Mathematics - Functions Question with Solution | TestHub

Given function $f(x)=\sqrt{\frac{2x^2-7x+5}{3x^2-5x-2}}$ Here, $\mathrm{f}(\mathrm{x})$ should be greater than or equal to 0 . So, $2x^2-7x+5=0$ $\Rightarrow 2x^2-5x-2x+5=0$ $\Rightarrow (x-1)(2x-5)=0$ $\Rightarrow x=1,\frac{5}{2}$ $3x^2-5x-2=0\Rightarrow 3x^2-6x+x-2=0$ $3x(x-2)+1(x-2)=0\Rightarrow (x-2)(3x+1)=0$ $\Rightarrow x=2,\frac{-1}{3}$When we include $x=\frac{-1}{3},2$ then $f(x)$ would give not define value so, we will exclude these values from the domain. When we take values between $\left(-\frac{1}{3},1\right)$ and So, domain is $\left(-\infty ,\frac{-1}{3}\right)\cup [1,2)\cup \left[\frac{5}{2},\infty \right)$