Unit 30 Exam Practice
Maclaurin Series
These questions are written in the style of IB examination questions, and they are harder than the review exercises. The marks for each part are shown in square brackets on the right, for example [3]. Paper 1 means no calculator; Paper 2 means a graphic display calculator (GDC) is allowed. Try each question under exam conditions first, then open the markscheme.
This whole unit is HL only AHL 5.19, so there is no SL set.
| Mark | Meaning |
|---|---|
| M | Method mark: a correct method is used, even with a small slip |
| A | Accuracy mark: a correct answer or line of working (usually needs the M mark before it) |
| R | Reasoning mark: a clear reason or justification is given |
| N | Mark for a correct final answer with no working shown |
| AG | “Answer given” in the question, so no extra mark is awarded for writing it |
The number after the letter is the number of marks, so A2 is worth 2 marks. In the exam, give answers exactly or correct to 3 significant figures (s.f.) unless the question says otherwise.
HL questions
These questions use the standard Maclaurin series, repeated differentiation, intervals of convergence, and new series built by substitution, multiplication and term-by-term integration, including limits and sums of series.
HL 1 Paper 1
It is given that the limit \[ \lim_{x \to 0} \frac{\mathrm{e}^{2x}\sin x - \ln(1 + ax) - bx^{2}}{x^{3}} \] exists, where \(a\) and \(b\) are constants.
Use Maclaurin series to find the value of \(a\), the value of \(b\) and the value of the limit. [8]
Total 8 marks
\(\mathrm{e}^{2x} = 1 + 2x + 2x^{2} + \dfrac{4}{3}x^{3} + \cdots\) A1
\(\mathrm{e}^{2x}\sin x = \left(1 + 2x + 2x^{2} + \cdots\right)\left(x - \dfrac{x^{3}}{6} + \cdots\right)\) M1
\(= x + 2x^{2} + \dfrac{11}{6}x^{3} + \cdots\) A1
\(\ln(1 + ax) = ax - \dfrac{a^{2}x^{2}}{2} + \dfrac{a^{3}x^{3}}{3} - \cdots\) A1
numerator \(= (1 - a)x + \left(2 + \dfrac{a^{2}}{2} - b\right)x^{2} + \cdots\) M1
the limit exists only if the coefficients of \(x\) and \(x^{2}\) are both \(0\) (otherwise the quotient is unbounded as \(x \to 0\)) R1
\(a = 1\) and \(b = \dfrac{5}{2}\) A1
limit \(=\) coefficient of \(x^{3}\) \(= \dfrac{11}{6} - \dfrac{a^{3}}{3} = \dfrac{11}{6} - \dfrac{1}{3} = \dfrac{3}{2}\) A1
[8 marks]
See 30.3.
HL 2 Paper 1
Let \(y = \ln(1 + \sin x)\), for \(-\dfrac{\pi}{2} < x < \dfrac{\pi}{2}\).
(a) Show that \(\mathrm{e}^{y}\dfrac{\mathrm{d}y}{\mathrm{d}x} = \cos x\). [2]
(b) By differentiating this result twice, find the Maclaurin series for \(y\) up to and including the term in \(x^{3}\). [5]
Total 7 marks
(a)
\(\mathrm{e}^{y} = 1 + \sin x\) M1
differentiating with respect to \(x\): \(\mathrm{e}^{y}\dfrac{\mathrm{d}y}{\mathrm{d}x} = \cos x\) A1 AG
[2 marks]
(b)
\(\mathrm{e}^{y}\left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)^{2} + \mathrm{e}^{y}\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} = -\sin x\) M1 A1
\(\mathrm{e}^{y}\left(y'^{3} + 3y'y'' + y'''\right) = -\cos x\) A1
at \(x = 0\): \(y = 0\), \(y' = 1\), \(y'' = -1\), \(y''' = -1 - 1 + 3 = 1\) A1
\(y = x - \dfrac{x^{2}}{2} + \dfrac{x^{3}}{6} + \cdots\) A1
Award M1 for an attempt at implicit differentiation using the product rule and the chain rule on \(\mathrm{e}^{y}\).
[5 marks]
See 30.1.
HL 3 Paper 1
Let \(f(x) = \ln\left(\dfrac{1 + 3x}{1 - 2x}\right)\), for \(-\dfrac{1}{3} < x < \dfrac{1}{2}\).
