Question

Find the value(s) of the parameter 'a' (a > 0) for each of which the area of the figure bounded by the straight line $y=\frac{a^2-ax}{1+a^4}$ & the parabola $y=\frac{x^2+2ax+3a^2}{1+a^4}$ is the greatest

• A. $a=2^{1/4}$
• B. $a=5^{1/4}$
• C. $a=7^{1/4}$
• D. $a=3^{1/4}$

Answer 1

Answer: (D)

$a=3^{1/4}$

$A={\int }_{-a}^{-2a}\frac{a^2-ax-(x^2+2ax+3a^2)}{1+a^4}dx$
$=\frac{3}{2}\frac{a^3}{1+a^4}$ Now f(a) $=\frac{3}{2}\frac{a^3}{1+a^4}\Rightarrow f^{\prime }\left(a\right)=0\Rightarrow (1+a^4)3a^2-a^{3}4a^3=0$
$\Rightarrow a_{min}=0,a_{min}=3^{1/4}$