Question

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

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

Answer 1

Answer: (C)

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