Question

If \( 5f(x) + 3f\left( \frac{1}{x} \right) = 2 - \frac{1}{x}, x \ne 0 \), then \( \int_1^2 f\left( \frac{1}{x} \right) dx = \)

• A. \( \frac{6 \log 2 - 7}{32} \)
• B. \( \frac{6 \log 2 - 17}{32} \)
• C. \( \frac{6 \log 2 - 1}{32} \)
• D. \( \frac{6 \log 2 - 7}{16} \)

Hint:

Use substitution and symmetry when functions are defined in terms of both \( x \) and \( \frac{1}{x} \).

Answer 1

The Correct Option is A

Given: \( 5f(x) + 3f\left( \frac{1}{x} \right) = 2 - \frac{1}{x} \)
Replace \( x \rightarrow \frac{1}{x} \):
\[ 5f\left( \frac{1}{x} \right) + 3f(x) = 2 - x \]
We now have a system of two equations. Solve these to eliminate one function.
Multiply the first equation by 5 and the second by 3: \[ 25f(x) + 15f\left( \frac{1}{x} \right) = 10 - \frac{5}{x}
15f\left( \frac{1}{x} \right) + 9f(x) = 6 - 3x \]
Subtracting: \[ 16f(x) = 4 + 3x - \frac{5}{x} \Rightarrow f(x) = \frac{1}{4} + \frac{3x}{16} - \frac{5}{16x} \]
Now substitute into: \[ \int_1^2 f\left( \frac{1}{x} \right) dx = \int_1^2 \left( \frac{1}{4} + \frac{3}{16x} - \frac{5x}{16} \right) dx \]
\[ = \left[ \frac{x}{4} + \frac{3}{16} \ln x - \frac{5x^2}{32} \right]_1^2 = \left( \frac{1}{2} + \frac{3}{16} \log 2 - \frac{20}{32} \right) - \left( \frac{1}{4} + 0 - \frac{5}{32} \right) \]
\[ = \frac{1}{4} + \frac{3}{16} \log 2 - \frac{15}{32} = \frac{6 \log 2 - 7}{32} \]