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JEE Main 202622 JanuaryMorning Shift
Exam
JEE Main
Year
2026
Date
22 January
Shift
Morning Shift
Subject
Mathematics
Chapter
Area Under Curves

JEE Main 2026 · 22 January · Morning Shift · Mathematics · Area Under Curves

Let the line x=-1 divide the area of the region \ (x, y): 1+x^ 2 ≤...

Let the line x=−1x=-1 divide the area of the region \left\{(x, y): 1+x^{2} \leq y \leq 3-x\right\}intheratioin the ratiom: n, \operatorname{gcd}(m, n)=1.Then. Thenm+n is equal toLet the line x=-1 divide the area of the region \ (x, y): 1+x^ 2 ≤ y ≤ 3-x\ in the ratio m: n, gcd(m, n)=1 . Then m+n

Options

  1. A27
  2. B25
  3. C28
  4. D26

Correct answer

A. 27

Step-by-step solution

Let the line x=-1 divide the area of the region \ (x, y): 1+x^ 2 ≤ y ≤ 3-x\ in the ratio m: n, gcd(m, n)=1 . Then m+n
mn=∫−11[(3−x)−(1+x2)] dx∫−21[(3−x)−(1+x2)] dx=207\dfrac{m}{n} = \dfrac{\int_{-1}^{1} [(3-x) - (1+x^2)] \, dx}{\int_{-2}^{1} [(3-x) - (1+x^2)] \, dx} = \dfrac{20}{7}
∴m+n=20+7\therefore m + n = 20 + 7
=27= 27
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