(a) Find the Maclaurin series for \(f(x)\) up to and including the term in \(x^{3}\). [3]
(b) Show that the coefficient of \(x^{n}\) in this series is \(\dfrac{2^{n} - (-3)^{n}}{n}\). [2]
(c) Find the set of values of \(x\) for which the series is valid. [3]
Total 8 marks
(a)
\(f(x) = \ln(1 + 3x) - \ln(1 - 2x)\) M1
\(\ln(1 + 3x) = 3x - \dfrac{9x^{2}}{2} + 9x^{3} - \cdots\) and \(\ln(1 - 2x) = -2x - 2x^{2} - \dfrac{8x^{3}}{3} - \cdots\) A1
\(f(x) = 5x - \dfrac{5}{2}x^{2} + \dfrac{35}{3}x^{3} + \cdots\) A1
[3 marks]
(b)
coefficient of \(x^{n}\) in \(\ln(1 + 3x)\): \(\dfrac{(-1)^{n-1}3^{n}}{n} = -\dfrac{(-3)^{n}}{n}\) M1
coefficient of \(x^{n}\) in \(-\ln(1 - 2x)\): \(\dfrac{2^{n}}{n}\) A1
adding: \(\dfrac{2^{n} - (-3)^{n}}{n}\) AG
[2 marks]
(c)
\(\ln(1 + 3x)\): \(-1 < 3x \leqslant 1\), so \(-\dfrac{1}{3} < x \leqslant \dfrac{1}{3}\) A1
\(\ln(1 - 2x)\): \(-1 < -2x \leqslant 1\), so \(-\dfrac{1}{2} \leqslant x < \dfrac{1}{2}\) A1
both series must be valid, so \(-\dfrac{1}{3} < x \leqslant \dfrac{1}{3}\) A1
Award A1A1A0 for \(|x| < \dfrac{1}{3}\) with both intervals correct.
[3 marks]
HL 4 Paper 2
Let \(g(x) = \dfrac{\ln(1 + x^{2})}{x}\) for \(x \neq 0\), and \(g(0) = 0\).
(a) Show that \[ \ln(1 + x^{2}) = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}x^{2n}}{n}, \] and find the set of values of \(x\) for which this series is valid. [4]
(b) Hence write down a series for \(g(x)\) in ascending powers of \(x\), up to and including the term in \(x^{5}\). [1]
(c) Use your answer to part (b) to show that \(\displaystyle\int_{0}^{\frac{1}{2}} g(x)\,\mathrm{d}x \approx \frac{17}{144}\). [4]
(d) Use your GDC to find the value of \(\displaystyle\int_{0}^{\frac{1}{2}} g(x)\,\mathrm{d}x\). Hence find the percentage error in the approximation \(\dfrac{17}{144}\). [3]
(e) Without further calculation, explain why the method of part (c) cannot be used to estimate \(\displaystyle\int_{0}^{2} g(x)\,\mathrm{d}x\). [2]
Total 14 marks
(a)
replace \(x\) by \(x^{2}\) in \(\ln(1 + x) = x - \dfrac{x^{2}}{2} + \dfrac{x^{3}}{3} - \cdots\) M1
\(\ln(1 + x^{2}) = x^{2} - \dfrac{x^{4}}{2} + \dfrac{x^{6}}{3} - \cdots = \displaystyle\sum_{n=1}^{\infty} \dfrac{(-1)^{n+1}x^{2n}}{n}\) A1 AG
valid for \(-1 < x^{2} \leqslant 1\) M1
\(x^{2} > -1\) for all \(x\), so \(-1 \leqslant x \leqslant 1\) A1
Do not award the final A1 for \(-1 < x < 1\).
[4 marks]
(b)
\(g(x) = x - \dfrac{x^{3}}{2} + \dfrac{x^{5}}{3} - \cdots\) A1
[1 mark]
(c)
\(\displaystyle\int_{0}^{\frac{1}{2}} \left(x - \dfrac{x^{3}}{2} + \dfrac{x^{5}}{3}\right)\mathrm{d}x\) M1
\(= \left[\dfrac{x^{2}}{2} - \dfrac{x^{4}}{8} + \dfrac{x^{6}}{18}\right]_{0}^{\frac{1}{2}}\) A1
\(= \dfrac{1}{8} - \dfrac{1}{128} + \dfrac{1}{1152}\) M1
\(= \dfrac{144 - 9 + 1}{1152} = \dfrac{136}{1152} = \dfrac{17}{144}\) A1 AG
[4 marks]
(d)
\(\displaystyle\int_{0}^{\frac{1}{2}} g(x)\,\mathrm{d}x = 0.117950\ldots\) (using the Integral feature on a GDC) A1
percentage error \(= \dfrac{\left|\frac{17}{144} - 0.117950\ldots\right|}{0.117950\ldots} \times 100\) M1
\(= 0.0894\%\) \((0.089365\ldots)\) A1
Award M1 for dividing by the GDC value. Do not award the final A1 for an answer found from the GDC value rounded to 3 s.f.
