Question 8

First, solve the linear differential equation $x \frac{dy}{dx} - y = -x^3$ by rewriting it as $\frac{dy}{dx} - \frac{1}{x} y = -x^2$. The integrating factor is $IF = e^{\int -\frac{1}{x} dx} = e^{-\ln x} = \frac{1}{x}$. Multiplying the equation by the IF gives $\frac{d}{dx}(\frac{y}{x}) = -x$. Integrating both sides results in $\frac{y}{x} = -\frac{x^2}{2} + C$. Given $f(1) = 0$, we have $0 = -\frac{1}{2} + C \implies C = \frac{1}{2}$. Thus, $f(x) = \frac{x}{2} - \frac{x^3}{2}$.

For local extrema, find $f'(x) = \frac{1}{2} - \frac{3x^2}{2} = \frac{1}{2}(1 - 3x^2)$. Setting $f'(x) = 0$ gives $x = \frac{1}{\sqrt{3}}$ (as $x > 0$). $f''(x) = -3x$, and at $x = \frac{1}{\sqrt{3}}$, $f''(x) < 0$, so $f$ has a local maximum at $x = \frac{1}{\sqrt{3}}$. Thus, (B) is true and (A) is false.

In the interval $(1, 2)$, $x > 1 \implies 3x^2 > 3 \implies 1 - 3x^2 < -2$. Since $f'(x) < 0$ on $(1, 2)$, the function is decreasing. Thus, (C) is false.

To check (D), set $f(x) = g(x)$: $\frac{x - x^3}{2} = 4x^3 - 5x^2 + \frac{3}{2}x$. Since $x > 0$, we divide by $x$ and simplify: $1 - x^2 = 8x^2 - 10x + 3 \implies 9x^2 - 10x + 2 = 0$. The discriminant $D = (-10)^2 - 4(9)(2) = 28 > 0$. The roots $x = \frac{10 \pm \sqrt{28}}{18}$ are both positive. Thus, there are exactly 2 elements in the set. (D) is true.