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

(B) $\frac{1}{2}<\alpha<1$; (C) $2\alpha^{4}-4\alpha^{2}+1=0$

(A) $0<\alpha \leq \frac{1}{2}$

(D) $\alpha^{4}+4\alpha^{2}-1=0$

JEE Advanced · 2017 · 21 May 2017 · Paper 2 · Area Under Curves

If the line

x = α

divides the area of region

R = {(x, y) \in R^{2} : x^{3} \leq y \leq x, 0 \leq x \leq 1}

into two equal parts, then

Answer: (B), (C)

The total area of region

R

is:

A = \int_{0}^{1} (x - x^{3}) d x = [\frac{x ^{2}}{2} - \frac{x ^{4}}{4}]_{0}^{1} = \frac{1}{2} - \frac{1}{4} = \frac{1}{4}

For the line

x = α

to divide this area equally:

\int_{0}^{α} (x - x^{3}) d x = \frac{1}{2} \cdot \frac{1}{4} = \frac{1}{8}

[\frac{x ^{2}}{2} - \frac{x ^{4}}{4}]_{0}^{α} = \frac{1}{8}

\frac{α ^{2}}{2} - \frac{α ^{4}}{4} = \frac{1}{8}

2 α^{2} - α^{4} = \frac{1}{2}

2 α^{4} - 4 α^{2} + 1 = 0

Solving:

α^{2} = \frac{4 \pm 16 - 8}{4} = \frac{4 \pm 2 2}{4} = \frac{2 \pm 2}{2}

Since

0 < α < 1

, we need

α^{2} = \frac{2 - 2}{2} \approx 0.293

, giving

α \approx 0.541

.

Therefore,

\frac{1}{2} < α < 1

JEE Advanced · 2017 · 21 May 2017 · Paper 2 · One or more correct options