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 $(\frac{7}{3},\infty )$

Given,
$(a-1)x^2-(a+1)x+a-1\ge 0$
or $a(x^2-x+1)-(x^2+x+1)\ge 0$
$\Rightarrow a\ge \frac{x^2+x+1}{x^2-x+1}$
$=1+\frac{2x}{x^2-x+1}$
$=1+\frac{2}{x+\frac{1}{x}-1}$ --(1)
Let $y=x+\frac{1}{x}$. Now, $y$ is increasing in $[2,\infty$).
Hence, $1+\frac{2}{x+\frac{1}{x}-1}\in (1,\frac{7}{3})$
For all $x\ge 2$, equation (1) should be true.
Hence, $a>\frac{7}{3}$.