The area enclosed by the curves x y + 4 y = 16 and x + y = 6 is equal to: [2024]

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Solution

Q.
The area enclosed by the curves $x y + 4 y = 16$ and $x + y = 6$ is equal to: [2024]

1 $28 - 30 \log_{e} 2$

2 $32 - 30 \log_{e} 2$

3 $30 - 28 \log_{e} 2$

4 $30 - 32 \log_{e} 2$

Ans.
(4)

We have, $y = \frac{16}{x + 4} and x + y = 6 \Rightarrow y = 6 - x$

For intersecting points:

$6 - x = \frac{16}{x + 4} \Rightarrow (6 - x) (x + 4) = 16$

$\Rightarrow x^{2} - 2 x - 8 = 0 \Rightarrow x^{2} - 4 x + 2 x - 8 = 0$

$\Rightarrow x (x - 4) + 2 (x - 4) = 0$

$\Rightarrow (x + 2) (x - 4) = 0 \Rightarrow x = - 2, 4$

$∴$ $y = 8, when x = - 2 and y = 2, when x = 4$

So, $Area = \int_{- 2}^{4} ((6 - x) - \frac{16}{x + 4}) d x = {[6 x - \frac{x^{2}}{2} - 16 \log_{e} (x + 4)]}_{- 2}^{4}$

$= 24 - 8 - 16 \log_{e} 8 + 12 + 2 + 16 \log_{e} 2 = 30 - 16 \log_{e} (\frac{8}{2})$

$= 30 - 16 \log_{e} 2^{2} = 30 - 32 \log_{e} 2$

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