Question

The smallest value of $x^2-3x+3$ in the interval $\left(-3,\frac{3}{2}\right)$ is

• A. $-20$
• B. $-15$
• C. $5$
• D. $\frac{3}{4}$

Answer 1

Answer: (D)

We have, $x^2-3x+3$
$={\left(x-\frac{3}{2}\right)}^2+3-\frac{9}{4}$
$={\left(x-\frac{3}{2}\right)}^2+\frac{3}{4}$
$\therefore$ smallest value $=\frac{3}{4}$,
which lies in the interval $\left(-3,\frac{3}{2}\right)$