Question

If $\frac{9x^2-4}{\sqrt{5x^2-1}}\le 3x+2,$ then $x\in$

• A. $[-\frac{2}{3},-\frac{1}{\sqrt{5}})\cup (\frac{1}{\sqrt{5}},\frac{5}{2}]$
• B. $[-\frac{2}{3},-\frac{1}{\sqrt{5}}]\cup [\frac{2}{3},\frac{5}{2}]$
• C. $[\frac{2}{3},\frac{5}{2}]$
• D. $[-\frac{2}{3},-\frac{1}{\sqrt{5}}]$

Answer 1

The correct option is A $[-\frac{2}{3},-\frac{1}{\sqrt{5}})\cup (\frac{1}{\sqrt{5}},\frac{5}{2}]$

Given $\frac{9x^2-4}{\sqrt{5x^2-1}}\le 3x+2$ to be valid
$5x^2-1>0$
$\Rightarrow x^2>\frac{1}{5}\Rightarrow (x-1/\sqrt{5})(x+1/\sqrt{5})>0$
$\therefore x\in (-\infty ,\frac{-1}{\sqrt{5}})\cup (\frac{1}{\sqrt{5}},\infty )\cdots \left(1\right)$

Now,
$\frac{9x^2-4}{\sqrt{5x^2-1}}-(3x+2)\le 0$
$\because \sqrt{5x^2-1}>0$
$\Rightarrow \left[\right(3x{)}^2-2^2]-(3x+2)\sqrt{5x^2-1}\le 0$
$\Rightarrow (3x+2)(3x-2-\sqrt{5x^2-1})\le 0$

Case $1:$ When $x<\frac{2}{3}$
$3x+2\ge 0,3x-2\le \sqrt{5x^2-1}\Rightarrow x\ge -\frac{2}{3},3x-2\le 0[\because \sqrt{5x^2-1}\ge 0]\therefore x\in [-\frac{2}{3},\frac{2}{3})\cdots \left(2\right)$

Case $2:$ When $x\ge \frac{2}{3}$
$3x+2\ge 0,3x-2\le \sqrt{5x^2-1}\Rightarrow x\ge -\frac{2}{3},(3x-2{)}^2\le 5x^2-1[\because \text{both sides are +ive}]\Rightarrow x\ge -\frac{2}{3},4x^2-12x+5\le 0\Rightarrow x\ge -\frac{2}{3},(2x-5\left)\right(2x-1)\le 0\Rightarrow x\ge -\frac{2}{3},x\in [\frac{1}{2},\frac{5}{2}]\therefore x\in [\frac{2}{3},\frac{5}{2}]\cdots (3)$

From $\left(1\right),\left(2\right)$ and $\left(3\right)$
$x\in [-\frac{2}{3},-\frac{1}{\sqrt{5}})\cup (\frac{1}{\sqrt{5}},\frac{5}{2}]$