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

Answer 1

Answer: (C)

Given,
$(a-1)x^2-(a+1)x+a-1\ge 0$
or $a\left(x^2-x+1\right)-\left(x^2+x+1\right)\ge 0$
$\Rightarrow a\ge \frac{x^2+x+1}{x^2-x+1}=1+\frac{2x}{x^2-x+1}$
$=1+\frac{2}{x+\frac{1}{x}-x}...$ (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 $x\ge 2$, Eq. (1) should be true. Hence, $a>7/3$.