Application of IntegralsmediumPYQ · JEE Main · 22 Jan 2026 · Shift 1 (Morning)Free

Application of Integrals: Let Line Divide Area Region Ratio (JEE Main 2026)

JEE Maths question with a full step-by-step solution.

Question
Let the line x=−1x=-1 divide the area of the region {(x,y):1+x2≤y≤3−x}\{(x,y):1+x^2\le y\le3-x\} in the ratio m:nm:n, gcd⁡(m,n)=1\gcd(m,n)=1. Then m+nm+n is
A2525
B2828
C2626
D2727correct
Solution
Step 1: 1+x2=3−x⇒x2+x−2=0⇒(x+2)(x−1)=0⇒x=−2,11+x^2=3-x\Rightarrow x^2+x-2=0\Rightarrow(x+2)(x-1)=0\Rightarrow x=-2,1. x=−1x=-1 splits [−2,1][-2,1] into [−2,−1][-2,-1] and [−1,1][-1,1]. Height =(3−x)−(1+x2)=2−x−x2=(3-x)-(1+x^2)=2-x-x^2. Step 2: A1=∫−11(2−x−x2) dx=[2x−x22−x33]−11A_1=\displaystyle\int_{-1}^{1}(2-x-x^2)\,dx=\Big[2x-\tfrac{x^2}{2}-\tfrac{x^3}{3}\Big]_{-1}^{1}. At x=1x=1: 2−12−13=12−3−26=762-\tfrac12-\tfrac13=\tfrac{12-3-2}{6}=\tfrac{7}{6}; at x=−1x=-1: −2−12+13=−12−3+26=−136-2-\tfrac12+\tfrac13=\tfrac{-12-3+2}{6}=-\tfrac{13}{6}. A1=76+136=206=103A_1=\tfrac{7}{6}+\tfrac{13}{6}=\tfrac{20}{6}=\tfrac{10}{3}. Step 3: A2=∫−2−1(2−x−x2) dx=[2x−x22−x33]−2−1A_2=\displaystyle\int_{-2}^{-1}(2-x-x^2)\,dx=\Big[2x-\tfrac{x^2}{2}-\tfrac{x^3}{3}\Big]_{-2}^{-1}. At x=−1x=-1: −136-\tfrac{13}{6}; at x=−2x=-2: −4−2+83=−6+83=−18+83=−206-4-2+\tfrac83=-6+\tfrac83=\tfrac{-18+8}{3}=-\tfrac{20}{6}. A2=−136+206=76A_2=-\tfrac{13}{6}+\tfrac{20}{6}=\tfrac{7}{6}. Step 4: mn=10/37/6=103⋅67=207\dfrac{m}{n}=\dfrac{10/3}{7/6}=\dfrac{10}{3}\cdot\dfrac{6}{7}=\dfrac{20}{7}, gcd⁡(20,7)=1⇒m=20, n=7\gcd(20,7)=1\Rightarrow m=20,\ n=7. Step 5: m+n=27m+n=27. Correct answer: (4)
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