Question

The smallest value of M such that $|x^2-3x+2|\le$ M for all $x\in [1,\frac{5}{2}]$ is

• A. $1$
• B. $\frac{1}{2}$
• C. $\frac{1}{4}$
• D. $\frac{3}{4}$

Answer 1

The correct option is C $\frac{1}{4}$

Let $y=x^2-3x+2$ $xϵ[1,\frac{5}{2}]$

$y^{\prime }=2x-3$

So y will decrease in $(-\infty ,\frac{3}{2})$ and increase in $(\frac{3}{2},\infty )$

Thus minimum value of $y$ in the given interval is at $x=\frac{3}{2}$

$y_{min}=\frac{9}{4}-3\times \frac{3}{2}+2=\frac{-1}{4}$

Thus $M=\frac{1}{4}$