[3 marks]
(e)
the series for \(\ln(1 + x^{2})\) (and so for \(g(x)\)) is only valid for \(-1 \leqslant x \leqslant 1\) R1
the interval \(0 \leqslant x \leqslant 2\) contains values \(x > 1\), where the series does not converge (its terms get larger) R1
[2 marks]
HL 5 Paper 1
Let \(f(x) = x\arctan x - \dfrac{1}{2}\ln(1 + x^{2})\), for \(x \in \mathbb{R}\).
(a) Show that \(f'(x) = \arctan x\). [3]
(b) Hence, by integrating the Maclaurin series for \(\arctan x\) term by term, show that \[ f(x) = \sum_{n=0}^{\infty} \frac{(-1)^{n}x^{2n+2}}{(2n + 1)(2n + 2)} \] for \(-1 \leqslant x \leqslant 1\). [4]
(c) Use the series in part (b) to find \(\displaystyle\lim_{x \to 0} \frac{2f(x) - x\sin x}{x^{6}}\). [4]
(d) Use the series in part (b) to find the exact value of \(\displaystyle\sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2n + 1)(2n + 2)\,3^{n}}\). [5]
Total 16 marks
(a)
\(\dfrac{\mathrm{d}}{\mathrm{d}x}\left(x\arctan x\right) = \arctan x + \dfrac{x}{1 + x^{2}}\) M1 A1
\(\dfrac{\mathrm{d}}{\mathrm{d}x}\left(\dfrac{1}{2}\ln(1 + x^{2})\right) = \dfrac{x}{1 + x^{2}}\) A1
\(f'(x) = \arctan x + \dfrac{x}{1 + x^{2}} - \dfrac{x}{1 + x^{2}} = \arctan x\) AG
[3 marks]
(b)
\(\arctan t = t - \dfrac{t^{3}}{3} + \dfrac{t^{5}}{5} - \cdots = \displaystyle\sum_{n=0}^{\infty} \dfrac{(-1)^{n}t^{2n+1}}{2n + 1}\) A1
\(f(x) = f(0) + \displaystyle\int_{0}^{x} \arctan t\,\mathrm{d}t\) M1
\(f(0) = 0 - \dfrac{1}{2}\ln 1 = 0\) A1
\(\displaystyle\int_{0}^{x} t^{2n+1}\,\mathrm{d}t = \dfrac{x^{2n+2}}{2n + 2}\), so \(f(x) = \displaystyle\sum_{n=0}^{\infty} \dfrac{(-1)^{n}x^{2n+2}}{(2n + 1)(2n + 2)}\) A1 AG
[4 marks]
(c)
\(2f(x) = x^{2} - \dfrac{x^{4}}{6} + \dfrac{x^{6}}{15} - \cdots\) A1
\(x\sin x = x^{2} - \dfrac{x^{4}}{6} + \dfrac{x^{6}}{120} - \cdots\) A1
\(2f(x) - x\sin x = \left(\dfrac{1}{15} - \dfrac{1}{120}\right)x^{6} + \cdots = \dfrac{7}{120}x^{6} + \cdots\) M1
limit \(= \dfrac{7}{120}\) A1
[4 marks]
(d)
put \(x = \dfrac{1}{\sqrt{3}}\), which is in \(-1 \leqslant x \leqslant 1\) R1
\(x^{2n+2} = \left(\dfrac{1}{3}\right)^{n+1}\), so \(f\left(\dfrac{1}{\sqrt{3}}\right) = \dfrac{1}{3}\displaystyle\sum_{n=0}^{\infty} \dfrac{(-1)^{n}}{(2n + 1)(2n + 2)\,3^{n}}\) M1 A1
\(f\left(\dfrac{1}{\sqrt{3}}\right) = \dfrac{1}{\sqrt{3}} \times \dfrac{\pi}{6} - \dfrac{1}{2}\ln\dfrac{4}{3}\) A1
sum \(= 3f\left(\dfrac{1}{\sqrt{3}}\right) = \dfrac{\sqrt{3}\,\pi}{6} - \dfrac{3}{2}\ln\dfrac{4}{3}\) A1
Accept equivalent forms, for example \(\dfrac{\pi}{2\sqrt{3}} + 3\ln\dfrac{\sqrt{3}}{2}\).
[5 marks]