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

Answer 2

We need to find the value of \(7a - 3\) given that the area of the region \( \left\{(x, y) : \frac{a}{x^2} \leq y \leq \frac{1}{x}, \, 1 \leq x \leq 2, \, 0

The region described involves solving an area problem between two curves: \( y = \frac{a}{x^2} \) and \( y = \frac{1}{x} \), over the interval \( 1 \leq x \leq 2 \).

The area between these two curves can be computed as:

\(A = \int_{1}^{2} \left(\frac{1}{x} - \frac{a}{x^2}\right) \, dx\)

Calculate the integrals separately:

• The integral of \(\frac{1}{x}\) over \([1, 2]\):

\(\int_{1}^{2} \frac{1}{x} \, dx = \left[\log x \right]_{1}^{2} = \log 2 - \log 1 = \log 2\)

• The integral of \(\frac{a}{x^2}\):

\(\int_{1}^{2} \frac{a}{x^2} \, dx = a \int_{1}^{2} x^{-2} \, dx = a \left[-\frac{1}{x}\right]_{1}^{2} = a \left(-\frac{1}{2} + 1\right) = a \left(\frac{1}{2}\right)\)

Thus, the total area A is given by:

\(A = \log 2 - a\left(\frac{1}{2}\right) = \log 2 - \frac{a}{2}\)

We are given that this area \( A \) is \(\log 2 - \frac{1}{7}\).

\(\log 2 - \frac{a}{2} = \log 2 - \frac{1}{7}\)

By equating the two expressions, we can solve for \(a\):

\(-\frac{a}{2} = -\frac{1}{7}\)

Solving for \(a\), we get:

\(\frac{a}{2} = \frac{1}{7}\) \(a = \frac{2}{7}\)

Finally, we need to find the value of \(7a - 3\):

\(7a - 3 = 7 \times \frac{2}{7} - 3 = 2 - 3 = -1\)

Thus, the value of \(7a - 3\) is \(-1\).

The correct answer is: -1