Question

The value of the function $f\left(x\right)=\frac{x^2-3x+2}{x^2+x-6}$ lies in the interval

• A. $(-\infty ,\infty )/\{\frac{1}{5},1\}$
• B. $(-\infty ,\infty )$
• C. $(-\infty ,\infty )/1$
• D. None of these

Answer 1

Answer: (B)

$(-\infty ,\infty )$

$f\left(x\right)$ is defined if $x^2+x-6\ne 0$

$\Rightarrow (x+3)(x-2)\ne 0\Rightarrow x\ne -3,2$

$\therefore$ Domain of $f\left(x\right)=(-\infty ,\infty )/\{-3,2\}$

Let $y=\frac{x^2-3x+2}{x^2+x-6}\Rightarrow x^{2}y+xy-6y=x^2-3x+2$

$\Rightarrow x^2(y-1)+x(y+3)-(6y+2)=0$

For $x$ to be real, ${(y+3)}^2+4(y-1)(6y+2)\ge 0$

$\Rightarrow 25y^2-10y+1\ge 0\Rightarrow {(5y-1)}^2\ge 0$

which is true for all real $y$.

$\therefore$ range of $f\left(x\right)=(-\infty ,\infty )$