Question

The range of the function $f\left(x\right)={\log }_2(3-2x-x^2)$ is

• A. $(-3,-1)$
• B. $[0,2]$
• C. $(-\infty ,2]$
• D. $R$

Answer 1

The correct option is C $(-\infty ,2]$

$f\left(x\right)={\log }_2(3-2x-x^2)$
For the function to be defined,
$3-2x-x^2>0\Rightarrow x^2+2x-3<0\Rightarrow (x+3)(x-1)<0\Rightarrow -3<x<1$
Hence, domain of $f$ is $(-3,1)$

Now, let $y={\log }_2(3-2x-x^2)$
$\Rightarrow 2^y=3-2x-x^2\Rightarrow x^2+2x+(2^y-3)=0$
Since, $x$ is real,
$\Delta \ge 0\Rightarrow 4-4(2^y-3)\ge 0\Rightarrow 1-2^y+3\ge 0\Rightarrow 2^y\le 4\Rightarrow y\le 2$
Therefore, the range of $f$ is $(-\infty ,2]$