Unit 3 Exam Practice
Partial Fractions
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 1.11, 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 partial fractions with two distinct linear factors, and combine them with sequences, series and sigma notation from Unit 2.
1 Paper 1
It is given that \[ \frac{3x + k}{(x + 1)(x - 2)} \equiv \frac{A}{x + 1} + \frac{B}{x - 2}, \] where \(k\), \(A\) and \(B\) are constants and \(A = 2B\).
Find the values of \(A\), \(B\) and \(k\). [5]
Total 5 marks
\(3x + k \equiv A(x - 2) + B(x + 1)\) M1
coefficients of \(x\): \(A + B = 3\) A1
\(A = 2B\) gives \(3B = 3\), so \(B = 1\) and \(A = 2\) A1
constant terms: \(k = -2A + B\) M1
\(k = -3\) A1
[5 marks]
See 3.1.
2 Paper 2
Let \(f(x) = \dfrac{13x - 4}{6x^{2} - x - 2}\), where \(x \neq \dfrac{2}{3}\) and \(x \neq -\dfrac{1}{2}\).
(a) Express \(f(x)\) in partial fractions. [4]
(b) Hence solve the equation \(f(x) = \dfrac{6}{2x + 1}\). [2]
(c) Solve the equation \(f(x) = x\) for \(x > \dfrac{2}{3}\). [2]
Total 8 marks
(a)
\(6x^{2} - x - 2 = (3x - 2)(2x + 1)\) A1
\(13x - 4 \equiv A(2x + 1) + B(3x - 2)\) M1
\(x = \dfrac{2}{3}\): \(\dfrac{14}{3} = \dfrac{7}{3}A\), so \(A = 2\) A1
\(x = -\dfrac{1}{2}\): \(-\dfrac{21}{2} = -\dfrac{7}{2}B\), so \(B = 3\), giving \(f(x) = \dfrac{2}{3x - 2} + \dfrac{3}{2x + 1}\) A1
[4 marks]
(b)
\(\dfrac{2}{3x - 2} + \dfrac{3}{2x + 1} = \dfrac{6}{2x + 1} \;\Rightarrow\; \dfrac{2}{3x - 2} = \dfrac{3}{2x + 1}\) M1
\(2(2x + 1) = 3(3x - 2)\), so \(x = \dfrac{8}{5}\) A1
[2 marks]
(c)
\(\dfrac{2}{3x - 2} + \dfrac{3}{2x + 1} = x\) (for example, using the intersection of the two graphs on a GDC) M1
\(x = 1.52\) \((1.52167\ldots)\) A1
Accept the equivalent equation \(6x^{3} - x^{2} - 15x + 4 = 0\) for M1. Do not award the A1 if the roots \(0.270\) or \(-1.62\) are also given.
[2 marks]
See 3.1.
3 Paper 1
It is given that \[ \frac{x + 11}{x^{2} + bx + c} \equiv \frac{3}{x - 1} + \frac{B}{x + q}, \] where \(b\), \(c\), \(B\) and \(q\) are constants.
Find the values of \(b\), \(c\), \(B\) and \(q\). [6]
Total 6 marks
the denominator must be \(x^{2} + bx + c \equiv (x - 1)(x + q)\) R1
\(x + 11 \equiv 3(x + q) + B(x - 1)\) M1
\(x = 1\): \(12 = 3(1 + q)\), so \(q = 3\) M1 A1
coefficients of \(x\): \(1 = 3 + B\), so \(B = -2\) A1
\((x - 1)(x + 3) = x^{2} + 2x - 3\), so \(b = 2\) and \(c = -3\) A1
[6 marks]
See 3.1.
4 Paper 1
Let \(f(x) = \dfrac{3 - 5x}{(1 - x)(1 - 3x)}\), where \(x \neq 1\) and \(x \neq \dfrac{1}{3}\).
(a) Express \(f(x)\) in partial fractions. [4]
(b) Using the sum to infinity of a geometric series, show that \[ f(x) = \sum_{r=0}^{\infty} \left(1 + 2 \times 3^{r}\right) x^{r} . \] State the set of values of \(x\) for which this is valid. [5]
(c) Write down the coefficient of \(x^{5}\) in this series. [1]
(d) Hence find the exact value of \(\displaystyle\sum_{r=0}^{\infty} \frac{1 + 2 \times 3^{r}}{5^{r}}\). [3]
Let \(c_r = 1 + 2 \times 3^{r}\) be the coefficient of \(x^{r}\).
