Question

The absolute minimum & maximum values of $f\left(x\right)=\frac{x^2-3x+4}{x^2+3x+4}$ are respectively

• A. $\frac{1}{7}$ and $7$
• B. $5$ & $7$
• C. $\frac{1}{7}$ & $1$
• D. None of these

Answer 1

The correct option is A $\frac{1}{7}$ and $7$

Let $y=\frac{x^2-3x+4}{x^2+3x+4}$
then $x^2(y-1)+3x(y+1)+4(y-1)=0$
this equation gives the values of $x$ for given values of $y$ but $y$ is the value when $x$ is real. So roots of this equation are real
$\Rightarrow$ $D\ge 0$ $\Rightarrow$ $9{(y+1)}^2-16{(y-1)}^2\ge 0$
$\Rightarrow$ $7y^2-50y+7\le 0$ $\Rightarrow$ $(7y-1)(y-7)\le 0$
$\Rightarrow$ $\frac{1}{7}\le y\le 7$
$\Rightarrow$ absolute minimum value is $\frac{1}{7}$ & absolute maximum is $7$
Ans: A