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Mathematics · Area under the Curves

JEE Main 2024 — 1 February, Shift 1 — Question 10

The area enclosed by the curves xy+4y=16x y+4 y=16 and x+y=6\mathrm{x}+\mathrm{y}=6 is equal to :

  1. Option A:

    28−30log⁡e228-30 \log _{e} 2

  2. Option B:

    30−28log⁡e230-28 \log _{e} 2

  3. Option C:

    30−32log⁡e230-32 \log _{e} 2

    Correct
  4. Option D:

    32−30log⁡e232-30 \log _{e} 2

Answer: C

Step-by-step solution

xy+4y=16$$\qquad$$$\begin{gathered}x+y=6\end{gathered}$$$\mathrm{y}(\mathrm{x}+4)=16 (2)

$x+y=6$

on solving,(1) & (2) we get

x=4,x=−2x=4,x=-2  Area =∫24((6x)(16x+4))dx\begin{aligned}\text { Area }= & \int_{2}^{4}\left((6x)\left(\frac{16}{x+4}\right)\right)dx&\end{aligned}

Area=30−32log⁡e2=30-32 \log _{e} 2

Solution figure

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Paper
1 February, Shift 1
Subject
Mathematics
Chapter
Area under the Curves
Topic
Area under the Curves
The area enclosed by the curves x y+4 y=16 and x + y =6 is equal to : | JEE Main 2024 PYQ with Solution · DhiX AI