JEE Advanced 2017 · Paper 2 · Mathematics Q13

MathematicsMultiple correct+4 / -2Question 49 of 54

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

0 < \alpha \le \frac{1}{2}

\frac{1}{2} < \alpha < 1

2\alpha^4 - 4\alpha^2 + 1 = 0

\alpha^4 + 4\alpha^2 - 1 = 0

Answer and worked solution

Official answer: B, C

Total area. $\displaystyle\int_0^1(x-x^3)\,dx=\frac14$ .

Half the area.
$\int_0^\alpha(x-x^3)\,dx=\frac{\alpha^2}2-\frac{\alpha^4}4=\frac18\implies2\alpha^4-4\alpha^2+1=0.$
(C) is true.

Size of α. $\alpha^2=\dfrac{2-\sqrt2}2\approx0.293$ , so $\alpha\approx0.54$ , which lies in $(\frac12,1)$ . (B) is true.

(B), (C)

Topics:Area between curves

JEE Advanced 2017 · Paper 2, Mathematics question 13