Question

Let $f:R\to R$ be defined as $f\left(x\right)=\frac{x^2-x+4}{x^2+x+4}$, then range of the function $f\left(x\right)$ is

• A. $[\frac{3}{5},\frac{5}{3}]$
• B. $(\frac{3}{5},\frac{5}{3})$
• C. $(-\infty ,\frac{3}{5})\cup (\frac{5}{3},\infty )$
• D. $[-\frac{5}{3},-\frac{3}{5}]$

Answer 1

The correct option is A $[\frac{3}{5},\frac{5}{3}]$

Let $y=\frac{x^2-x+4}{x^2+x+4}$
$\Rightarrow x^{2}y+xy+4y=x^2-x+4$
$\Rightarrow (y-1)x^2+(y+1)x+4y-4=0$

For $x$ to be real, discriminant of the above quadratic equation should be greater than or equal to $0$.

$\Rightarrow (y+1{)}^2-4(y-1\left)\right(4y-4)\ge 0$
$\Rightarrow (y+1{)}^2-16(y-1{)}^2\ge 0$

$\Rightarrow (y+1+4y-4)(y+1-4y+4)\ge 0$
$\Rightarrow (5y-3)(5-3y)\ge 0$

$\therefore y\in [\frac{3}{5},\frac{5}{3}]$