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JEE Main 202528 JanuaryEvening Shift
Exam
JEE Main
Year
2025
Date
28 January
Shift
Evening Shift
Subject
Mathematics
Chapter
Area Under Curves

JEE Main 2025 · 28 January · Evening Shift · Mathematics · Area Under Curves

The area of the region bounded by the curves x(1+y^2)=1 and y^2=2 x...

The area of the region bounded by the curves x(1+y2)=1andx\left(1+y^2\right)=1 andy^2=2 x$ is:

Options

  1. A2(π2−13)2\left(\frac{\pi}{2}-\frac{1}{3}\right)
  2. Bπ2−13\frac{\pi}{2}-\frac{1}{3}
  3. Cπ4−13\frac{\pi}{4}-\frac{1}{3}
  4. D12(π2−13)\frac{1}{2}\left(\frac{\pi}{2}-\frac{1}{3}\right)

Correct answer

B. π2−13\frac{\pi}{2}-\frac{1}{3}

Step-by-step solution

x(1+y2)=1y2=2x\begin{aligned} & x\left(1+y^2\right)=1 \\ & y^2=2 x \end{aligned} From equation (1) \& (2) x(1+2x)=1⇒2x2+x−1=0⇒x=12,x=−1 (Reject) ⇒y2=2(12)⇒y=±1\begin{aligned} x(1+2 x)=1 & \Rightarrow 2 x^2+x-1=0 \\ & \Rightarrow x=\frac{1}{2}, x=-1 \text{ (Reject) } \\ & \Rightarrow y^2=2\left(\frac{1}{2}\right) \\ & \Rightarrow y= \pm 1 \end{aligned} Area bounded =∫−11(11+y2−y22)dy=(tan⁡−1y−y36)∣−11=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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