Question

If $f\left(x\right)=\frac{1}{(x-1)(x-2)}$ and $g\left(x\right)=\frac{1}{x^2}$,then set of points in domain of $fog\left(x\right)$ at which $fog\left(x\right)$ discontinuous

• A. $\{-1,0,1,\frac{1}{\sqrt{2}}\}$
• B. $\varnothing$
• C. $\{0,1\}$
• D. $\{0,1,\frac{1}{\sqrt{2}}\}$

Answer 1

The correct option is A $\{-1,0,1,\frac{1}{\sqrt{2}}\}$

$fog\left(x\right)=\frac{1}{(g\left(x\right)-1)(g\left(x\right)-2)}$

$=\frac{1}{(\frac{1}{x^2}-1)(\frac{1}{x^2}-2)}$

$=\frac{x^4}{(1-x^2)(1-2x^2)}$

discontinuous at $1-x^2=0$ and $1-2x^2=0$

$x=\pm 1$ and $x=\sqrt{\frac{1}{2}}$

$x\in [-1,0,1,\frac{1}{\sqrt{2}}]$

Hence, the answer is $\{-1,0,1,\frac{1}{\sqrt{2}}\}.$