If the area of the region { ( x , y ) : a x 2 ? y ? 1 x , 1 ? x ? 2 , 0 < a < 1 } is ( log e 2 ) - 1 7 then the value of 7 a - 3 is equal to : [2024]

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Solution

Q.
If the area of the region ${(x, y) : \frac{a}{x^{2}} \leq y \leq \frac{1}{x}, 1 \leq x \leq 2, 0 < a < 1}$ is $(\log_{e} 2) - \frac{1}{7}$ then the value of $7 a - 3$ is equal to : [2024]

1 2

2 - 1

3 1

4 0

Ans.
(2)

Area of region ${(x, y) : \frac{a}{x^{2}} \leq y \leq \frac{1}{x}, 1 \leq x \leq 2, 0 < a < 1} = \int_{1}^{2} (\frac{1}{x} - \frac{a}{x^{2}}) d x$

$\Rightarrow (\ln 2) - \frac{1}{7} = {[\ln x + \frac{a}{x}]}_{1}^{2} = \ln 2 + \frac{a}{2} - (\ln (1 + a))$

$= \ln 2 - \frac{a}{2} \Rightarrow \frac{a}{2} = \frac{1}{7} \Rightarrow a = \frac{2}{7}$

$So, 7 a - 3 = \frac{2}{7} \times 7 - 3 = - 1$

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