(e) Show that \(c_{r+1} = 3c_r - 2\) for all \(r \geq 0\). [2]
Total 15 marks
(a)
\(3 - 5x \equiv A(1 - 3x) + B(1 - x)\) M1
\(x = 1\): \(-2 = -2A\), so \(A = 1\) A1
\(x = \dfrac{1}{3}\): \(\dfrac{4}{3} = \dfrac{2}{3}B\), so \(B = 2\) A1
\(f(x) = \dfrac{1}{1 - x} + \dfrac{2}{1 - 3x}\) A1
[4 marks]
(b)
\(\dfrac{1}{1 - x}\) is the sum to infinity of a geometric series with \(u_1 = 1\) and \(r = x\) M1
\(\dfrac{1}{1 - x} = \displaystyle\sum_{r=0}^{\infty} x^{r}\), valid for \(|x| < 1\) A1
\(\dfrac{2}{1 - 3x} = \displaystyle\sum_{r=0}^{\infty} 2(3x)^{r} = \sum_{r=0}^{\infty} 2 \times 3^{r} x^{r}\), valid for \(|3x| < 1\) A1
adding: \(f(x) = \displaystyle\sum_{r=0}^{\infty} \left(1 + 2 \times 3^{r}\right) x^{r}\) A1 AG
both series must converge, so \(|x| < \dfrac{1}{3}\) R1
[5 marks]
(c)
\(1 + 2 \times 3^{5} = 487\) A1
[1 mark]
(d)
the sum is \(f\left(\dfrac{1}{5}\right)\), which is valid because \(\dfrac{1}{5} < \dfrac{1}{3}\) R1
\(f\left(\dfrac{1}{5}\right) = \dfrac{3 - 1}{\left(\frac{4}{5}\right)\left(\frac{2}{5}\right)}\) M1
\(= \dfrac{25}{4}\) A1
[3 marks]
(e)
\(3c_r - 2 = 3\left(1 + 2 \times 3^{r}\right) - 2 = 3 + 2 \times 3^{r+1} - 2\) M1
\(= 1 + 2 \times 3^{r+1} = c_{r+1}\) A1 AG
[2 marks]
5 Paper 1
Let \(u_r = \dfrac{6}{(3r - 1)(3r + 2)}\), for \(r \in \mathbb{Z}^{+}\).
(a) Express \(u_r\) in partial fractions. [3]
(b) Hence show that \(\displaystyle\sum_{r=1}^{n} u_r = \frac{3n}{3n + 2}\). [4]
(c) Find the value of \(\displaystyle\sum_{r=1}^{\infty} u_r\). [2]
(d) Find the smallest value of \(n\) for which \(\displaystyle\sum_{r=1}^{n} u_r > 0.99\). [3]
(e) Find the exact value of \(\displaystyle\sum_{r=11}^{30} u_r\). [3]
Total 15 marks
(a)
\(6 \equiv A(3r + 2) + B(3r - 1)\) M1
\(r = \dfrac{1}{3}\): \(6 = 3A\), so \(A = 2\) A1
\(r = -\dfrac{2}{3}\): \(6 = -3B\), so \(B = -2\), giving \(u_r = \dfrac{2}{3r - 1} - \dfrac{2}{3r + 2}\) A1
[3 marks]
(b)
\(\displaystyle\sum_{r=1}^{n} u_r = \left(\dfrac{2}{2} - \dfrac{2}{5}\right) + \left(\dfrac{2}{5} - \dfrac{2}{8}\right) + \left(\dfrac{2}{8} - \dfrac{2}{11}\right) + \cdots + \left(\dfrac{2}{3n - 1} - \dfrac{2}{3n + 2}\right)\) M1 A1
all the middle terms cancel, so the sum is \(1 - \dfrac{2}{3n + 2}\) A1
\(= \dfrac{3n + 2 - 2}{3n + 2} = \dfrac{3n}{3n + 2}\) A1 AG
Award M1 for writing out at least the first two terms and the last term.
[4 marks]
(c)
as \(n\) gets larger, \(\dfrac{2}{3n + 2}\) gets closer to \(0\) R1
\(\displaystyle\sum_{r=1}^{\infty} u_r = 1\) A1
[2 marks]
(d)
\(\dfrac{3n}{3n + 2} > \dfrac{99}{100}\) M1
\(300n > 297n + 198\), so \(n > 66\) A1
\(n = 67\) A1
Do not award the final A1 for \(n = 66\).
[3 marks]
(e)
\(\displaystyle\sum_{r=11}^{30} u_r = \sum_{r=1}^{30} u_r - \sum_{r=1}^{10} u_r\) M1
\(= \dfrac{90}{92} - \dfrac{30}{32} = \dfrac{45}{46} - \dfrac{15}{16}\) A1
\(= \dfrac{15}{368}\) A1
Accept the direct method: \(\dfrac{2}{32} - \dfrac{2}{92}\).
[3 marks]