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1. [Maximum mark: 28]

The following question explores features of a family of curves. The family is then linked to a homogeneous differential equation.

Consider the curve given by $y = \frac{x(x^2 - 16)}{x^2 + 16}$.

(a) (i) Sketch the curve of $y$ for $-10 \le x \le 10$.

(ii) State the coordinates of the points where the curve crosses the $x$-axis.

(iii) State the coordinates of the local maximum point and the coordinates of the local minimum point.

(b) State whether the function $f(x) = \frac{x(x^2 - 16)}{x^2 + 16}$ is odd, even or neither. Justify your answer.

Now consider the general curve given by $y = \frac{x(x^2 - A)}{x^2 + A}$, where $A$ is a positive constant and $x \in \mathbb{R}$.

(c) Given $f(x) = \frac{x(x^2 - A)}{x^2 + A}$, prove that $f'(\sqrt{A})$ is independent of $A$.

(d) (i) Show that $x - \frac{2Ax}{x^2 + A} \equiv \frac{x(x^2 - A)}{x^2 + A}$.

(ii) Hence, determine the equation of the oblique asymptote to the curve.

(iii) Write down the coordinates of a point on the curve where the oblique asymptote is parallel to the tangent to the curve at that point.

Now consider the differential equation $x^2 \frac{dy}{dx} = x(x + y) - y^2$, where $x \neq 0, y \neq \pm x$.

Using the substitution $y = vx$, the differential equation can be written as $x \frac{dv}{dx} = 1 - v^2$.

(e) Using partial fractions, show that $\int \frac{1}{1 - v^2} dv = \frac{1}{2} \ln \left| \frac{A(1 + v)}{1 - v} \right|$, where $A$ is a positive constant.

(f) Hence, show that a solution to the original differential equation may be expressed in the form $x^2 = \left| \frac{A(x + y)}{x - y} \right|$, where $A$ is a positive constant.

Now consider only the case where $\frac{A(x + y)}{x - y} > 0$.

(g) Show that a solution to the original differential equation is $y = \frac{x(x^2 - A)}{x^2 + A}$.