Question

If $x\in R,$ then the range of $f\left(x\right)=\sin \sqrt{2x^2+4x+3}$ is

• A. $[0,1]$
• B. $[-1,1]$
• C. $[1,\infty )$
• D. $\{-1,1\}$

Answer 1

The correct option is B $[-1,1]$

Given : $f\left(x\right)=\sin \sqrt{2x^2+4x+3}$
For domain,
$2x^2+4x+3\ge 0$
$\Rightarrow 2(x^2+2x+\frac{3}{2})\ge 0$
$\Rightarrow 2\left(\right(x+1{)}^2+\frac{1}{2})\ge 0$ which is true for all real $x.$
$\therefore$ Domain of $f$ is $R$

Let $y=2x^2+4x+3$
$\Rightarrow y\in [1,\infty )$
$\Rightarrow \sqrt{y}\in [1,\infty )$
We know, $\sin g\left(x\right)$ is periodic with period $2\pi$
$\Rightarrow -1\le \sin x\le 1$ for $x\in [a,a+k\pi ],a\in R$ and $k\ge 2$
$\therefore$ Range of $f\left(x\right)$ is $[-1,1]$