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Q.
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
Relations and Functions - Part 2
Solution:
$f(x)=\frac{x-\frac{1}{x}}{x^2+\frac{1}{x^2}-1}=\frac{x-\frac{1}{x}}{\left(x-\frac{1}{x}\right)^2+1}$(divide by $x ^2$ )
Put $\left(x-\frac{1}{x}\right)=t ; \quad t \neq 0$
$f(t)=\frac{t}{t^2+1}=\frac{1}{t+\frac{1}{t}}$
So, $\frac{-1}{2} \leq f ( t ) \leq \frac{1}{2}$
$\therefore$ Range is $\left[\frac{-1}{2}, \frac{1}{2}\right] $
$\Rightarrow$ Number of integers is one