Question

For what value of '$a$' is the area bounded by the curve $y=a^2{x}^2+ax+1$ and the straight lines $y=0,x=0$ & $x=1$ is least ?

• A. $a=3/4$
• B. $a=-3/4$
• C. $a=2/3$
• D. $a=-2/3$

Answer 1

The correct option is B $a=-3/4$

The required area will be
$={\int }_0^{1}y.dx$
$={\int }_0^1{a}^2{x}^2+ax+1.dx$
$=\frac{a^2}{3}+\frac{a}{2}+1$
Hence
$f\left(a\right)=\frac{a^2}{3}+\frac{a}{2}+1$
Now
$f^{\prime }\left(a\right)=\frac{2a}{3}+\frac{1}{2}$
$=0$
Or
$a=\frac{-3}{4}$.