Question

Consider the equation of curve $y=x^2-3x+3$ and $x\ne 3.$
Then which of the following is CORRECT?

[2 marks]

• A. Minimum exists at $x=\frac{3}{2}$
• B. Maximum exists at $x=\frac{3}{2}$
• C. Neither maximum nor minimum exists at $x=\frac{3}{2}$
• D. Maximum exists at $x=\frac{1}{2}$

Answer 1

The correct option is A Minimum exists at $x=\frac{3}{2}$

$y=x^2-3x+3$
$\Rightarrow \frac{dy}{dx}=2x-3$
For critical points : $\frac{dy}{dx}=0$
$\Rightarrow 2x-3=0$
$\Rightarrow x=\frac{3}{2}$
Sign scheme for $\frac{dy}{dx}:$
By first derivative test, we conclude that $x=\frac{3}{2}$ is the point of minimum.