JEE MAIN - Mathematics (2024 - 6th April Evening Shift - No. 10)
If the area of the region $$\left\{(x, y): \frac{\mathrm{a}}{x^2} \leq y \leq \frac{1}{x}, 1 \leq x \leq 2,0<\mathrm{a}<1\right\}$$ is $$\left(\log _{\mathrm{e}} 2\right)-\frac{1}{7}$$ then the value of $$7 \mathrm{a}-3$$ is equal to :
1
0
2
$$-$$1
Explanation
$$\left\{(x, y): \frac{a}{x^2} \leq y \leq \frac{1}{x}, 1 \leq x \leq 2,0< a<1\right\}$$
$$ \Rightarrow \int\limits_1^2 {\left( {{1 \over x} - {a \over {{x^2}}}} \right)dx = \left| {\ln |x| + {a \over x}} \right|_1^2} $$
$$\begin{aligned} & \left(\ln 2+\frac{a}{2}\right)-(\ln 1+a)=\ln 2-\frac{a}{2} \\ & \Rightarrow \quad \frac{a}{2}=\frac{1}{7} \\ & \Rightarrow \quad a=\frac{2}{7} \\ & \quad 7 a-3=\frac{2}{7} \times 7-3=-1 \end{aligned}$$
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