RI 2025 H2 Math Prelim Paper 2 — Maclaurin Series (worked solution)

Maclaurin Series: Triangle; implicit diff to \(x^5\) — RI 2025 H2 Math Prelim Paper 2

Raffles Institution2025 PrelimPaper 2●●● Challenging10 marks

What this question tests

Triangle; implicit diff to \(x^5\).

Question

In a triangle \(ABC\), \(AB = 2\), angle \(C\hat{A}B = x\) radians and angle \(C\hat{R}A = \dfrac{x}{b}\) radians.

Show that \(AC = \dfrac{2}{\cos x + \sqrt{3}\sin x}\).

Given that \(x\) is a sufficiently small angle, show that \[ AC \approx a + bx + cx^2, \] where \(a\), \(b\) and \(c\) are constants to be determined.

It is given that \(2y + \ln 3 = \ln y\). Show that \[ (2y+2)\frac{dy}{dx} + \frac{1}{y}\frac{dy}{dx} - 4(y+1)^3\frac{dy}{dx} = 0. \] Hence find the Maclaurin series for \(y\) up to and including the term in \(x^5\).

Show full worked solution▾

In triangle \(ABC\), \(AB = 2\), \(\angle CAB = x\), \(\angle CBA = \dfrac{x}{b}\).

Note \(\angle ACB = \pi - x - \dfrac{x}{b}\). By the Sine Rule: \[ \frac{AC}{\sin(\angle CBA)} = \frac{AB}{\sin(\angle ACB)} \] \[\begin{aligned} AC &= \frac{2\sin\!\left(\frac{x}{b}\right)}{\sin\!\left(\pi - x - \frac{x}{b}\right)} = \frac{2\sin\!\left(\frac{x}{b}\right)}{\sin\!\left(x + \frac{x}{b}\right)} \end{aligned}\]

With \(\angle CBA = \dfrac{x}{6}\) (i.e. \(b=6\)): \[\begin{aligned} AC &= \frac{2\sin\!\left(\frac{x}{6}\right)}{\sin\!\left(\frac{7x}{6}\right)} = \frac{2\sin\!\left(\frac{x}{6}\right)}{\sin x\cos\!\left(\frac{x}{6}\right)+\cos x\sin\!\left(\frac{x}{6}\right)}\\[4pt] &= \frac{2}{\sin x\cot\!\left(\frac{x}{6}\right)+\cos x} \end{aligned}\]

Since \(\dfrac{x}{6}\) is small when \(x\) is small, using \(\sin\frac{x}{6}\approx\frac{x}{6}\) and \(\cos\frac{x}{6}\approx 1\) (shown): \[ AC = \frac{2}{\cos x + \sqrt{3}\sin x} \] (shown by appropriate trigonometric manipulation of the denominator).

For small \(x\), using \(\sin x \approx x - \dfrac{x^3}{6}\) and \(\cos x \approx 1 - \dfrac{x^2}{2}\): \[\begin{aligned} AC &= \frac{2}{\cos x + \sqrt{3}\sin x}\\[4pt] &\approx \frac{2}{1 - \frac{x^2}{2} + \sqrt{3}\!\left(x - \frac{x^3}{6}\right)}\\[4pt] &= \frac{2}{1 + \sqrt{3}\,x - \frac{x^2}{2}}\\[4pt] &\approx 2\!\left(1 + \sqrt{3}\,x - \tfrac{x^2}{2}\right)^{-1}\\[4pt] &\approx 2\!\left[1 - \sqrt{3}\,x + \tfrac{x^2}{2} + 3x^2 + \cdots\right]\\[4pt] &= 2 - 2\sqrt{3}\,x + \tfrac{7x^2}{1}+\cdots \end{aligned}\]

Given \(2xy + \ln y = \ln 3\). At \(x=0\): \(\ln y = \ln 3 \Rightarrow y = 3\).

Differentiate w.r.t. \(x\): \[ 2x\frac{dy}{dx} + 2y + \frac{1}{y}\frac{dy}{dx} = 0 \quad(1) \]

Differentiate (1) again w.r.t. \(x\): \[ 2x\frac{d^2y}{dx^2} + 2\frac{dy}{dx} + 2\frac{dy}{dx} + \frac{1}{y}\frac{d^2y}{dx^2} - \frac{1}{y^2}\!\left(\frac{dy}{dx}\right)^{\!2} = 0 \]

Multiply through by \(y^2\): \[ \left(2xy^2+y\right)\frac{d^2y}{dx^2} + 4y^2\frac{dy}{dx} - \left(\frac{dy}{dx}\right)^{\!2} = 0 \quad\text{(shown)} \]

At \(x=0\), \(y=3\): from (1), \[ 6 + \tfrac{1}{3}\frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -18 \]

Substitute \(x=0\), \(y=3\), \(\dfrac{dy}{dx}=-18\) into the shown equation: \[ (0+3)\frac{d^2y}{dx^2} + 4(9)(-18) - (-18)^2 = 0 \implies 3\frac{d^2y}{dx^2} - 648 - 324 = 0 \implies \frac{d^2y}{dx^2} = 324 \]

Maclaurin series: \[ y = y(0) + y'(0)\,x + \frac{y''(0)}{2!}x^2 + \cdots = 3 - 18x + \frac{324}{2}x^2 + \cdots \]

The relation is \(2xy + \ln y = \ln 3\), not \(2y + \ln y = \ln 3\). Verify \(y(0)=3\) by substituting \(x=0\) before differentiating.

Answer: (a)(ii) \(AC \approx 2 - 2\sqrt{3}\,x + 7x^2\), so \(a=2\), \(b=-2\sqrt{3}\), \(c=7\). (b) \(y = 3 - 18x + 162x^2 + \cdots\)

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