Question

If the line $x=\alpha$ divides the area of region $R=\left\{\right(x,y)\in R^2:x^3\le y\le x,0\le x\le 1\}$ into two equal parts, then

• A. $0<\alpha \le \frac{1}{2}$
• B. $\frac{1}{2}<\alpha <1$
• C. $2{\alpha }^4-4{\alpha }^2+1=0$
• D. ${\alpha }^4+4{\alpha }^2-1=0$

Answer 1

Answer: (B), (C)

The correct options are
B $\frac{1}{2}<\alpha <1$
C $2{\alpha }^4-4{\alpha }^2+1=0$


Area of the region $R$ is shaded in the figure.
$\text{Area,}A=\underset{0}{\overset{1}{\int }}(x-x^3)dx={[\frac{x^2}{2}-\frac{x^4}{4}]}_0^1=\frac{1}{4}$
Since $x=\alpha$ divides the area in two equal parts,
$\text{ar}\left(OPQ\right)=\frac{A}{2}=\frac{1}{8}$
$\frac{A}{2}=\underset{0}{\overset{\alpha }{\int }}(x-x^3)dx={[\frac{x^2}{2}-\frac{x^4}{4}]}_0^{\alpha }=\frac{{\alpha }^2}{2}-\frac{{\alpha }^4}{4}=\frac{1}{8}$
$\Rightarrow 2{\alpha }^4-4{\alpha }^2+1=0$
$\Rightarrow ({\alpha }^2-1{)}^2=\frac{1}{2}\Rightarrow {\alpha }^2=\frac{\sqrt{2}-1}{\sqrt{2}}$
$\Rightarrow \frac{1}{2}<\alpha <1$