Question

The complete set of values of $a$ for which the inequality $(a-1)x^2-(a+1)x+(a-1)\ge 0$ is true for all $x\ge 2$ is

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

Answer 1

The correct option is D $[\frac{7}{3},\infty )$

Since, the given inequality is true for all $x\ge 2$, hence required conditions are $f\left(2\right)\ge 0$ and $a-1\ge 0$

$f\left(2\right)\ge 0\Rightarrow (a-1)4-(a+1)2+(a-1)\ge 0$
$\Rightarrow 3a-7\ge 0\Rightarrow a\in [\frac{7}{3},\infty )...\left(1\right)$

$a-1\ge 0\Rightarrow a\ge 1...\left(2\right)$

From $\left(1\right)$ and $\left(2\right)$
$a\in [\frac{7}{3},\infty )$

Alternate Solution:
$(a-1)x^2-(a+1)x+(a-1)\ge 0$
$\Rightarrow a\ge \frac{x^2+x+1}{x^2-x+1}$
$\Rightarrow a\ge 1+\frac{2x}{x^2-x+1}$
$\Rightarrow a\ge 1+\frac{2}{x+\frac{1}{x}-1}$
$\because x\ge 2$
$\Rightarrow x+\frac{1}{x}\ge \frac{5}{2}$
$\Rightarrow x+\frac{1}{x}-1\ge \frac{3}{2}$
$\therefore a\ge 1+\frac{2}{\frac{3}{2}}$
$\Rightarrow a\in [\frac{7}{3},\infty )$