Question

Let the minimum value of real quadratic expression $a{x}^2-bx+\frac{1}{2a},a>0$be$y_0$. If $y_0$occursat$x=k$and $k=2y_0$, then possible value(s) of $b$ are

• A. $2$
• B. $-2$
• C. $1$
• D. $-1$

Answer 1

Answer: (B), (C)

$-2$
$1$

$a{x}^2-bx+\frac{1}{2a}$
$=a\left(\right(x-\frac{b}{2a}{)}^2-\frac{b^2}{4a^2})+\frac{1}{2a}$
$=a{[x-\frac{b}{2a}]}^2+\frac{2-b^2}{4a}$
Now, the minimum value will occur when the square becomes $0$.
$\therefore k=\frac{b}{2a}$and$y_0=\frac{2-b^2}{4a}$
Given $k=2y_0$
$\Rightarrow \frac{b}{2a}=2\frac{2-b^2}{4a}$
$\Rightarrow b^2+b-2=0$
$b=-2,1$.