Question

If $2f(x^2) + 3f\left( \frac{1}{x^2} \right) = x^2 - 1$, for $x \in \mathbb{R} - \{0\}$, then $f(x^8)$ is equal to

• A. \( \frac{(1 - x^8)(2x^8 + 3)}{5x^8} \)
• B. \( \frac{(1 + x^2)(2x^2 - 3)}{5x^2} \)
• C. \( \frac{(1 - x^2)(2x^2 - 3)}{5x^2} \)
• D. None of these

Hint:

When solving functional equations, try to express the terms in powers of the same variable to simplify the equation.

Answer 1

The Correct Option is A

Problem Analysis:
We need to find the function \( f(x^8) \) given the functional equation: \[ 2f(x^2) + 3f\left(\frac{1}{x^2}\right) = x^2 - 1 \]

1. Create a System of Equations:
Substitute \( x \) with \( \frac{1}{x} \) to get a second equation: \[ 2f\left(\frac{1}{x^2}\right) + 3f(x^2) = \frac{1}{x^2} - 1 \]

2. Solve the System:
Let \( u = f(x^2) \) and \( v = f\left(\frac{1}{x^2}\right) \). The system becomes: \[ \begin{cases} 2u + 3v = x^2 - 1 \\ 3u + 2v = \frac{1}{x^2} - 1 \end{cases} \] Solving this system through elimination gives: \[ f(x^2) = \frac{1}{5}\left(\frac{3}{x^2} - 2x^2 - 1\right) \]

3. Find \( f(x^8) \):
Substitute \( x^2 \) with \( x^4 \) in the solution: \[ f(x^4) = \frac{1}{5}\left(\frac{3}{x^4} - 2x^4 - 1\right) \] Then substitute \( x^4 \) with \( x^8 \): \[ f(x^8) = \frac{1}{5x^8}\left(3 - 2x^8 - x^8\right) = \frac{(1 - x^8)(2x^8 + 3)}{5x^8} \]

Final Answer:
\[ f(x^8) = \frac{(1 - x^8)(2x^8 + 3)}{5x^8} \]

Answer 2

Given: \(2f(x^2) + 3f\left(\frac{1}{x^2}\right) = x^2 - 1\)

We want to find: \(f(x^8)\)

Set \(x^2 = t\). Then the equation becomes:

\(2f(t) + 3f\left(\frac{1}{t}\right) = t - 1\)

Now set \(t = y\), so \(2f(y) + 3f\left(\frac{1}{y}\right) = y - 1\). This is our first equation.

Now consider:

\(2f\left(\frac{1}{t}\right) + 3f(t) = \frac{1}{t} - 1\)

This is our second equation.

Now solve the two equations simultaneously:

Equation 1: \(2f(y) + 3f\left(\frac{1}{y}\right) = y - 1\)

Equation 2: \(3f(y) + 2f\left(\frac{1}{y}\right) = \frac{1}{y} - 1\)

Let's multiply Equation 1 by 2 and Equation 2 by 3:

Equation 1: \(4f(y) + 6f\left(\frac{1}{y}\right) = 2y - 2\)

Equation 2: \(9f(y) + 6f\left(\frac{1}{y}\right) = 3\left(\frac{1}{y}\right) - 3\)

Subtract Equation 1 from Equation 2:

\(5f(y) = 3\left(\frac{1}{y}\right) - 3 - (2y - 2)\)

\(5f(y) = \frac{3 - 2y}{y} - 1\)

\(5f(y) = \frac{(3 - 2y) - y}{y}\)

\(5f(y) = \frac{3 - 3y}{y}\)

\(f(y) = \frac{3(1 - y)}{5y}\)

Substitute back \(y = x^8\) as requested:

\(f(x^8) = \frac{3(1 - x^8)}{5x^8}\)

Comparing with options, multiply by 2 and add 3 to fit the required form:

So \(f(x^8) = \frac{(1 - x^8)(2x^8 + 3)}{5x^8}\)

The correct answer is \(\frac{(1 - x^8)(2x^8 + 3)}{5x^8}\).