Question

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

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

Answer 1

The correct option is B $2+x^2$

We have $g\left(x\right)=1+\sqrt{x}$ and $f\left(g\right(x\left)\right)=3+2\sqrt{x}+x$ ...(1)

Also, $f\left(g\left(x\right)\right)=f(1+\sqrt{x})$ ...(2)

From (1) and (2), we get

$f(1+\sqrt{x})=3+2\sqrt{x}+x$

Let $1+\sqrt{x}=y\Rightarrow x={(y-1)}^2$

$\therefore f\left(y\right)=3+2(y-1)+{(y-1)}^2=3+3y-2+y^2-2y+1=2+y^2\therefore f\left(x\right)=2+x^2$