Question

If $\frac{x^{2}+2x+7}{2x+3} < 6, x\,\in\,R,$ then

• A. $x >11$ or $x < -\frac{3}{2}$
• B. $x > 11$ or $x < -1$
• C. $-\frac{3}{2} < x < -1$
• D. $-1 < x < 11$ or $x < -\frac{3}{2}$

Answer 1

Answer: (D)

$-1 < x < 11$ or $x < -\frac{3}{2}$

Answer 2

Given, $\frac{x^{2}+2 x+7}{2 x+3}<6$
$\Rightarrow \frac{x^{2}+2 x+7}{2 x+3}-6<0$
$\Rightarrow \frac{x^{2}-10 x-11}{2 x+3}<0$
$\Rightarrow \frac{(x-11)(x+1)}{2 x+3}<0$
$\Rightarrow \frac{(x-11)(x+1)(2 x+3)}{(2 x+3)^{2}}<0$
$\Rightarrow (x-11)(x+1)(2 x+3) < 0$
$\Rightarrow x \in\left(-\infty,-\frac{3}{2}\right) \cup(-1,11)$