Question

Let $f:R\to R$ be defined as $f\left(x\right)=\frac{x^2-x+4}{x^2+x+4}$. Then the 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 $y=\frac{x^2-x+4}{x^2+x+4}$

$\Rightarrow (y-1)x^2+(y+1)x+4y-4=0$

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

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

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

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

$\Rightarrow (y+1-4y+4)(y+1+4y-4)\ge 0$ $[\because a^2-b^2=(a+b\left)\right(a-b\left)\right]$

$\Rightarrow (3y-5)(5y-3)\le 0$

$\Rightarrow \frac{3}{5}\le y\le \frac{5}{3}$

Hence, option A.