The value of a such that the area bounded by the curve y=x2+2ax+3a2 , the coordinate axes and the line x=1 , attains its least value is equal to

Question

The value of a such that the area bounded by the curve y=x2+2ax+3a2 , the coordinate axes and the line x=1 , attains its least value is equal to (A) (1/6) (B) -(1/6) (C) (1/3) (D) -(1/3)

Tardigrade Admin · Accepted Answer

y=x2+2ax+3a2 is positive for all x∈ R Area =∫ limits 01(x2 + 2 a x + 3 a2)dx =(1/3)+2a((1/2))+3a2 =(1/3)+a+3a2 Area =((9 a2 + 3 a + 1/3)) sq. units Which is minimum at a=-(1/6)

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

$\frac{1}{6}$

$- \frac{1}{6}$

$\frac{1}{3}$

$- \frac{1}{3}$

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

Q. The value of a such that the area bounded by the curve y=x2+2ax+3a2 , the coordinate axes and the line x=1 , attains its least value is equal to

61​

−61​

31​

−31​