Verified Solution • 2025–26 Session moderate

QJEEM25Mabe54: JEE Main 2025 Mathematics - Class 12 JEEM25Mabe54 Solution: The area of the region bounded by the curves and is:

The area of the region bounded by the curves x(1+y2)=1x(1+y^2)=1 and y2=2xy^2=2x is:

Multiple Choice Options

  • A π4−13\frac{\pi}{4} - \frac{1}{3}
  • B π2−13\frac{\pi}{2} - \frac{1}{3}
  • C 2(π2−13)2\left(\frac{\pi}{2} - \frac{1}{3}\right)
  • D 12(π2−13)\frac{1}{2}\left(\frac{\pi}{2} - \frac{1}{3}\right)

Step-by-Step Verified Solution

amp;x(1+y2)=1..... (1)amp;y2=2x..... (2)\begin{aligned}& x\left(1+y^2\right)=1 \quad\text{..... (1)}\\& y^2=2 x\quad\text{..... (2)}\end{aligned}

 From equation (1) & (2) x(1+2x)=1⇒2x2+x−1=0⇒x=12,x=−1 (Reject) ⇒y2=2(12)⇒y=±1

JEE Main 2025 (Online) 28th January Evening Shift Mathematics - Area Under The Curves Question 7 English Explanation

 Area bounded amp;=∫−11(11+y2−y22)dyamp;=(tan⁡−1y−y36)∣−11amp;=π2−13\begin{aligned}\text { Area bounded } & =\int_{-1}^1\left(\frac{1}{1+y^2}-\frac{y^2}{2}\right) d y \\& =\left.\left(\tan ^{-1} y-\frac{y^3}{6}\right)\right|_{-1} ^1 \\& =\frac{\pi}{2}-\frac{1}{3}\end{aligned}

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