Question

If \( \displaystyle \int_{3}^{b} \frac{x - 1}{2x - x^2} \, dx = \frac{1}{2} \), then \( (b - 1)^2 = \)

• A. \( 2 \)
• B. \( \sqrt{2} \)
• C. \( 1 + \frac{3}{e} \)
• D. \( \sqrt{\frac{3}{e} - 1} \)

Hint:

Use partial fractions and logarithmic integration for rational functions. Completing the square helps in extracting specific forms.

Answer 1

The Correct Option is C

We are given: \[ \int_{3}^{b} \frac{x - 1}{2x - x^2} \, dx = \frac{1}{2} \] First simplify the integrand: \[ \frac{x - 1}{2x - x^2} = \frac{x - 1}{x(2 - x)} \] Use partial fractions: \[ \frac{x - 1}{x(2 - x)} = \frac{x - 1}{-x(x - 2)} = \frac{A}{x} + \frac{B}{x - 2} \] Solve: \[ \frac{x - 1}{-x(x - 2)} = \frac{A}{x} + \frac{B}{x - 2} \Rightarrow x - 1 = -A(x - 2) - Bx \Rightarrow x - 1 = -Ax + 2A - Bx \Rightarrow x - 1 = -x(A + B) + 2A \] Matching coefficients: \[ -A - B = 1, \quad 2A = -1 \Rightarrow A = -\frac{1}{2}, \quad B = -\frac{1}{2} \] So, \[ \int_{3}^{b} \frac{x - 1}{2x - x^2} dx = \int_{3}^{b} \left(-\frac{1}{2x} - \frac{1}{2(x - 2)}\right) dx \] Integrate: \[ = -\frac{1}{2} \left[ \ln |x| + \ln |x - 2| \right]_3^b = -\frac{1}{2} \ln \left( \frac{b(b - 2)}{3(1)} \right) = \frac{1}{2} \ln \left( \frac{3}{b(b - 2)} \right) \] We are told this equals \( \frac{1}{2} \), so: \[ \frac{1}{2} \ln \left( \frac{3}{b(b - 2)} \right) = \frac{1}{2} \Rightarrow \ln \left( \frac{3}{b(b - 2)} \right) = 1 \Rightarrow \frac{3}{b(b - 2)} = e \Rightarrow b(b - 2) = \frac{3}{e} \Rightarrow b^2 - 2b = \frac{3}{e} \] Complete the square: \[ b^2 - 2b + 1 = \frac{3}{e} + 1 \Rightarrow (b - 1)^2 = \boxed{1 + \frac{3}{e}} \]