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SL Paper 1

Mia baked a very large apple pie that she cuts into slices to share with her friends. The smallest slice is cut first. The volume of each successive slice of pie forms a geometric sequence.

The second smallest slice has a volume of $30 {cm}^{3}$ . The fifth smallest slice has a volume of $240 {cm}^{3}$ .

Find the common ratio of the sequence.

[2]

Find the volume of the smallest slice of pie.

[2]

The apple pie has a volume of $61 425 {cm}^{3}$ .

Find the total number of slices Mia can cut from this pie.

[2]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure. It appeared in a paper that permitted the use of a calculator, and so might not be suitable for all forms of practice.

$u_{1} r = 30$ and $u_{1} r^{4} = 240,$ (M1)

Note: Award (M1) for both the given terms expressed in the formula for $u_{n}$ .

OR

$30 r^{3} = 240 (r^{3} = 8)$ (M1)

Note: Award (M1) for a correct equation seen.

$(r =) 2$ (A1) (C2)

[2 marks]

$u_{1} \times 2 = 30$ OR $u_{1} \times 2^{4} = 240$ (M1)

Note: Award (M1) for their correct substitution in geometric sequence formula.

$(u_{1} =) 15$ (A1)(ft) (C2)

Note: Follow through from part (a).

[2 marks]

$\frac{15 (2^{n} - 1)}{2 - 1} = 61425$ (M1)

Note: Award (M1) for correctly substituted geometric series formula equated to $61425$ .

$(n =) 12$ (slices) (A1)(ft) (C2)

Note: Follow through from parts (a) and (b).

[2 marks]

Examiners report

[N/A]

Solve the equation $2 \ln x = \ln 9 + 4$ . Give your answer in the form $x = p e^{q}$ where $p, q \in ℤ^{+}$ .

Markscheme

* This sample question was produced by experienced DP mathematics senior examiners to aid teachers in preparing for external assessment in the new MAA course. There may be minor differences in formatting compared to formal exam papers.

METHOD 1

$2 \ln x - \ln 9 = 4$

uses $m \ln x = \ln x^{m}$ (M1)

$\ln x^{2} - \ln 9 = 4$

uses $\ln a - \ln b = \ln \frac{a}{b}$ (M1)

$\ln \frac{x^{2}}{9} = 4$

$\frac{x^{2}}{9} = e^{4}$ A1

$x^{2} = 9 e^{4} \Rightarrow x = \sqrt{9 e^{4}} (x > 0)$ A1

$x = 3 e^{2} (p = 3, q = 2)$ A1

METHOD 2

expresses $4$ as $4 \ln e$ and uses $\ln x^{m} = m \ln x$ (M1)

$2 \ln x = 2 \ln 3 + 4 \ln e (\ln x = \ln 3 + 2 \ln e)$ A1

uses $2 \ln e = \ln e^{2}$ and $\ln a + \ln b = \ln a b$ (M1)

$\ln x = \ln (3 e^{2})$ A1

$x = 3 e^{2} (p = 3, q = 2)$ A1

METHOD 3

expresses $4$ as $4 \ln e$ and uses $m \ln x = \ln x^{m}$ (M1)

$\ln x^{2} = \ln 3^{2} + \ln e^{4}$ A1

uses $\ln a + \ln b = \ln a b$ (M1)

$\ln x^{2} = \ln (3^{2} e^{4})$

$x^{2} = 3^{2} e^{4} \Rightarrow x = \sqrt{3^{2} e^{4}} (x > 0)$ A1

so $x = 3 e^{2} (x > 0) (p = 3, q = 2)$ A1

[5 marks]

Examiners report

[N/A]

Consider the series $\ln x + p \ln x + \frac{1}{3} \ln x + \dots$ , where $x \in ℝ, x > 1$ and $p \in ℝ, p \neq 0$ .

Consider the case where the series is geometric.

Now consider the case where the series is arithmetic with common difference $d$ .

Show that $p = \pm \frac{1}{\sqrt{3}}$ .

