Question

The range of $\frac{3x^2+9x+17}{3x^2+9x+7}$ is

• A. $[1,41]$
• B. $(0,10]$
• C. $[11,\infty )$
• D. $(1,41]$

Answer 1

The correct option is D $(1,41]$

$3x^2+9x+7=3{(x+\frac{3}{2})}^2+\frac{1}{4}\ne 0\forall x\in R$
$\therefore$ Domain is $R$.

Let $y=\frac{3x^2+9x+17}{3x^2+9x+7}$
$\Rightarrow y=1+\frac{10}{3x^2+9x+7}$

Now, $\frac{1}{4}\le 3x^2+9x+7<\infty$
$\Rightarrow 0<\frac{10}{3x^2+9x+7}\le 40$
$\Rightarrow 1<1+\frac{10}{3x^2+9x+7}\le 41$
$\Rightarrow 1<y\le 41$

Hence, $R\left(f\right)=(1,41]$