Question

Let, the minimum value of real quadratic expression $a{x}^2+bx+\frac{1}{2a},a>0,$ is equal to $y_0.$ If $y_0$ occurs at $x=k$ and $k=2y_0,$ then set of all possible value of $b$ is

• A. $2,-1$
• B. $-1,-2$
• C. $2,1$
• D. $-2,1$

Answer 1

The correct option is A $2,-1$

Minimum value of parabola occurs at its vertex =$(\frac{-b}{2a},\frac{4ac-b^2}{4a})$

$\Rightarrow k=\frac{-b}{2a}$

$\Rightarrow y_0=\frac{4\left(a\right)\left(\frac{1}{2a}\right)-b^2}{4a}$

$\Rightarrow y_0=\frac{2-b^2}{4a}=k/2$

$\Rightarrow \frac{2-b^2}{4a}=\frac{-b}{4a}$

$\Rightarrow b^2-b-2=0$

$\Rightarrow b=2$ or $b=-1$