Question

Graph of $f(x)=a{x}^2+bx+c,a,b,c\in R,a\ne 0$ is shown in the adjacent figure. If area of rectangle $OABC=18$ sq. units ( $B$ is the vertex of the parabola) and sum of the roots of the equation $f(x)$ $=0$ is 12 , then

• A. least value of ac is $-\frac{1}{16}$
• B. least value of ac is $\frac{1}{2}$
• C. if $ac=\frac{1}{2}$ then value of $b$ is -1
• D. if $ac=\frac{1}{2}$ then value of $b$ is 2 .

Answer 1

Answer: (C)

$\frac{-b}{a}=12$
$\Rightarrow \left(\frac{-D}{4a}\right)\cdot \left(\frac{-b}{2a}\right)=18$
$\Rightarrow \left(\frac{-D}{4a}\right)\cdot \left(\frac{12}{2}\right)=18$
$\Rightarrow \frac{-\left(b^2-4ac\right)}{a}=12$
$\Rightarrow -b^2+4ac=-b$
$\Rightarrow ac=\frac{b^2-b}{4}$
$\Rightarrow a{c}_{\min }=\frac{{\left(b-\frac{1}{2}\right)}^2-\frac{1}{4}}{4}=\frac{-1}{16}$
which is not possible as a and $c$ both are positive.
$\text{ If }ac=\frac{1}{2}$
$-b^2+2=-b$
$\Rightarrow b^2-b-2=0$
$b=-1,2$
$b\ne 2\Rightarrow a=\frac{-1}{6},a>0$