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JEE Advanced 2017Paper 2
Exam
JEE Advanced
Year
2017
Paper
Paper 2
Subject
Mathematics
Chapter
Area Under Curves

JEE Advanced 2017 · Paper 2 · Mathematics · Area Under Curves

If the line x = α divides the area of the region R = (x, y) ∈ R ^ 2...

If the line x=αx = \alpha divides the area of the region R=(x,y)∈R2:x3≤y≤x,0≤x≤1R = { \left(x, y\right) \in {R}^{2}: {x}^{3} \le y \le x, 0 \le x \le 1 } into two equal parts, thenIf 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,

Options

  1. A12<α<1\dfrac{1}{2} < \alpha < 1
  2. Bα4+4α2−1=0{\alpha}^{4} + 4{\alpha}^{2} - 1 = 0
  3. C0<α≤120 < \alpha \le \dfrac{1}{2}
  4. D2α4−4α2+1=02{\alpha}^{4} - 4{\alpha}^{2} + 1 = 0

Correct answer

A. 12<α<1\dfrac{1}{2} < \alpha < 1

Step-by-step solution

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,

A=∫01(x−x3)dx=14A = \mathop{∫}_{0}^{1}\left(x - {x}^{3}\right)dx = \dfrac{1}{4}

Area of curvilinear triangle OPQ=A2=18OPQ = \dfrac{A}{2} = \dfrac{1}{8}

⇒∫01(x−x3)dx=18⇒2α4−4α2+1=0⇒ \mathop{∫}_{0}^{1}\left(x - {x}^{3}\right)dx = \dfrac{1}{8} ⇒ 2{\alpha}^{4} - 4{\alpha}^{2} + 1 = 0

⇒(α2−1)2=12⇒α2=2−12⇒ {\left({\alpha}^{2} - 1\right)}^{2} = \dfrac{1}{2} ⇒ {\alpha}^{2} = \dfrac{\sqrt{2} - 1}{\sqrt{2}}

so α lies between 0.5 and 1
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