Question

The maximum possible value of $f\left(x\right)=\frac{1-x^2}{x^2+4}$ is

• A. $\frac{1}{4}$
• B. $\frac{1}{3}$
• C. $\frac{1}{2}$
• D. 0.0

Answer 1

The correct option is A $\frac{1}{4}$

$\mathrm{Let}y=f\left(x\right)=\frac{1-x^2}{x^2+4}\Rightarrow y\left(x^2+4\right)=1-x^2\Rightarrow {\mathrm{yx}}^2+4y=1-x^2\Rightarrow {\mathrm{yx}}^2+x^2=1-4y\Rightarrow x^2\left(y+1\right)=1-4y\Rightarrow x^2=\frac{1-4y}{y+1}\mathrm{We}\mathrm{know}\mathrm{that}{x}^2\ge 0.\Rightarrow \frac{1-4y}{y+1}\ge 0\Rightarrow \frac{4y-1}{y+1}\le 0$
$\therefore -1<x\le \frac{1}{4}asx\ne -1$
The maximum value of f(x) is $\frac{1}{4}$