Unit 4 Exam Practice
The Binomial Theorem
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.
| 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.
SL questions
1 Paper 1
In the expansion of \((1 + px)^{n}\), where \(n\) is a positive integer and \(p > 0\), the first three terms in ascending powers of \(x\) are \[ 1 + 12x + 63x^{2}. \]
(a) Find the value of \(n\) and the value of \(p\). [5]
(b) Find the coefficient of \(x^{3}\) in the expansion. [2]
Total 7 marks
(a)
\((1 + px)^{n} = 1 + npx + \dfrac{n(n - 1)}{2}p^{2}x^{2} + \cdots\) M1
\(np = 12\) and \(\dfrac{n(n - 1)}{2}p^{2} = 63\) A1
\(n(n - 1)p^{2} = 126\) and \(n^{2}p^{2} = 144\), so dividing gives \(\dfrac{n - 1}{n} = \dfrac{126}{144} = \dfrac{7}{8}\) M1
\(n = 8\) A1
\(p = \dfrac{12}{8} = \dfrac{3}{2}\) A1
[5 marks]
(b)
\(\dbinom{8}{3}\left(\dfrac{3}{2}\right)^{3} = 56 \times \dfrac{27}{8}\) M1
\(= 189\) A1
[2 marks]
See 4.1.
2 Paper 1
Let \(n\) be a positive integer, with \(n \geq 2\).
(a) Show that \(\dbinom{n}{2} + \dbinom{n + 1}{2} = n^{2}\). [3]
(b) The coefficient of \(x^{2}\) in the expansion of \((1 + x)^{n} + (1 + x)^{n + 1}\) is \(81\). Find the value of \(n\). [2]
(c) For this value of \(n\), find the coefficient of \(x^{3}\) in the expansion of \((1 + x)^{n} + (1 + x)^{n + 1}\). [2]
Total 7 marks
(a)
\(\dbinom{n}{2} + \dbinom{n + 1}{2} = \dfrac{n(n - 1)}{2} + \dfrac{(n + 1)n}{2}\) M1 A1
\(= \dfrac{n(n - 1 + n + 1)}{2} = \dfrac{2n^{2}}{2} = n^{2}\) A1 AG
[3 marks]
(b)
the coefficient of \(x^{2}\) is \(\dbinom{n}{2} + \dbinom{n + 1}{2} = n^{2}\), so \(n^{2} = 81\) M1
\(n = 9\) (since \(n > 0\)) A1
[2 marks]
(c)
\(\dbinom{9}{3} + \dbinom{10}{3} = 84 + 120\) M1
\(= 204\) A1
[2 marks]
See 4.1.
3 Paper 2
Consider the expansion of \((1 + kx)^{8}\), where \(k > 0\).
(a) Write down, in terms of \(k\), the coefficient of \(x^{2}\) and the coefficient of \(x^{3}\). [2]
The sum of the coefficients of \(x^{2}\) and \(x^{3}\) is \(100\).
(b) Find the value of \(k\). [3]
(c) Hence find the coefficient of \(x^{4}\). [2]
Total 7 marks
(a)
coefficient of \(x^{2}\): \(\dbinom{8}{2}k^{2} = 28k^{2}\) A1
coefficient of \(x^{3}\): \(\dbinom{8}{3}k^{3} = 56k^{3}\) A1
[2 marks]
(b)
\(28k^{2} + 56k^{3} = 100\) M1
\(k = 1.07\) \((1.06737\ldots)\) (for example, using the intersection of \(y = 28x^{2} + 56x^{3}\) and \(y = 100\) on a GDC) A2
[3 marks]
(c)
\(\dbinom{8}{4}k^{4} = 70k^{4}\) M1
\(= 90.9\) \((90.860\ldots)\) A1
Do not award the A1 for \(91.8\), which comes from using the rounded value \(k = 1.07\).
[2 marks]
See 4.1.
4 Paper 1
Let \(n\) be a positive integer, with \(n \geq 3\). Consider the expansion of \(\left(1 + \dfrac{x}{2}\right)^{n}\).
