Question

If the area of the region \left\\{(x, y) : \frac{a}{x^2} \leq y \leq \frac{1}{x}, \, 1 \leq x \leq 2, \, 0<a<1 \right\\} is $(\log 2) - \frac{1}{7},$ then the value of $7a - 3$ is equal to:

• A. 2
• B. 0
• C. -1
• D. 1

Answer 1

Answer: (C)

-1

Answer 2

The area of the region is given by:
$\text{Area} = \int_1^2 \left( \frac{1}{x} - \frac{a}{x^2} \right) dx$
Evaluating this integral:
$= \left[ \ln x + \frac{a}{x} \right]_1^2$
$= \ln 2 + \frac{a}{2} - a = \log_2 2 - \frac{1}{7}$
Equating and solving for $a$:
$-\frac{a}{2} = -\frac{1}{7}$
$a = \frac{2}{7}$
Now, calculating $7a - 3$:
$7a = 2$
$7a - 3 = -1$