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

Given the equation:

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

Solving for \(f(x)\) yields:

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

Substituting \(f(x)\) into the equation for \(y\):

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

Simplifying the expression for \(y\):

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

Differentiating \(y\) with respect to \(x\):

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

Further simplification yields:

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

For \(y\) to be strictly increasing, the condition \(\frac{dy}{dx} > 0\) must hold, which implies:

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

The critical points for the inequality \(x^3 - x - 1 > 0\) are:

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

Consequently, \(y\) is strictly increasing on the intervals:

\[ x \in \left(-\frac{1}{\sqrt{5}}, 0\right) \cup \left(\frac{1}{\sqrt{5}}, \infty\right) \]