1}{u^{2}+u-2}=\frac{1}{\left(u+2\right)\left(u-1\right)},$
which is discontinous at $u = - 2, 1$
when 0$u=-2,$ then $\frac{1}{x-1}=-2 \Rightarrow x=\frac{1}{2}$
when $u=-2,$ then $\frac{1}{x-1}=1 \Rightarrow x=2$
Hence given composite function is discontinous at three points, $x=1, \frac{1}{2}$ and $2.$