Question

The range of $f\left(x\right)=\frac{x^4}{1+x^8}$ is

• A. $[0,\infty )$
• B. $[0,\frac{1}{2}]$
• C. $[0,1]$
• D. $(-\infty ,\infty )$

Answer 1

The correct option is B $[0,\frac{1}{2}]$

$\begin{array}{c}Wehave\\ f\left(x\right)=\frac{x^4}{1+x^8}\\ y=\frac{t}{1+t^2}t\ge 0\\ y=\frac{1}{t+\frac{1}{t}}\{\begin{array}{c}fort=0\\ y=0\end{array}\\ y\le \frac{1}{2}\\ y\in \left[0,\frac{1}{2}\right]\\ Hence,theoptionBisthecorrectanswer.\end{array}$