Verified Solution • 2025–26 Session moderate

QJEEM20Ma0588: JEE Main 2020 Mathematics - Class 12 JEEM20Ma0588 Solution: Consider a region R = {(x, y) R : x 2 y 2x}. if a line y =…

Consider a region R = {(x, y) ∈\in R : x2 ≤\le y ≤\le 2x}. if a line y = α\alpha divides the area of region R into two equal parts, then which of the following is true?

Multiple Choice Options

  • A 3 α\alpha 2 - 8 α\alpha + 8 = 0
  • B α\alpha 3 - 6 α\alpha 3/2 - 16 = 0
  • C 3 α\alpha 2 - 8 α\alpha 3/2 + 8 = 0
  • D α\alpha 3 - 6 α\alpha 2 + 16 = 0

Step-by-Step Verified Solution

JEE Main 2020 (Online) 2nd September Evening Slot Mathematics - Area Under The Curves Question 102 English Explanation

y ≥\ge x2 ⇒\Rightarrow upper region of y = x2

y ≤\le 2x ⇒\Rightarrow lower region of y = 2x

According to question, area of OABC = 2 ×\times area of OAC

⇒\Rightarrow ∫04(y−y2)dy\int\limits^{4}_{0} \left( \sqrt{y} -\frac{y}{2} \right) dy = 2 ∫0α(y−y2)dy\int\limits^{\alpha }_{0} \left( \sqrt{y} -\frac{y}{2} \right) dy

⇒[23y32−y24]04=2[23y32−y24]0α\Rightarrow \left[ {{2 \over 3}{y^{{3 \over 2}}} - {{{y^2}} \over 4}} \right]_0^4 = 2\left[ {{2 \over 3}{y^{{3 \over 2}}} - {{{y^2}} \over 4}} \right]_0^\alpha

⇒163−4=2[23(α)32−α24]\Rightarrow {{16} \over 3} - 4 = 2\left[ {{2 \over 3}{{\left( \alpha \right)}^{{3 \over 2}}} - {{{\alpha ^2}} \over 4}} \right]

⇒43=2[23α32−α24]\Rightarrow {4 \over 3} = 2\left[ {{2 \over 3}{\alpha ^{{3 \over 2}}} - {{{\alpha ^2}} \over 4}} \right]

⇒23=23α32−α24\Rightarrow {2 \over 3} = {2 \over 3}{\alpha ^{{3 \over 2}}} - {{{\alpha ^2}} \over 4}

⇒8=8α32−3α2\Rightarrow 8 = 8{\alpha ^{{3 \over 2}}} - 3{\alpha ^2}

⇒3α2−8α32+8=0\Rightarrow 3{\alpha ^2} - 8{\alpha ^{{3 \over 2}}} + 8 = 0

Related Learning Resources & Exam Preparation

JEE Main PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring