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

To address the problem, we will analyze the provided functional equation and evaluate the given limit condition. The objective is to determine the value of \( \alpha + 2\beta \).

The functional equation is:

\(f(x) - 6f\left(\frac{1}{x}\right) = \frac{35}{3x} - \frac{5}{2}\)

Substituting \( x \) with \( \frac{1}{x} \) yields:

\(f\left(\frac{1}{x}\right) - 6f(x) = \frac{35x}{3} - \frac{5}{2}\)

These equations are labeled as:

Equation (1): \(f(x) - 6f\left(\frac{1}{x}\right) = \frac{35}{3x} - \frac{5}{2}\)

Equation (2): \(f\left(\frac{1}{x}\right) - 6f(x) = \frac{35x}{3} - \frac{5}{2}\)

Solving these equations simultaneously by adding Equation (1) and Equation (2):

\(f(x) - 6f\left(\frac{1}{x}\right) + f\left(\frac{1}{x}\right) - 6f(x) = \frac{35}{3x} - \frac{5}{2} + \frac{35x}{3} - \frac{5}{2}\)

Simplification results in:

\(-5f(x) - 5f\left(\frac{1}{x}\right) = \frac{35}{3x} + \frac{35x}{3} - 5\)

Rearranging to solve for \( f(x) \):

\(f(x) + f\left(\frac{1}{x}\right) = - \frac{1}{5} \left( \frac{35}{3x} + \frac{35x}{3} - 5 \right)\)

Further simplification yields:

\(f(x) + f\left(\frac{1}{x}\right) = - \frac{7}{3x} - \frac{7x}{3} + 1\)

Considering the limit condition:

\(\lim_{x \to 0} \left( \frac{1}{\alpha x} + f(x) \right) = \beta\)

For \( \beta \) to be finite as \( x \to 0 \), the term \( \frac{1}{\alpha x} \) must approach zero. This implies that \( \alpha \) must be \( -\frac{7}{3} \) to cancel the singularity from \( f(x) \).

Thus, under the condition that \( \lim_{x \to 0} \left( \frac{1}{\alpha x} + f(x) \right) = 0 \), we have \( 2\beta = 1 \).

Solving these conditions, we find \(\alpha = -\frac{7}{3}\) and \( \beta = 1+ \frac{7}{3} = \frac{10}{3} \).

Therefore, \( \alpha + 2\beta = -\frac{7}{3} + 2\left(\frac{10}{3}\right) = -\frac{7}{3} + \frac{20}{3} = \frac{13}{3} \).