Question

* If $f\left(x\right)=1+x^2$ and $f\left[g\left(x\right)\right]=1+x^2-2x^3+x^4$, then determine the function g(x) along with range.

• A. $(-\infty ,\frac{1}{2}).$
• B. $(-\infty ,\frac{1}{4}).$
• C. $(-\infty ,\frac{1}{8}).$
• D. $(-\infty ,\frac{-1}{4}).$

Answer 1

The correct option is B $(-\infty ,\frac{1}{4}).$

$f\left[g\left(x\right)\right]=1+{\left[g\left(x\right)\right]}^2=1+x^2-2x^3+x^4$
$\therefore {\left[g\left(x\right)\right]}^2=x^2(x^2-2x+1)=x^2{(x-1)}^2$
$\therefore g\left(x\right)=\pm \left\{x(x-1)\right\}>\pm \{x^2-x\}$
$=\pm \{(x^2-x+\frac{1}{4})-\frac{1}{4}\}$
$=\pm \{{(x-\frac{1}{2})}^2-\frac{1}{4}\}$
$\therefore g\left(x\right)={(x-\frac{1}{2})}^2-\frac{1}{4}$ ...(1)or $g\left(x\right)=\frac{1}{4}-{(x-\frac{1}{2})}^2$ ...(2)In either case domain of $g\left(x\right)$ is R.In form (1) of $g\left(x\right)$ it is always $\ge -\frac{1}{4}$
$\therefore$ Range is $[-\frac{1}{4},\infty ].$ In the form (2) of $g\left(x\right)$ it is always $\le \frac{1}{4}$
$\therefore$ Range is $(-\infty ,\frac{1}{4}).$