[2]

a.i.

Given that $p > 0$ and $S_{\infty} = 3 + \sqrt{3}$ , find the value of $x$ .

[3]

a.ii.

Show that $p = \frac{2}{3}$ .

[3]

b.i.

Write down $d$ in the form $k \ln x$ , where $k \in ℚ$ .

[1]

b.ii.

The sum of the first $n$ terms of the series is $- 3 \ln x$ .

Find the value of $n$ .

[6]

b.iii.

Markscheme

EITHER

attempt to use a ratio from consecutive terms M1

$\frac{p \ln x}{\ln x} = \frac{\frac{1}{3} \ln x}{p \ln x}$ OR $\frac{1}{3} \ln x = (\ln x) r^{2}$ OR $p \ln x = \ln x (\frac{1}{3 p})$

Note: Candidates may use $\ln x^{1} + \ln x^{p} + \ln x^{\frac{1}{3}} \dots$ and consider the powers of $x$ in geometric sequence

Award M1 for $\frac{p}{1} = \frac{\frac{1}{3}}{p}$ .

$r = p$ and $r^{2} = \frac{1}{3}$ M1

THEN

$p^{2} = \frac{1}{3}$ OR $r = \pm \frac{1}{\sqrt{3}}$ A1

$p = \pm \frac{1}{\sqrt{3}}$ AG

Note: Award M0A0 for $r^{2} = \frac{1}{3}$ or $p^{2} = \frac{1}{3}$ with no other working seen.

[2 marks]

a.i.

$\frac{\ln x}{1 - \frac{1}{\sqrt{3}}} (= 3 + \sqrt{3})$ (A1)

$\ln x = 3 - \frac{3}{\sqrt{3}} + \sqrt{3} - \frac{\sqrt{3}}{\sqrt{3}}$ OR $\ln x = 3 - \sqrt{3} + \sqrt{3} - 1 (\Rightarrow \ln x = 2)$ A1

$x = e^{2}$ A1

[3 marks]

a.ii.

METHOD 1

attempt to find a difference from consecutive terms or from $u_{2}$ M1

correct equation A1

$p \ln x - \ln x = \frac{1}{3} \ln x - p \ln x$ OR $\frac{1}{3} \ln x = \ln x + 2 (p \ln x - \ln x)$

Note: Candidates may use $\ln x^{1} + \ln x^{p} + \ln x^{\frac{1}{3}} + \dots$ and consider the powers of $x$ in arithmetic sequence.

Award M1A1 for $p - 1 = \frac{1}{3} - p$

$2 p \ln x = \frac{4}{3} \ln x (\Rightarrow 2 p = \frac{4}{3})$ A1

$p = \frac{2}{3}$ AG

METHOD 2

attempt to use arithmetic mean $u_{2} = \frac{u_{1} + u_{3}}{2}$ M1

$p \ln x = \frac{\ln x + \frac{1}{3} \ln x}{2}$ A1

$2 p \ln x = \frac{4}{3} \ln x (\Rightarrow 2 p = \frac{4}{3})$ A1

$p = \frac{2}{3}$ AG

METHOD 3

attempt to find difference using $u_{3}$ M1

$\frac{1}{3} \ln x = \ln x + 2 d (\Rightarrow d = - \frac{1}{3} \ln x)$

$u_{2} = \ln x + \frac{1}{2} (\frac{1}{3} \ln x - \ln x)$ OR $p \ln x - \ln x = - \frac{1}{3} \ln x$ A1

$p \ln x = \frac{2}{3} \ln x$ A1

$p = \frac{2}{3}$ AG

[3 marks]

b.i.

$d = - \frac{1}{3} \ln x$ A1

[1 mark]

b.ii.

