Question

The graph of $f(x)=x^2$ and $g(x)=c{x}^3$ intersect at two points. If the area of the region over the interval $\left[0,\frac{1}{c}\right]$ is equal to $\frac{2}{3}$, then the value of $\left(\frac{1}{c}+\frac{1}{c^2}\right)$, is

• A. 20
• B. 2
• C. 6
• D. 12

Answer 1

Answer: (C)

Obviously for $c\in (0,1),f(x)$ lies obove the $g(x)$
also $x^2=c{x}^3\Rightarrow x=0$ or $x=\frac{1}{c}$
hence $\underset{0}{\overset{1/c}{\int }}\left(x^2-c{x}^3\right)dx=\frac{1}{12c^3}$

or $\frac{1}{12c^3}=\frac{2}{3}\Rightarrow c^3=\frac{1}{8}\Rightarrow c=\frac{1}{2}$
Hence, $\left(\frac{1}{c}+\frac{1}{c^2}\right)=2+4=6$