Question

If $g\left(x\right)=2x+1$ and $f\circ g\left(x\right)=4x^2+4x+2$, then range of $f\left(x\right)$ is

• A. $R$
• B. $(6,\infty )$
• C. $[1,\infty )$
• D. $[6,\infty )$

Answer 1

Answer: (C)

$[1,\infty )$

$fog\left(x\right)=4x^2+4x+2$

$⟹f\left(g\right(x\left)\right)=(4x^2+4x+1)+1$

$⟹f\left(g\right(x\left)\right)=(2x+1{)}^2+1$

$⟹f\left(g\right(x\left)\right)=\left(g\right(x){)}^2+1$

Replacing $g\left(x\right)$ by $x$, we get-

$f\left(x\right)=x^2+1$

Since $\forall x\in R$

we have $x^2\ge 0$

$⟹x^2+1\ge 1$

$⟹f\left(x\right)\ge 1$

Hence range of $f\left(x\right)$ is $[1,\infty )$