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1995 Paper 1 Q1
- Find the real values of \(x\) for which \[ x^{3}-4x^{2}-x+4\geqslant0. \]
- Find the three lines in the \((x,y)\) plane on which \[ x^{3}-4x^{2}y-xy^{2}+4y^{3}=0. \]
- On a sketch shade the regions of the \((x,y)\) plane for which \[ x^{3}-4x^{2}y-xy^{2}+4y^{3}\geqslant0. \]
Solution:
- \(\,\) \begin{align*} && 0 & \leq x^3 - 4x^2 - x + 4 \\ &&&= (x-1)(x^2-3x-4) \\ &&&= (x-1)(x-4)(x+1) \\ \Leftrightarrow && x &\in [-1, 1] \cup [4, \infty) \end{align*}
- \(\,\) \begin{align*} && 0 &= x^{3}-4x^{2}y-xy^{2}+4y^{3} \\ && 0 &= (x-y)(x-4y)(x+y) \end{align*} Therefore the lines are \(y = x, 4y = x, y=-x\).
- (quickest way to see this is to check the \(x\) or \(y\)-axis)
1987 Paper 1 Q3
By substituting \(y(x)=xv(x)\) in the differential equation \[ x^{3}\frac{\mathrm{d}v}{\mathrm{d}x}+x^{2}v=\frac{1+x^{2}v^{2}}{\left(1+x^{2}\right)v}, \] or otherwise, find the solution \(v(x)\) that satisfies \(v=1\) when \(x=1\). What value does this solution approach when \(x\) becomes large?
Solution: Let \(y = xv\) then \(y' = v + xv'\) and so \(x^2y' = x^2v + x^3v'\) Our differential equation is now: \begin{align*} && x^2 y' &= \frac{1+y^2}{(1+x^2)\frac{y}{x}} \\ \Rightarrow && xy' &= \frac{(1+y^2)}{(1+x^2)y} \\ \Rightarrow && \frac{y}{1+y^2} \frac{\d y}{\d x} &= \frac{1}{x(1+x^2)} \\ \Rightarrow && \frac{y}{1+y^2} \frac{\d y}{\d x} &= \frac{1}{x} - \frac{x}{1+x^2} \\ \Rightarrow && \frac12 \ln(1+y^2) &= \ln x - \frac12 \ln(1+x^2) + C \\ \Rightarrow && \frac12 \ln (1 + y^2) &= \frac12 \ln \l \frac{x^2}{1+x^2}\r + C \\ \Rightarrow && 1+y^2 &= \frac{Dx^2}{1+x^2} \\ \Rightarrow && D &= 4 \quad \quad: (x = 1, v = 1, y = 1) \\ \Rightarrow && 1 + x^2v^2&= \frac{4x^2}{1+x^2}\\ \Rightarrow && v^2 &= \frac{3x^2-1}{x^2(1+x^2)} \\ \Rightarrow && v &= \sqrt{\frac{3x^2-1}{x^2(1+x^2)}} \\ \end{align*} As \(x \to \infty\), \(v \to 0\)