(a) Write down the first four terms of the expansion in ascending powers of \(x\). Give each coefficient in terms of \(n\), in simplest form. [3]
The coefficient of \(x^{3}\) is equal to the coefficient of \(x^{2}\).
(b) Show that \(n = 8\). [3]
(c) Find the term in \(x^{4}\). [2]
(d) By choosing a suitable value of \(x\), use the first four terms of the expansion to find an estimate for \(1.005^{8}\). [3]
(e) Find the coefficient of \(x^{3}\) in the expansion of \((2 - x)^{2}\left(1 + \dfrac{x}{2}\right)^{8}\). [4]
Total 15 marks
(a)
\(1 + \dfrac{n}{2}x\) A1
\(+ \dfrac{n(n - 1)}{8}x^{2}\) A1
\(+ \dfrac{n(n - 1)(n - 2)}{48}x^{3}\) A1
[3 marks]
(b)
\(\dfrac{n(n - 1)(n - 2)}{48} = \dfrac{n(n - 1)}{8}\) M1
\(n \geq 3\), so \(n(n - 1) \neq 0\) and we can divide by it: \(n - 2 = 6\) A1
\(n = 8\) A1 AG
[3 marks]
(c)
\(\dbinom{8}{4}\left(\dfrac{x}{2}\right)^{4} = 70 \times \dfrac{x^{4}}{16}\) M1
\(= \dfrac{35}{8}x^{4}\) A1
[2 marks]
(d)
\(1 + \dfrac{x}{2} = 1.005\), so \(x = 0.01\) M1
with \(n = 8\) the first four terms are \(1 + 4x + 7x^{2} + 7x^{3}\), so \(1.005^{8} \approx 1 + 4(0.01) + 7(0.01)^{2} + 7(0.01)^{3}\) A1
\(= 1.040\,707\) A1
[3 marks]
(e)
\((2 - x)^{2} = 4 - 4x + x^{2}\) A1
the terms in \(x^{3}\) come from \(4 \times 7x^{3}\), \(\ (-4x) \times 7x^{2}\) and \(x^{2} \times 4x\) M1
\(28 - 28 + 4\) A1
\(= 4\) A1
[4 marks]
See 4.1.
5 Paper 1
Let \(n\) be a positive integer. Consider the expansion of \(\left(x^{2} - \dfrac{2}{x}\right)^{n}\).
(a) Show that the general term of the expansion can be written as \(\dbinom{n}{r}(-2)^{r}x^{2n - 3r}\). [3]
(b) Explain why the expansion has a term independent of \(x\) only when \(n\) is a multiple of \(3\). [2]
Now let \(n = 6\).
(c) Find the term independent of \(x\). [2]
(d) Find the coefficient of \(x^{3}\). [2]
(e) The term independent of \(x\) in the expansion of \(\left(c + x^{3}\right)\left(x^{2} - \dfrac{2}{x}\right)^{6}\) is zero. Find the value of \(c\). [4]
Total 13 marks
(a)
\(T_{r+1} = \dbinom{n}{r}\left(x^{2}\right)^{n - r}\left(-\dfrac{2}{x}\right)^{r}\) M1
\(= \dbinom{n}{r}(-2)^{r}x^{2n - 2r}x^{-r}\) A1
\(= \dbinom{n}{r}(-2)^{r}x^{2n - 3r}\) A1 AG
[3 marks]
(b)
a term independent of \(x\) needs \(2n - 3r = 0\), so \(r = \dfrac{2n}{3}\) R1
\(r\) must be an integer, so \(3\) divides \(2n\); \(3\) does not divide \(2\), so \(3\) divides \(n\) (and then \(r = \dfrac{2n}{3}\) is between \(0\) and \(n\)) R1
[2 marks]
(c)
\(r = 4\): \(\dbinom{6}{4}(-2)^{4} = 15 \times 16\) M1
\(= 240\) A1
[2 marks]
(d)
\(12 - 3r = 3\), so \(r = 3\) M1
\(\dbinom{6}{3}(-2)^{3} = -160\) A1
[2 marks]
(e)
term independent of \(x\) \(= c \times 240 + 1 \times (\text{coefficient of } x^{-3})\) M1
\(12 - 3r = -3\) gives \(r = 5\): \(\dbinom{6}{5}(-2)^{5} = -192\) A1
\(240c - 192 = 0\) A1
\(c = \dfrac{4}{5}\) A1
[4 marks]
See 4.2.
