If the line x=α divides the area of the region R={x,y∈R^2:x^3≤y≤x, 0≤x≤1} into two equal parts, then

Correct answer: (A) 1/2<α<1; (D) 2α^4-4α^2+1=0. A= ∫01x-x^3dx=1/4 Area of curvilinear triangle OPQ=A/2=1/8 ⇒ ∫01x-x^3dx=1/8 ⇒ 2α^4-4α^2+1=0 ⇒ α^2-12=1/2 ⇒ α^2=(√2-1)/(√2) so α lies between 0.5 and 1

Q31 / 54 · JEE Advanced 2017 (Paper 2) · Mathematics

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

Correct answer
A
$\frac{1}{2} < α < 1$
B
$α^{4} + 4 α^{2} - 1 = 0$
C
$0 < α \leq \frac{1}{2}$
Correct answer
D
$2 α^{4} - 4 α^{2} + 1 = 0$

Solution

$A = \int_{0}^{1} (x - x^{3}) d x = \frac{1}{4}$

Area of curvilinear triangle $O P Q = \frac{A}{2} = \frac{1}{8}$

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

$\Rightarrow {(α^{2} - 1)}^{2} = \frac{1}{2} \Rightarrow α^{2} = \frac{\sqrt{2} - 1}{\sqrt{2}}$

so α lies between 0.5 and 1

Hence, the correct answer is options: 'A', 'D'.

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If the line x=α divides the area of the region R={x,y∈R2:x3≤y≤x, 0≤x≤1} into two equal parts, then

Solution

Relevant Questions

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