Question

If \( 5f(x) + 4f\left(\frac{1}{x}\right) = x^2 - 2 \), for all \( x \neq 0 \), and \( y = 9x^2f(x) \), then \( y \) is strictly increasing in:

• A. \( (0, \frac{1}{\sqrt{5}}) \cup (\frac{1}{\sqrt{5}}, \infty) \)
• B. \( (-\frac{1}{\sqrt{5}}, 0) \cup (\frac{1}{\sqrt{5}}, \infty) \)
• C. \( (-\frac{1}{\sqrt{5}}, 0) \cup (0, \frac{1}{\sqrt{5}}) \)
• D. \( (-\infty, -\frac{1}{\sqrt{5}}) \cup (0, \frac{1}{\sqrt{5}}) \)

Answer 1

The Correct Option is B

We are given the equation:

\[ 5f(x) + 4 \left(\frac{1}{x}\right) = x^2 - 2 \]

Solving for \(f(x)\):

\[ f(x) = \frac{x^2 - 2 - \frac{4}{x}}{5} \]

Now, substitute \(f(x)\) into the equation for \(y\):

\[ y = 9x^2 f(x) = 9x^2 \left(\frac{x^2 - 2 - \frac{4}{x}}{5}\right) \]

Simplifying:

\[ y = \frac{9x^4 - 18x^2 - 36x}{5} \]

Now, differentiate \(y\) with respect to \(x\):

\[ \frac{dy}{dx} = \frac{1}{5} \left(36x^3 - 36x - 36\right) \]

Simplifying:

\[ \frac{dy}{dx} = \frac{36}{5} \left(x^3 - x - 1\right) \]

For \(y\) to be strictly increasing, we need \(\frac{dy}{dx} > 0\), which implies:

\[ x^3 - x - 1 > 0 \]

Solving the inequality \(x^3 - x - 1 > 0\), we find that the critical points are:

\[ x = \pm \frac{1}{\sqrt{5}} \]

Thus, \(y\) is strictly increasing in the intervals:
\[ x \in \left(-\frac{1}{\sqrt{5}}, 0\right) \cup \left(\frac{1}{\sqrt{5}}, \infty\right) \]

Answer 2

Given the functional equation \( 5f(x) + 4f\left(\frac{1}{x}\right) = x^2 - 2 \) for all \( x \neq 0 \), and \( y = 9x^2f(x) \), determine the interval(s) where \( y \) is strictly increasing.

Concept Used:

To solve for \( f(x) \), we can form a system by replacing \( x \) with \( \frac{1}{x} \) in the given equation. Then solve for \( f(x) \) and compute \( y \), and study its derivative.

Step-by-Step Solution:

Step 1: Set up the system of equations.

Given: \( 5f(x) + 4f\left(\frac{1}{x}\right) = x^2 - 2 \) ...(1)

Replace \( x \) by \( \frac{1}{x} \):
\( 5f\left(\frac{1}{x}\right) + 4f(x) = \frac{1}{x^2} - 2 \) ...(2)

Step 2: Solve for \( f(x) \).

Multiply (1) by 5: \( 25f(x) + 20f\left(\frac{1}{x}\right) = 5x^2 - 10 \) ...(3)

Multiply (2) by 4: \( 20f\left(\frac{1}{x}\right) + 16f(x) = \frac{4}{x^2} - 8 \) ...(4)

Subtract (4) from (3):
\( (25f(x) - 16f(x)) + (20f(1/x) - 20f(1/x)) = 5x^2 - 10 - \left(\frac{4}{x^2} - 8\right) \)
\( 9f(x) = 5x^2 - 10 - \frac{4}{x^2} + 8 \)
\( 9f(x) = 5x^2 - \frac{4}{x^2} - 2 \)

Thus:
\( f(x) = \frac{5x^2 - \frac{4}{x^2} - 2}{9} = \frac{5x^2}{9} - \frac{4}{9x^2} - \frac{2}{9} \)

Step 3: Compute \( y = 9x^2f(x) \).

\( y = 9x^2 \left( \frac{5x^2}{9} - \frac{4}{9x^2} - \frac{2}{9} \right) \)
\( y = 9x^2 \cdot \frac{5x^2}{9} - 9x^2 \cdot \frac{4}{9x^2} - 9x^2 \cdot \frac{2}{9} \)
\( y = 5x^4 - 4 - 2x^2 \)

So \( y = 5x^4 - 2x^2 - 4 \).

Step 4: Find where \( y \) is strictly increasing.

Differentiate:
\( \frac{dy}{dx} = 20x^3 - 4x = 4x(5x^2 - 1) \)

Set \( \frac{dy}{dx} = 0 \):
\( 4x(5x^2 - 1) = 0 \) ⇒ \( x = 0 \) or \( x = \pm \frac{1}{\sqrt{5}} \)

Step 5: Analyze sign of \( \frac{dy}{dx} \).

\( \frac{dy}{dx} = 4x(5x^2 - 1) \)

For \( x < -\frac{1}{\sqrt{5}} \):
\( x < 0 \), \( 5x^2 - 1 > 0 \) ⇒ \( \frac{dy}{dx} = (-)(+) = - \) (decreasing)

For \( -\frac{1}{\sqrt{5}} < x < 0 \):
\( x < 0 \), \( 5x^2 - 1 < 0 \) ⇒ \( \frac{dy}{dx} = (-)(-) = + \) (increasing)

For \( 0 < x < \frac{1}{\sqrt{5}} \):
\( x > 0 \), \( 5x^2 - 1 < 0 \) ⇒ \( \frac{dy}{dx} = (+)(-) = - \) (decreasing)

For \( x > \frac{1}{\sqrt{5}} \):
\( x > 0 \), \( 5x^2 - 1 > 0 \) ⇒ \( \frac{dy}{dx} = (+)(+) = + \) (increasing)

Step 6: State intervals of strict increase.

\( y \) is strictly increasing on \( \left(-\frac{1}{\sqrt{5}}, 0\right) \) and \( \left(\frac{1}{\sqrt{5}}, \infty\right) \).

Therefore, \( y \) is strictly increasing in \( \mathbf{\left(-\frac{1}{\sqrt{5}}, 0\right) \cup \left(\frac{1}{\sqrt{5}}, \infty\right)} \).