Question

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

• A. $(-\infty ,1)$
• B. $(1,\frac{7}{3})$
• C. $[\frac{7}{3},\infty )$
• D. None of these

Answer 1

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

$(a-1)x^2-(a+1)x+a-1\ge 0$
$\Rightarrow a(x^2-x+1)-(x^2+x+1)\ge 0$
$\Rightarrow a\ge \frac{x^2+x+1}{x^2-x+1}$
$=1+\frac{2x}{x^2-x+1}$
$=1+\frac{2}{x+\frac{1}{x}-1}$

Let $y=x+\frac{1}{x}$
$y$ is increasing in $[2,\infty )$
$\therefore 1+\frac{2}{x+\frac{1}{x}-1}\in (1,\frac{7}{3}]$
$\therefore a\ge \frac{7}{3}$