Question

The value of $a$ such that the area bounded by the curve $y=x^{2}+2ax+3a^{2}$ , the coordinate axes and the line $x=1$ , attains its least value is equal to

• A. $\frac{1}{6}$
• B. $-\frac{1}{6}$
• C. $\frac{1}{3}$
• D. $-\frac{1}{3}$

Answer 1

Answer: (B)

$y=x^{2}+2ax+3a^{2}$ is positive for all $x\in R$
Area $=\int\limits _{0}^{1}\left(x^{2} + 2 a x + 3 a^{2}\right)dx$
$=\frac{1}{3}+2a\left(\frac{1}{2}\right)+3a^{2}$
$=\frac{1}{3}+a+3a^{2}$
Area $=\left(\frac{9 a^{2} + 3 a + 1}{3}\right)$ sq. units
Which is minimum at $a=-\frac{1}{6}$