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:

• A

  0
• B

  2
• Correct answer
  C

  -1
• D

  1

Solution

$\begin{aligned} & \text { area } \int_1^2\left(\frac{1}{x}-\frac{a}{x^2}\right) d x \\ & {\left[\ell n x+\frac{a}{x}\right]_1^2} \\ & \ell \operatorname{nn} 2+\frac{a}{2}-a=\log _e 2-\frac{1}{7} \\ & \frac{-a}{2}=-\frac{1}{7} \\ & a=\frac{2}{7} \\ & 7 a=2 \\ & 7 a-3=-1\end{aligned}$

Hence, the correct answer is option: 'C'.