Question

If $f\left(x\right)=\frac{1}{(x-1)(x-2)}andg\left(x\right)=\frac{1}{x^2}$, then points of discontinuity of f(g(x)) are

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

Answer 1

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

$f\left(g\right(x\left)\right)=\frac{1}{(\frac{1}{x^2}-1)(\frac{1}{x^2}-2)}=\frac{x^4}{(1-x^2)(1-2x^2)}\Rightarrow f\left(g\right(x\left)\right)isdiscontinuousatx=\pm 1,x=\pm \frac{1}{\sqrt{2}}andx=0$
Since, g(x) is discontinuous at x=0