Maclaurin Series: \(e^x\sin(x+\pi)\); binomial match — DHS 2025 H2 Math Prelim Paper 1
What this question tests
Question
- Find the first three non-zero terms in the Maclaurin series for \(e^x\sin(x+\pi)\).
- It is given that the three terms found in part (a) are equal to the first three terms in the series expansion of \(ax(1+bx)^c\) for small \(x\), where \(a\), \(b\) and \(c\) are constants. Find the exact values of \(a\), \(b\) and \(c\). Use these values to find the coefficient of \(x^4\) in the expansion of \(ax(1+bx)^c\), giving your answer as a simplified rational number.
Show full worked solution▾
(a) Using \(\sin(x+\pi) = -\sin x\): \[\begin{aligned} e^x\sin(x+\pi) &= e^x(-\sin x) = -e^x\sin x \\ &= -\!\left(1 + x + \frac{x^2}{2!} + \cdots\right)\!\left(x - \frac{x^3}{3!} + \cdots\right) \\ &= -\!\left(x - \frac{x^3}{6} + x^2 + \cdots\right) \\ &= -x - x^2 - \frac{x^3}{3} + \cdots \end{aligned}\]
\(e^x\sin(x+\pi) = -x - x^2 - \dfrac{x^3}{3} + \cdots\)(b) Expanding \(ax(1+bx)^c\): \[\begin{aligned} ax(1+bx)^c &= ax\!\left(1 + cbx + \frac{c(c-1)}{2!}(bx)^2 + \cdots\right) \\ &= ax + abcx^2 + \frac{ab^2c(c-1)}{2}x^3 + \cdots \end{aligned}\]
Comparing with \(-x - x^2 - \dfrac{x^3}{3} + \cdots\): \[ a = -1, \quad abc = -1, \quad \frac{ab^2c(c-1)}{2} = -\frac{1}{3} \]
From \(abc = -1\): \(bc = 1 \Rightarrow b = \dfrac{1}{c}\).
From the third equation with \(a=-1\), \(b=\frac{1}{c}\): \[ \frac{(-1)\cdot\frac{1}{c^2}\cdot c(c-1)}{2} = -\frac{1}{3} \implies \frac{c-1}{2c} = \frac{1}{3} \implies 3c-3 = 2c \implies c = 3 \]
Therefore \(b = \dfrac{1}{3}\).
\(a = -1\), \(b = \dfrac{1}{3}\), \(c = 3\)
So \(ax(1+bx)^c = -x\!\left(1+\dfrac{1}{3}x\right)^3\).
The \(x^4\) term in \(-x(1+\frac{1}{3}x)^3\): we need the \(x^3\) term in \(-(1+\frac{1}{3}x)^3\): \[ -x \cdot \binom{3}{3}\!\left(\frac{x}{3}\right)^3 = -x \cdot \frac{x^3}{27} = -\frac{x^4}{27} \]