Question

Let \( f : \mathbb{R} - \{0\} \to \mathbb{R} \) be a function such that \[ f(x) - 6f\left(\frac{1}{x}\right) = \frac{35}{3x} - \frac{5}{2}. \] If \( \lim_{x \to 0} \left( \frac{1}{\alpha x} + f(x) \right) = \beta \), then \( \alpha, \beta \in \mathbb{R} \), and \( \alpha + 2\beta \) is equal to:

• A. 3
• B. 5
• C. 4
• D. 6

Hint:

When dealing with functional equations, always try to express the function in terms of simpler variables. Additionally, carefully analyze limits and cancel terms to find the necessary parameters.

Answer 1

The Correct Option is C

Step 1: Find \( f(x) \) from the given functional equation We are given the functional equation: \[ f(x) - 6f\left(\frac{1}{x}\right) = \frac{35}{3x} - \frac{5}{2} \] Now, replace \( x \) with \( \frac{1}{x} \) in the above equation: \[ f\left(\frac{1}{x}\right) - 6f(x) = \frac{35}{\frac{3}{x}} - \frac{5}{2} \quad \Rightarrow \quad f\left(\frac{1}{x}\right) - 6f(x) = \frac{35x}{3} - \frac{5}{2} \] Now we have two equations: 
Equation 1: \[ f(x) - 6f\left(\frac{1}{x}\right) = \frac{35}{3x} - \frac{5}{2} \] 
Equation 2: \[ f\left(\frac{1}{x}\right) - 6f(x) = \frac{35x}{3} - \frac{5}{2} \] Multiply Equation 2 by 6: \[ 6f\left(\frac{1}{x}\right) - 36f(x) = \frac{70x}{3} - 15 \] Add this modified Equation 2 to Equation 1: \[ f(x) - 6f\left(\frac{1}{x}\right) + 6f\left(\frac{1}{x}\right) - 36f(x) = \frac{35}{3x} - \frac{5}{2} + \frac{70x}{3} - 15 \] Simplifying: \[ -35f(x) = \frac{35}{3x} + \frac{70x}{3} - \frac{35}{2} \] Solving for \( f(x) \): \[ f(x) = -\frac{1}{3x} - \frac{2x}{3} + \frac{1}{2} \] Thus, the expression for \( f(x) \) is: \[ f(x) = -\frac{1}{3x} - \frac{2x}{3} + \frac{1}{2} \] 
Step 2: Evaluate the Limit We are given that: \[ \lim_{x \to 0} \left( \frac{1}{\alpha x} + f(x) \right) = \beta \] Substitute the expression for \( f(x) \): \[ \lim_{x \to 0} \left( \frac{1}{\alpha x} - \frac{1}{3x} - \frac{2x}{3} + \frac{1}{2} \right) = \beta \] Simplify: \[ \lim_{x \to 0} \left( \frac{3 - \alpha}{3\alpha x} - \frac{2x}{3} + \frac{1}{2} \right) = \beta \] For the limit to exist as \( x \to 0 \), the term \( \frac{3 - \alpha}{3\alpha x} \) must not go to infinity. Therefore, we must have: \[ 3 - \alpha = 0 \quad \Rightarrow \quad \alpha = 3 \] Now the limit becomes: \[ \lim_{x \to 0} \left( - \frac{2x}{3} + \frac{1}{2} \right) = \beta \] As \( x \to 0 \), the term \( -\frac{2x}{3} \) approaches 0. Therefore: \[ \beta = \frac{1}{2} \] Step 3: Calculate \( \alpha + 2\beta \) We found \( \alpha = 3 \) and \( \beta = \frac{1}{2} \). \[ \alpha + 2\beta = 3 + 2\left(\frac{1}{2}\right) = 3 + 1 = 4 \] Thus, the correct answer is \( \alpha + 2\beta = 4 \).