METHOD 1

$S_{n} = \frac{n}{2} [2 \ln x + (n - 1) \times (- \frac{1}{3} \ln x)]$

attempt to substitute into $S_{n}$ and equate to $- 3 \ln x$ (M1)

$\frac{n}{2} [2 \ln x + (n - 1) \times (- \frac{1}{3} \ln x)] = - 3 \ln x$

correct working with $S_{n}$ (seen anywhere) (A1)

$\frac{n}{2} [2 \ln x - \frac{n}{3} \ln x + \frac{1}{3} \ln x]$ OR $n \ln x - \frac{n (n - 1)}{6} \ln x$ OR $\frac{n}{2} (\ln x + (\frac{4 - n}{3}) \ln x)$

correct equation without $\ln x$ A1

$\frac{n}{2} (\frac{7}{3} - \frac{n}{3}) = - 3$ OR $n - \frac{n (n - 1)}{6} = - 3$ or equivalent

Note: Award as above if the series $1 + p + \frac{1}{3} + \dots$ is considered leading to $\frac{n}{2} (\frac{7}{3} - \frac{n}{3}) = - 3$ .

attempt to form a quadratic $= 0$ (M1)

$n^{2} - 7 n - 18 = 0$

attempt to solve their quadratic (M1)

$(n - 9) (n + 2) = 0$

$n = 9$ A1

METHOD 2

listing the first $7$ terms of the sequence (A1)

$\ln x + \frac{2}{3} \ln x + \frac{1}{3} \ln x + 0 - \frac{1}{3} \ln x - \frac{2}{3} \ln x - \ln x + \dots$

recognizing first $7$ terms sum to $0$ M1

$8$ ^th term is $- \frac{4}{3} \ln x$ (A1)

$9$ ^th term is $- \frac{5}{3} \ln x$ (A1)

sum of $8$ ^th and $9$ ^th term $= - 3 \ln x$ (A1)

$n = 9$ A1

[6 marks]

b.iii.

Examiners report

Many candidates were able to identify the key relationship between consecutive terms for both geometric and arithmetic sequences. Substitution into the infinity sum formula was good with solving involving the natural logarithm done quite well. The complexity of the equation formed using 𝑆𝑛 was a stumbling block for some candidates. Those who factored out and cancelled the ln𝑥 expression were typically successful in solving the resulting quadratic.

a.i.

[N/A]

a.ii.

[N/A]

b.i.

[N/A]

b.ii.

[N/A]

b.iii.

The $n$ ^th term of an arithmetic sequence is given by $u_{n} = 15 - 3 n$ .

State the value of the first term, $u_{1}$ .

[1]

Given that the $n$ ^th term of this sequence is $- 33$ , find the value of $n$ .

[2]

Find the common difference, $d$ .

[2]

Markscheme

$u_{1} = 12$ A1

[1 mark]

$15 - 3 n = - 33$ (A1)

$n = 16$ A1

[2 marks]

valid approach to find $d$ (M1)

$u_{2} - u_{1} = 9 - 12$ OR recognize gradient is $- 3$ OR attempts to solve $- 33 = 12 + 15 d$

$d = - 3$ A1

[2 marks]

Examiners report

A large majority of candidates earned full marks for this question. In part (a), a surprising number of candidates did not substitute $n = 1$ into the given expression, erroneously stating $u_{1} = 15$ . Many of these candidates were able to earn follow-through marks in later parts of the question. In part (b), algebraic errors led a few candidates to find inappropriate values for $n$ , such as $n = - 6$ .

[N/A]

In an arithmetic sequence, the first term is 3 and the second term is 7.

Find the common difference.

[2]

Find the tenth term.

[2]

Find the sum of the first ten terms of the sequence.

[2]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

attempt to subtract terms (M1)

eg $\,\,\,\,\,$ $d = {u_2} - {u_1},{\text{ }}7 - 3$

$d = 4$ A1 N2

[2 marks]

correct approach (A1)

eg $\,\,\,\,\,$ ${u_{10}} = 3 + 9(4)$

${u_{10}} = 39$ A1 N2

[2 marks]

correct substitution into sum (A1)

eg $\,\,\,\,\,$ ${S_{10}} = 5(3 + 39),{\text{ }}{S_{10}} = \frac{{10}}{2}(2 \times 3 + 9 \times 4)$

${S_{10}} = 210$ A1 N2

[2 marks]

Examiners report

[N/A]

Consider $\left( {\begin{array}{*{20}{c}} {11} \\ a \end{array}} \right) = \frac{{11{\text{!}}}}{{a{\text{!}}\,9{\text{!}}}}$ .

