9. Given that \( y=x^{2} \sqrt{9-2 x^{2}} \), (i) state the range of values of \( x \) for \( y \) to be valid, (ii) show that \( y \frac{\mathrm{~d} y}{\mathrm{~d} x}+6 x^{3}\left(x^{2}-3\right)=0 \).

Let's address each part of the problem step by step. ### Given: \[ y = x^{2} \sqrt{9 - 2x^{2}} \] --- ### Part (i): Determine the Range of \( x \) for \( y \) to Be Valid For \( y \) to be a real number, the expression under the square root must be non-negative: \[ 9 - 2x^{2} \geq 0 \] **Solve for \( x \):** \[ 9 \geq 2x^{2} \\ \frac{9}{2} \geq x^{2} \\ x^{2} \leq \frac{9}{2} \] Taking the square root of both sides: \[ |x| \leq \frac{3}{\sqrt{2}} \] **Conclusion:** \[ x \in \left[ -\frac{3}{\sqrt{2}}, \frac{3}{\sqrt{2}} \right] \] So, the range of \( x \) for which \( y \) is real and valid is all real numbers \( x \) such that \( |x| \leq \frac{3}{\sqrt{2}} \). --- ### Part (ii): Show That \( y \frac{dy}{dx} + 6x^{3}(x^{2} - 3) = 0 \) **Step 1: Differentiate \( y \) with Respect to \( x \).** Given: \[ y = x^{2} \sqrt{9 - 2x^{2}} \] Let’s compute \( \frac{dy}{dx} \) using the product rule: \[ \frac{dy}{dx} = \frac{d}{dx}\left( x^{2} \right) \cdot \sqrt{9 - 2x^{2}} + x^{2} \cdot \frac{d}{dx}\left( \sqrt{9 - 2x^{2}} \right) \] Compute each derivative: \[ \frac{d}{dx}\left( x^{2} \right) = 2x \] \[ \frac{d}{dx}\left( \sqrt{9 - 2x^{2}} \right) = \frac{1}{2} (9 - 2x^{2})^{-1/2} \cdot (-4x) = -\frac{2x}{\sqrt{9 - 2x^{2}}} \] Substitute back: \[ \frac{dy}{dx} = 2x \sqrt{9 - 2x^{2}} + x^{2} \left( -\frac{2x}{\sqrt{9 - 2x^{2}}} \right) \\ = 2x \sqrt{9 - 2x^{2}} - \frac{2x^{3}}{\sqrt{9 - 2x^{2}}} \] **Step 2: Compute \( y \frac{dy}{dx} \).** Multiply \( y \) by \( \frac{dy}{dx} \): \[ y \frac{dy}{dx} = x^{2} \sqrt{9 - 2x^{2}} \left( 2x \sqrt{9 - 2x^{2}} - \frac{2x^{3}}{\sqrt{9 - 2x^{2}}} \right) \] Simplify each term: \[ = x^{2} \left( 2x (9 - 2x^{2}) \right) - x^{2} \left( \frac{2x^{3}}{1} \right) \\ = 2x^{3} (9 - 2x^{2}) - 2x^{5} \\ = 18x^{3} - 4x^{5} - 2x^{5} \\ = 18x^{3} - 6x^{5} \\ = 6x^{3} (3 - x^{2}) \\ = -6x^{3} (x^{2} - 3) \] **Step 3: Verify the Given Equation.** We have: \[ y \frac{dy}{dx} = -6x^{3} (x^{2} - 3) \] Adding \( 6x^{3}(x^{2} - 3) \) to both sides: \[ y \frac{dy}{dx} + 6x^{3} (x^{2} - 3) = 0 \] **Conclusion:** \[ y \frac{dy}{dx} + 6x^{3}(x^{2} - 3) = 0 \quad \text{is verified.} \] --- ### Final Answer: **(i)** All real numbers \( x \) with absolute value at most \( \dfrac{3}{\sqrt{2}} \); that is, \( |x| \le \dfrac{3}{\sqrt{2}} \). **(ii)** After expanding, \( y\, dy/dx + 6x^{3}(x^{2} -3) = 0 \) holds.