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 $x=\frac{3}{2}$, then c is equal to

• A. $4$
• B. $3$
• C. $2$
• D. $1$

Answer 1

The correct option is C $2$

Consider $g\left(x\right)$
$g\left(x\right)=-x^2+bx+2$
Now
$g^{\prime }\left(x\right)=-2x+b$
For maxima, $g^{\prime }\left(x\right)=0$
Or
$-2x+b=0$
$x=\frac{b}{2}$
However it is given that the maximum value occurs at $x=\frac{3}{2}$.
Hence
$\frac{b}{2}=\frac{3}{2}$
$b=3$.
Now consider
$f\left(x\right)$$=x^2+bx+c$
$=x^2+3x+c$
$f^{\prime }\left(x\right)=2x+3$
Hence for minima,
$2x+3=0$
Or
$x=\frac{-3}{2}$.
Thus the minimum value occurs at $x=\frac{-3}{2}$
Now
$f\left(\frac{-3}{2}\right)=\frac{9}{4}-\frac{9}{2}+c$
$=\frac{9}{4}-\frac{18}{4}+c$
$=\frac{-9}{4}+c$
Now it is given that the minimum value is $\frac{1}{4}$.
Hence
$\frac{-9}{4}+c=\frac{-1}{4}$
$c=\frac{9-1}{4}$
$=\frac{8}{4}$
$=2.$