Mathematics - Area Under Curves Question with Solution | TestHub

The point of intersection of (i) and (ii) is given by:

$$\begin{array}{rl}y& =\frac{a^2-ax}{1+a^4}\\ y& =\frac{x^2+2ax+3a^2}{1+a^4}\end{array}$$

Equating the two expressions for $y$:

$$\frac{a^2-ax}{1+a^4}=\frac{x^2+2ax+3a^2}{1+a^4}$$This simplifies to:

$$(x+a)(x+2a)=0$$

Thus, $x=-a,-2a$.

The required area is:

$\begin{array}{c}S=\frac{1}{1+a^4}{\int }_{-2a}^{-a}\\ \left[-(x+a)(x+2a)\right]dx\\ =\frac{a^3}{6(1+a^4)}.\end{array}$

Let $f(a)=\frac{a^3}{6(1+a^4)}$.

For $f(a)$ to be maximum, $f^{\prime }(a)=0$.

$$3+3a^4-4a^4=0$$