Question

Let $f\left(x\right)=\sqrt{x^2-3x+2}$ and $g\left(x\right)=\sqrt{x}$ be two given functions . If $S$ be the domain of $f\circ g$ and $T$ be the domain of $g\circ f$, then

• A. $S=T$
• B. $S\cap T=\phi$
• C. $S\cap T$ is a singleton
• D. $S\cap T$ is an interval

Answer 1

The correct option is D $S\cap T$ is an interval

$f\left(x\right)=\sqrt{x^2-3x+2},g\left(x\right)=\sqrt{x}f\circ g\left(x\right)=f\left(\sqrt{x}\right)=\sqrt{x-3\sqrt{x}+2},x-3\sqrt{x}+2\ge 0,x>0\Rightarrow x+2\ge 3\sqrt{x}\Rightarrow x^2+4+4x-9x\ge 0\Rightarrow x^2-5x+4\ge 0(x-1)(x-4)\ge 0S=x\in (0,1]\cup [4,\infty )g\circ f\left(x\right)=g\left(\sqrt{x^2-3x+2}\right)=\sqrt{\sqrt{x^2-3x+2}}\Rightarrow x^2-3x+2\ge 0\Rightarrow (x-1)(x-2)\ge 0T\Rightarrow x\in (-\infty ,1]\cup [2,\infty )\therefore S\cap T=(0,1]\cup [4,\infty )$