Question

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

• A. $a=\frac{-3}{4}$
• B. $a=\frac{-7}{4}$
• C. $a=\frac{-1}{4}$
• D. $a=\frac{-5}{4}$

Answer 1

The correct option is A $a=\frac{-3}{4}$

Required area
$={\int }_0^{1}y.dx$
$={\int }_0^1{a}^2{x}^2+ax+1.dx$
$=[a^2\frac{x^3}{3}+a\frac{x^2}{2}+x{]}_0^1$
$A=\frac{a^2}{3}+\frac{a}{2}+1$
$\frac{dA}{da}=\frac{2a}{3}+\frac{1}{2}=0$
$\Rightarrow 4a=-3$
$\Rightarrow a=\frac{-3}{4}$