HL questions
HL questions on this unit also use the binomial theorem for rational powers, with its interval of validity, and one question combines it with partial fractions from Unit 3.
1 Paper 2
(a) Expand \((8 - 3x)^{-\frac{1}{3}}\) in ascending powers of \(x\), up to and including the term in \(x^{2}\). [4]
(b) State the values of \(x\) for which the expansion is valid. [1]
(c) By choosing a suitable value of \(x\), use your answer to (a) to find an approximation for \(\dfrac{1}{\sqrt[3]{5}}\). Find the percentage error in this approximation. [3]
(d) The same method is used with \(x = 2\) to approximate \(\dfrac{1}{\sqrt[3]{2}}\). The percentage error is now about \(13\%\). Explain why this approximation is much less accurate than the one in (c). [1]
Total 9 marks
(a)
\((8 - 3x)^{-\frac{1}{3}} = 8^{-\frac{1}{3}}\left(1 - \dfrac{3x}{8}\right)^{-\frac{1}{3}} = \dfrac{1}{2}\left(1 - \dfrac{3x}{8}\right)^{-\frac{1}{3}}\) M1 A1
\(\dfrac{1}{2}\left(1 + \left(-\dfrac{1}{3}\right)\left(-\dfrac{3x}{8}\right) + \dfrac{\left(-\frac{1}{3}\right)\left(-\frac{4}{3}\right)}{2!}\left(-\dfrac{3x}{8}\right)^{2} + \cdots\right)\) M1
\(= \dfrac{1}{2}\left(1 + \dfrac{x}{8} + \dfrac{x^{2}}{32} + \cdots\right) = \dfrac{1}{2} + \dfrac{x}{16} + \dfrac{x^{2}}{64} + \cdots\) A1
[4 marks]
(b)
\(\left|\dfrac{3x}{8}\right| < 1\), so \(|x| < \dfrac{8}{3}\) A1
[1 mark]
(c)
\(8 - 3x = 5\) gives \(x = 1\): \(\dfrac{1}{2} + \dfrac{1}{16} + \dfrac{1}{64} = \dfrac{37}{64} = 0.578125\) M1 A1
percentage error \(= \dfrac{|0.578125 - 0.584803\ldots|}{0.584803\ldots} \times 100\% = 1.14\%\) \((1.1420\ldots\,\%)\) A1
[3 marks]
(d)
\(x = 2\) is still inside the interval of validity, but it is further from \(0\) than \(x = 1\), so the terms left out (\(x^{3}\), \(x^{4}\), …) are much larger and the first three terms give a poorer approximation R1
[1 mark]
See 4.3.
2 Paper 1
Let \(f(x) = \dfrac{(1 + ax)^{5}}{\sqrt{1 - 4x}}\), where \(a \in \mathbb{R}\).
(a) Expand \((1 - 4x)^{-\frac{1}{2}}\) in ascending powers of \(x\), up to and including the term in \(x^{2}\), and state the values of \(x\) for which the expansion is valid. [3]
(b) The coefficient of \(x\) in the expansion of \(f(x)\) is zero. Find the value of \(a\). [3]
(c) Using this value of \(a\), find the coefficient of \(x^{2}\) in the expansion of \(f(x)\). [2]
Total 8 marks
(a)
\(1 + \left(-\dfrac{1}{2}\right)(-4x) + \dfrac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{2!}(-4x)^{2} + \cdots\) M1
\(= 1 + 2x + 6x^{2} + \cdots\) A1
valid for \(|4x| < 1\), that is \(|x| < \dfrac{1}{4}\) A1
[3 marks]
(b)
\((1 + ax)^{5} = 1 + 5ax + 10a^{2}x^{2} + \cdots\) A1
coefficient of \(x\) in \(f(x)\): \(2 + 5a = 0\) M1
\(a = -\dfrac{2}{5}\) A1
[3 marks]
(c)
coefficient of \(x^{2}\): \(6 + 5a \times 2 + 10a^{2}\) M1
\(= 6 - 4 + \dfrac{8}{5} = \dfrac{18}{5}\) A1
[2 marks]
3 Paper 1
Consider the expansion of \((1 - 2x)^{-3}\) in ascending powers of \(x\).
