Question

Number of integers in the range of the function $f(x)=\frac{x\left(x^2-1\right)}{x^4-x^2+1}$ is equal to

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

Answer: (A)

$f(x)=\frac{x-\frac{1}{x}}{x^2+\frac{1}{x^2}-1}=\frac{x-\frac{1}{x}}{{\left(x-\frac{1}{x}\right)}^2+1}$(divide by $x^2$ )
Put $\left(x-\frac{1}{x}\right)=t;t\ne 0$
$f(t)=\frac{t}{t^2+1}=\frac{1}{t+\frac{1}{t}}$
So, $\frac{-1}{2}\le f(t)\le \frac{1}{2}$
$\therefore$ Range is $\left[\frac{-1}{2},\frac{1}{2}\right]$
$\Rightarrow$ Number of integers is one