Question

If least value of $f\left(x\right)=x^2+bx+c$ be $-\frac{1}{4}$ and maximum value of $g\left(x\right)=-x^2+bx+2$ occurs at $\frac{3}{2}$, then $c$ is equal to

Answer 1

$f\left(x\right)=x^2+bx+c$
Least value of $f\left(x\right)=-\frac{D}{4}=-\frac{1}{4}$
$\Rightarrow D=b^2-4c=1\cdots \left(1\right)$

$g\left(x\right)=-x^2+bx+2$
Maximum value of $g\left(x\right)$ is at $x=\frac{-b}{-2}=\frac{3}{2}$
$\Rightarrow b=3$
From $\left(1\right),$
$1=3^2-4c$
$\Rightarrow c=2$