Question

If \[ 3f(x) - 2f\left(\frac{1}{x}\right) = x, \] then \( f'(2) \) is:

• A. \( 1 \)
• B. \( \frac{1}{2} \)
• C. \( 2 \)
• D. \( \frac{7}{2} \)

Hint:

For functional equations, assume a linear form and solve using coefficient comparison.

Answer 1

The Correct Option is B

Step 1: Finding \( f(x) \) Assume \( f(x) = ax + b \). Substituting: \[ 3(ax + b) - 2(a/x + b) = x. \] Expanding and equating coefficients: \[ 3ax + 3b - 2a/x - 2b = x. \] Solving, we get: \[ a = \frac{1}{2}, \quad b = 0. \] Thus, \[ f(x) = \frac{x}{2}. \] Step 2: Finding \( f'(x) \) \[ f'(x) = \frac{1}{2}, \quad f'(2) = \frac{1}{2}. \]

Answer 2

Given:
\[ 3 f(x) - 2 f\left(\frac{1}{x}\right) = x, \] find \( f'(2) \).

Step 1: Differentiate both sides with respect to \( x \):
\[ 3 f'(x) - 2 \frac{d}{dx} \left[ f\left(\frac{1}{x}\right) \right] = 1 \]

Step 2: Use chain rule on the second term:
\[ \frac{d}{dx} f\left(\frac{1}{x}\right) = f'\left(\frac{1}{x}\right) \cdot \frac{d}{dx}\left(\frac{1}{x}\right) = f'\left(\frac{1}{x}\right) \cdot \left(-\frac{1}{x^2}\right) \]

Step 3: Substitute back:
\[ 3 f'(x) - 2 \left[ f'\left(\frac{1}{x}\right) \cdot \left(-\frac{1}{x^2}\right) \right] = 1 \] \[ 3 f'(x) + \frac{2}{x^2} f'\left(\frac{1}{x}\right) = 1 \]

Step 4: Replace \( x \) by \( \frac{1}{x} \) in the original equation:
\[ 3 f\left(\frac{1}{x}\right) - 2 f(x) = \frac{1}{x} \] Differentiate both sides w.r.t. \( x \):
\[ 3 \frac{d}{dx} f\left(\frac{1}{x}\right) - 2 f'(x) = -\frac{1}{x^2} \] Using chain rule again:
\[ 3 \cdot f'\left(\frac{1}{x}\right) \cdot \left(-\frac{1}{x^2}\right) - 2 f'(x) = -\frac{1}{x^2} \] \[ -\frac{3}{x^2} f'\left(\frac{1}{x}\right) - 2 f'(x) = -\frac{1}{x^2} \] Multiply both sides by -1:
\[ \frac{3}{x^2} f'\left(\frac{1}{x}\right) + 2 f'(x) = \frac{1}{x^2} \]

Step 5: We have the system:
\[ 3 f'(x) + \frac{2}{x^2} f'\left(\frac{1}{x}\right) = 1 \] \[ 2 f'(x) + \frac{3}{x^2} f'\left(\frac{1}{x}\right) = \frac{1}{x^2} \]

Step 6: Solve for \( f'(x) \) and \( f'\left(\frac{1}{x}\right) \).
Multiply the first equation by 3 and second by 2:
\[ 9 f'(x) + \frac{6}{x^2} f'\left(\frac{1}{x}\right) = 3 \] \[ 4 f'(x) + \frac{6}{x^2} f'\left(\frac{1}{x}\right) = \frac{2}{x^2} \] Subtract second from first:
\[ (9 - 4) f'(x) = 3 - \frac{2}{x^2} \implies 5 f'(x) = 3 - \frac{2}{x^2} \] \[ f'(x) = \frac{3 - \frac{2}{x^2}}{5} = \frac{3}{5} - \frac{2}{5 x^2} \]

Step 7: Find \( f'(2) \):
\[ f'(2) = \frac{3}{5} - \frac{2}{5 \times 4} = \frac{3}{5} - \frac{2}{20} = \frac{3}{5} - \frac{1}{10} = \frac{6}{10} - \frac{1}{10} = \frac{5}{10} = \frac{1}{2} \]

Therefore,
\[ \boxed{ \frac{1}{2} } \]