Question

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

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is A $1$

$\begin{array}{c}f\left(x\right)=\frac{x\left(x^2-1\right)}{x^4-x^2+1}\\ =\frac{x\left(x^2-1\right)}{x^2\left[{\left(x-\frac{1}{x}\right)}^2+1\right]}\\ =\frac{\left[x-\frac{1}{x}\right]}{{\left(x-\frac{1}{x}\right)}^2+1}\\ d\left(t\right)=\frac{t}{t^2+1}\\ differentiatef\left(t\right)tofindmaximumandminimum\\ f^{\prime }\left(t\right)=\frac{1-t^2}{{\left(t^2+1\right)}^2}=0\\ 1-t^2=0\Rightarrow t=\pm 1\\ f\left(1\right)=\frac{-1}{{\left(-1\right)}^2+1}=0.5\\ f\left(-1\right)=\frac{-1}{{\left(-1\right)}^2+1}=-0.5\\ Rangeoff\left(x\right)is\left[-0.5,0.5\right]\\ thereisonlyoneintegerb/w-0.5and0.5thatiszero.\end{array}$