Question

The value of the parameter `a' such that the area bounded by $y=a^2{x}^2+ax+1$, positive coordinate axes and the line x=1 attains its least value, is equal to

• A. $-\frac{1}{4}$
• B. $-\frac{1}{2}$
• C. $-\frac{3}{4}$
• D. $-1$

Answer 1

The correct option is C $-\frac{3}{4}$

$A={\int }_0^1(a^2{x}^2+ax+1)dx$
$=\frac{a^2}{3}+\frac{a}{2}+1$ which is minimum for $a=-\frac{3}{4}$