Question

The set of values of $a$ for which
$(\sqrt{a}-1)x^2-(\sqrt{a}+1)x+\sqrt{a}-1\ge 0$
is true for all $x\ge 2$

• A. $[\frac{49}{9},\infty )$
• B. $(1,\frac{7}{3})$
• C. $(1,\frac{49}{9})$
• D. $(1,\infty )$

Answer 1

The correct option is A $[\frac{49}{9},\infty )$

Given,
$(\sqrt{a}-1)x^2-(\sqrt{a}+1)x+\sqrt{a}-1\ge 0$
$\Rightarrow \sqrt{a}(x^2-x+1)-(x^2+x+1)\ge 0$
$[\because (x^2-x+1)>0,(x^2+x+1)>0,\forall x\in R]$
$\Rightarrow \sqrt{a}\ge \frac{x^2+x+1}{x^2-x+1}$
$\Rightarrow \sqrt{a}\ge 1+\frac{2x}{x^2-x+1}\Rightarrow \sqrt{a}\ge 1+\frac{2}{x+1/x-1}$
Now
$\frac{3}{2}\le x+\frac{1}{x}-1<\infty ,\forall x\ge 2\Rightarrow 0<\frac{1}{x+\frac{1}{x}-1}\le \frac{2}{3},\forall x\ge 2\Rightarrow 1+0<1+\frac{2}{x+\frac{1}{x}-1}\le 1+\frac{4}{3},\forall x\ge 2$
$\therefore 1+\frac{2}{x+\frac{1}{x}-1}\in (1,\frac{7}{3}]\Rightarrow \sqrt{a}\ge \frac{7}{3}\Rightarrow a\in [\frac{49}{9},\infty )$