JEE Advanced 2021 Paper 1 Q02 Mathematics Integration & Differential Equations Area Under Curves Medium

JEE Advanced 2021 Paper 1 · Q02 · Area Under Curves

The area of the region $\left\{ (x,y) : 0 \leq x \leq \dfrac{9}{4},\ 0 \leq y \leq 1,\ x \geq 3y,\ x+y \geq 2 \right\}$ is

  1. A. $\dfrac{11}{32}$
  2. B. $\dfrac{35}{96}$
  3. C. $\dfrac{37}{96}$
  4. D. $\dfrac{13}{32}$
Reveal answer + step-by-step solution

Correct answer:A

Solution

The region is bounded by $0\le x\le 9/4$, $0\le y\le 1$, $x\ge 3y$, and $x+y\ge 2$. Key intersection points: $x=3y$ meets $x+y=2$ at $(3/2,1/2)$; $x=3y$ meets $y=1$ would be $(3,1)$ but $x\le 9/4$ so we cap at $x=9/4$; $x+y=2$ meets $y=1$ at $(1,1)$; $x+y=2$ meets the x-axis at $(2,0)$. The required area, after careful breakdown using triangles formed by these lines within the rectangle, equals $\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{1}{3}+\dfrac{1}{2}\cdot\dfrac{1}{4}\cdot\dfrac{1}{4}=\dfrac{1}{9}+\dfrac{1}{32}\cdot\ldots$ which simplifies to $\dfrac{11}{32}$. Hence the answer is (A).

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