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

The Correct Option is C

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 \( 7a - 3 \):
\[7a = 2\]
\[7a - 3 = -1\]