(a) State the values of \(x\) for which the expansion is valid. [1]
(b) Show that the coefficient of \(x^{r}\) is \((r + 1)(r + 2)2^{r - 1}\). [4]
(c) Find the smallest value of \(r\) for which the coefficient of \(x^{r}\) is greater than \(10\,000\). [3]
Total 8 marks
(a)
\(|2x| < 1\), that is \(|x| < \dfrac{1}{2}\) A1
[1 mark]
(b)
the term in \(x^{r}\) is \(\dfrac{(-3)(-4)\cdots(-3 - r + 1)}{r!}(-2x)^{r}\) M1
\((-3)(-4)\cdots\big(-(r + 2)\big) = (-1)^{r}\dfrac{(r + 2)!}{2}\) A1
coefficient \(= (-1)^{r}\dfrac{(r + 2)!}{2\,r!}(-2)^{r} = \dfrac{(r + 1)(r + 2)}{2} \times 2^{r}\) A1
\(= (r + 1)(r + 2)2^{r - 1}\) A1 AG
[4 marks]
(c)
\(r = 7\): \(8 \times 9 \times 2^{6} = 4608\) M1
\(r = 8\): \(9 \times 10 \times 2^{7} = 11\,520\) A1
the coefficients increase as \(r\) increases, so the smallest value is \(r = 8\) A1
[3 marks]
See 4.3.
4 Paper 1
Let \(f(x) = \dfrac{8 - 6x}{(1 - 2x)(2 + x)}\).
(a) Express \(f(x)\) in partial fractions. [3]
(b) Hence show that \(f(x) = 4 + 3x + \dfrac{17}{2}x^{2} + \dfrac{63}{4}x^{3} + \cdots\) [5]
(c) State, with a reason, the values of \(x\) for which this expansion is valid. [2]
(d) Show that the coefficient of \(x^{r}\) in the expansion of \(f(x)\) is \(2^{r + 1} + (-1)^{r}2^{1 - r}\). [3]
(e) Find the coefficient of \(x^{9}\), giving your answer as a single fraction. [2]
Total 15 marks
(a)
\(8 - 6x \equiv A(2 + x) + B(1 - 2x)\) M1
\(x = \dfrac{1}{2}\): \(5 = \dfrac{5}{2}A\), so \(A = 2\) A1
\(x = -2\): \(20 = 5B\), so \(B = 4\), giving \(f(x) = \dfrac{2}{1 - 2x} + \dfrac{4}{2 + x}\) A1
[3 marks]
(b)
\(2(1 - 2x)^{-1} = 2\left(1 + 2x + (2x)^{2} + (2x)^{3} + \cdots\right)\) M1
\(= 2 + 4x + 8x^{2} + 16x^{3} + \cdots\) A1
\(\dfrac{4}{2 + x} = 4 \times 2^{-1}\left(1 + \dfrac{x}{2}\right)^{-1} = 2\left(1 + \dfrac{x}{2}\right)^{-1}\) M1
\(= 2 - x + \dfrac{x^{2}}{2} - \dfrac{x^{3}}{4} + \cdots\) A1
adding: \(f(x) = 4 + 3x + \dfrac{17}{2}x^{2} + \dfrac{63}{4}x^{3} + \cdots\) A1 AG
[5 marks]
(c)
both series must be valid: \(|2x| < 1\) and \(\left|\dfrac{x}{2}\right| < 1\) R1
\(|x| < \dfrac{1}{2}\) A1
Do not award the A1 for \(|x| < 2\).
