Q48 / 160 · WBJEE 2015 · Mathematics

Let $f: R \rightarrow R$ be defined as $f(x)=\frac{x^{2}-x+4}{x^{2}+x+4}$. Then, range of the function $f(x)$ is

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

Solution

Let $y=\frac{x^{2}-x+4}{x^{2}+x+4}$ $\Rightarrow \quad x^{2} y+x y+4 y=x^{2}-x+4$ $\Rightarrow(y-1) x^{2}+(y+1) x+4 y-4=0$ For $x$ to be real, discriminant of the above quadratic equation should be greater than equal to 0 $\\ \Rightarrow \quad(y+1)^{2}-4(y-1)(4 y-4) \geq 0$ $\Rightarrow \quad(y+1)^{2}-16(y-1)^{2} \geq$ $\Rightarrow(y+1+4 y-4)(y+1-4 y+4) \geq$ $(5 y-3)(5-3 y) \geq 0$ $\therefore \quad y \in\left[\frac{3}{5}, \frac{5}{3}\right]$

Hence, the correct answer is option: 'A'.

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