Question

If $g\left(x\right)=1+\sqrt{x}$ and $f\left\{g\right(x\left)\right\}=3+2\sqrt{x}+x$, then $f\left(x\right)$ is equal to

• A. $1+2x^2$
• B. $2+x^2$
• C. $1+x$
• D. $2+x$

Answer 1

Answer: (B)

$2+x^2$

Given, $g\left(x\right)=1+\sqrt{x}$
and $f\left\{g\right(x\left)\right\}=3+2\sqrt{x}+x$ ..... $\left(i\right)$
$\Rightarrow f(1+\sqrt{x})=3+2\sqrt{x}+x$
Put $1+\sqrt{x}=y\Rightarrow x=(y-1{)}^2$
$\therefore f\left(y\right)=3+2(y-1)+(y-1{)}^2=3+2y-2+y^2-2y+1=2+y^2$
$\therefore f\left(x\right)=2+x^2$