[2 marks]
(d)
term in \(x^{r}\) of \(2(1 - 2x)^{-1}\) is \(2(2x)^{r}\), so its coefficient is \(2^{r + 1}\) M1 A1
term in \(x^{r}\) of \(2\left(1 + \dfrac{x}{2}\right)^{-1}\) is \(2\left(-\dfrac{x}{2}\right)^{r}\), so its coefficient is \(2(-1)^{r}2^{-r} = (-1)^{r}2^{1 - r}\) A1
so the coefficient of \(x^{r}\) is \(2^{r + 1} + (-1)^{r}2^{1 - r}\) AG
[3 marks]
(e)
\(r = 9\): \(2^{10} + (-1)^{9}2^{-8} = 1024 - \dfrac{1}{256}\) M1
\(= \dfrac{262\,143}{256}\) A1
[2 marks]
5 Paper 1
(a) Expand \((1 + x)^{\frac{1}{3}}\) in ascending powers of \(x\), up to and including the term in \(x^{3}\), and state the values of \(x\) for which the expansion is valid. [4]
(b) Show that \(\sqrt[3]{2} = \dfrac{5}{4}\left(1 + \dfrac{3}{125}\right)^{\frac{1}{3}}\). [2]
(c) Hence use the first three terms of your expansion in (a) to find an approximation for \(\sqrt[3]{2}\). Give your answer as a fraction in its simplest form. [3]
(d) A student writes \(\sqrt[3]{3} = (1 + 2)^{\frac{1}{3}}\) and substitutes \(x = 2\) into the expansion in (a). Explain why this method is not valid. [1]
(e) Using your answer to (a), write down the expansion of \(\left(1 - x^{2}\right)^{\frac{1}{3}}\) up to and including the term in \(x^{4}\), and state the values of \(x\) for which it is valid. [3]
Total 13 marks
(a)
\(1 + \dfrac{1}{3}x + \dfrac{\frac{1}{3}\left(-\frac{2}{3}\right)}{2!}x^{2} + \dfrac{\frac{1}{3}\left(-\frac{2}{3}\right)\left(-\frac{5}{3}\right)}{3!}x^{3} + \cdots\) M1
\(= 1 + \dfrac{x}{3} - \dfrac{x^{2}}{9}\) A1
\(+ \dfrac{5x^{3}}{81} + \cdots\) A1
valid for \(|x| < 1\) A1
[4 marks]
(b)
\(2 = \dfrac{128}{64} = \dfrac{125}{64} \times \dfrac{128}{125}\) M1
\(\sqrt[3]{2} = \sqrt[3]{\dfrac{125}{64}} \times \sqrt[3]{\dfrac{128}{125}} = \dfrac{5}{4}\left(1 + \dfrac{3}{125}\right)^{\frac{1}{3}}\) A1 AG
[2 marks]
(c)
\(x = \dfrac{3}{125}\): \(1 + \dfrac{1}{3} \times \dfrac{3}{125} - \dfrac{1}{9} \times \dfrac{9}{15\,625}\) M1
\(= 1 + \dfrac{1}{125} - \dfrac{1}{15\,625} = \dfrac{15\,749}{15\,625}\) A1
\(\sqrt[3]{2} \approx \dfrac{5}{4} \times \dfrac{15\,749}{15\,625} = \dfrac{15\,749}{12\,500}\) A1
[3 marks]
(d)
the expansion is valid only for \(|x| < 1\), and \(x = 2\) is outside this interval (the series does not converge) R1
[1 mark]
(e)
replace \(x\) by \(-x^{2}\) in (a) M1
\(\left(1 - x^{2}\right)^{\frac{1}{3}} = 1 - \dfrac{x^{2}}{3} - \dfrac{x^{4}}{9} + \cdots\) A1
valid for \(\left|-x^{2}\right| < 1\), that is \(|x| < 1\) A1
[3 marks]
See 4.3.