Maclaurin Series: \(\ln(a+x)\); composition; integral — VJC 2025 H2 Math Prelim Paper 2
What this question tests
Question
It is given that \(\mathrm{f}(x) = \ln(a + x)\), \(x \in \mathbb{R}\), \(x > -a\), where \(a\) is a constant.
- (a) Using the standard series from the List of Formulae (MF27), find the series expansion for \(\mathrm{f}(x)\), up to and including the term in \(x^3\).
- (b) Hence, or otherwise, show that the series expansion of \(\sin\!\left[\mathrm{f}(x)\right]\), up to and including the term in \(x^3\), is given by \(x - \dfrac{1}{2}x^2 + \dfrac{1}{6}x^3\).
- (c) Deduce the Maclaurin series for \(\cos\!\left[\mathrm{f}(x)\right]\) up to and including the term in \(x^2\).
- (d) Find \(\displaystyle\int_{1}^{3}\!\left(x - \tfrac{1}{2}x^2 + \tfrac{1}{6}x^3\right) \mathrm{d}x\). Without the use of a calculator or any further calculation, explain whether this value is a good approximation to the value of \(\displaystyle\int_{1}^{3}\sin\!\left[\mathrm{f}(x)\right]\mathrm{d}x\).
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(a) Using \(\ln(1+u) = u - \dfrac{u^2}{2} + \dfrac{u^3}{3} - \cdots\) with \(u = \dfrac{x}{a}\): \[\begin{aligned} \ln(a + x) &= \ln\!\left[a\!\left(1 + \dfrac{x}{a}\right)\right]\\ &= \ln a + \ln\!\left(1 + \dfrac{x}{a}\right)\\ &= \ln a + \left[\dfrac{x}{a} - \dfrac{1}{2}\!\left(\dfrac{x}{a}\right)^{\!2} + \dfrac{1}{3}\!\left(\dfrac{x}{a}\right)^{\!3} + \cdots\right]\\ &= \ln a + \dfrac{x}{a} - \dfrac{x^2}{2a^2} + \dfrac{x^3}{3a^3} + \cdots \end{aligned}\]
(b) With \(a = 1\): \(\mathrm{f}(x) = \ln(1+x) = x - \dfrac{x^2}{2} + \dfrac{x^3}{3} + \cdots\)
Using \(\sin u = u - \dfrac{u^3}{3!} + \cdots\) and keeping terms up to \(x^3\): \[\begin{aligned} \sin\!\left[\ln(1+x)\right] &= \left(x - \dfrac{x^2}{2} + \dfrac{x^3}{3} + \cdots\right) - \dfrac{1}{6}\!\left(x - \dfrac{x^2}{2} + \cdots\right)^{\!3} + \cdots\\ &= x - \dfrac{x^2}{2} + \dfrac{x^3}{3} - \dfrac{x^3}{6} + \cdots\\ &= x - \dfrac{x^2}{2} + \dfrac{x^3}{6} + \cdots \quad \text{(shown)} \end{aligned}\]
(c) Differentiating both sides of \(\sin\!\left[\ln(1+x)\right] = x - \dfrac{1}{2}x^2 + \dfrac{1}{6}x^3 + \cdots\) with respect to \(x\): \[\begin{aligned} \dfrac{1}{1+x}\cos\!\left[\ln(1+x)\right] &= 1 - x + \dfrac{1}{2}x^2 + \cdots\\ \cos\!\left[\ln(1+x)\right] &= (1+x)\!\left(1 - x + \dfrac{1}{2}x^2 + \cdots\right)\\ &= 1 - \dfrac{1}{2}x^2 + \cdots \end{aligned}\]
(d) Computing the integral: \[\begin{aligned} \int_{1}^{3}\!\left(x - \dfrac{x^2}{2} + \dfrac{x^3}{6}\right)\mathrm{d}x &= \left[\dfrac{x^2}{2} - \dfrac{x^3}{6} + \dfrac{x^4}{24}\right]_{1}^{3}\\ &= \left(\dfrac{9}{2} - \dfrac{27}{6} + \dfrac{81}{24}\right) - \left(\dfrac{1}{2} - \dfrac{1}{6} + \dfrac{1}{24}\right)\\ &= 3 \end{aligned}\]
For \(x \in (1,3] \not\subseteq (-1,1]\), the series expansion \(\sin\!\left[\ln(1+x)\right] = \sin\!\left(x - \dfrac{x^2}{2} + \dfrac{x^3}{3} + \cdots\right)\) is not valid. Hence the value is not a good approximation.