JEE Mains · Maths · STD 12 - 8. Application and integration
The area enclosed by the curves \(x y+4 y=16\) and \(x+y=6\) is equal to :
- A \(28-30 \log _e 2\)
- B \(30-28 \log _e 2\)
- C \(30-32 \log _e 2\)
- D \(32-30 \log _e 2\)
Answer & Solution
Correct Answer
(C) \(30-32 \log _e 2\)
Step-by-step Solution
Detailed explanation
\( x y+4 y=16 \) \( y(x+4)=16 \ldots \) (1) \(x+y=6 \) \( x+y=6\) \(....(2)\) on solving, \((1) \& (2)\) we get \(x=4, x=-2\) Area = \( \int_{-2}^4\left((6-x)-\left(\frac{16}{x+4}\right)\right) d x\) \( =30-32 \ln 2\)
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