Question

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

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

Answer 1

Answer: (A), (D)

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

Area $(x=0\to x=\alpha )=$ Area of $△OAB-$ Area of $△OCB$

$=\frac{1}{2}\alpha \cdot \alpha -{\int }_0^{\alpha }{x}^{2}dx$

$=\frac{{\alpha }^2}{2}-\frac{{\alpha }^2}{4}$

Area $(0=\alpha \to x=1)=$ Area of $\square BAEF-$ Area of $\square BCEF$

$\frac{1}{2}(\alpha +1)(1-\alpha )-{\int }_{\alpha }^1{x}^{3}dx$

$\frac{1-{\alpha }^2}{2}-(\frac{1}{4}-\frac{{\alpha }^2}{4})$

According to the question:

$\frac{{\alpha }^2}{2}-\frac{{\alpha }^4}{4}=\frac{1}{2}-\frac{{\alpha }^2}{2}-\frac{1}{4}+\frac{{\alpha }^4}{4}$

$\Rightarrow 2{\alpha }^4-4{\alpha }^2+1=0$ ... Option A

$f\left(a\right)=2{\alpha }^2-4{\alpha }^2+1=0$

At $\alpha =0,f\left(a\right)=1$

At $\alpha =1,f\left(\alpha \right)=-1$

At $\alpha =\frac{1}{2},f\left(a\right)=\frac{1}{8}>0$

Therefore, Root lies in $\alpha \in (\frac{1}{2},1)$ .. Option D.