Find the value of $a$ .

[2]

Hence or otherwise find the coefficient of the term in ${x^9}$ in the expansion of ${\left( {x + 3} \right)^{11}}$ .

[4]

Markscheme

valid approach (M1)

eg $11 - a = 9$ , $\frac{{11{\text{!}}}}{{9{\text{!}}\left( {11 - 9} \right){\text{!}}}}$

$a = 2$ A1 N2

[2 marks]

valid approach for expansion using $n = 11$ (M1)

eg $\left( {\begin{array}{*{20}{c}} {11} \\ r \end{array}} \right){x^{11 - r}}{3^r}$ , ${a^{11}}{b^0} + \left( {\begin{array}{*{20}{c}} {11} \\ 1 \end{array}} \right){a^{10}}{b^1} + \left( {\begin{array}{*{20}{c}} {11} \\ 2 \end{array}} \right){a^9}{b^2} + \ldots$

evidence of choosing correct term A1

eg $\left( {\begin{array}{*{20}{c}} {11} \\ 2 \end{array}} \right){3^2}$ , $\left( {\begin{array}{*{20}{c}} {11} \\ 2 \end{array}} \right){x^9}{3^2}$ , $\left( {\begin{array}{*{20}{c}} {11} \\ 9 \end{array}} \right){3^2}$

correct working for binomial coefficient (seen anywhere, do not accept factorials) A1

eg $55$ , $\left( {\begin{array}{*{20}{c}} {11} \\ 2 \end{array}} \right) = 55$ , $55 \times {3^2}$ , $\left( {55 \times 9} \right){x^9}$ , $\frac{{11 \times 10}}{2} \times 9$

$495$ A1 N2

Note: If there is clear evidence of adding instead of multiplying, award A1 for the correct working for binomial coefficient, but no other marks. For example, $55{x^9} \times {3^2}$ would earn M0A0A1A0.

Do not award final A1 for a final answer of $495{x^9}$ , even if $495$ is seen previously. If no working shown, award N1 for $495{x^9}$ .

[4 marks]

Examiners report

[N/A]

Juan buys a bicycle in a sale. He gets a discount of 30% off the original price and pays 560 US dollars (USD).

To buy the bicycle, Juan takes a loan of 560 USD for 6 months at a nominal annual interest rate of 75%, compounded monthly. Juan believes that the total amount he will pay will be less than the original price of the bicycle.

Calculate the original price of the bicycle.

[2]

Calculate the difference between the original price of the bicycle and the total amount Juan will pay.

[4]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$\frac{{560}}{{70}} \times 100$ (or equivalent) (M1)

Note: Award (M1) for dividing 560 by 0.7 or equivalent.

$= 800{\text{ (USD)}}$ (A1) (C2)

[2 marks]

$560{\left( {1 + \frac{{75}}{{12 \times 100}}} \right)^{12 \times \frac{1}{2}}}$ (M1)(A1)

Note: Award (M1) for substitution into interest formula, (A1) for their correct substitution.

${\text{N}} = \frac{1}{2}$

${\text{I% }} = 75$

${\text{PV}} = ( \pm )560$

${\text{P/Y}} = 1$

${\text{C/Y}} = 12$ (A1)(M1)

Note: Award (A1) for ${\text{C/Y}} = 12$ seen, (M1) for all other entries correct.

${\text{N}} = 6$

${\text{I% }} = 75$

${\text{PV}} = ( \pm )560$

${\text{P/Y}} = 12$

${\text{C/Y}} = 12$ (A1)(M1)

Note: Award (A1) for ${\text{C/Y}} = 12$ seen, (M1) for all other entries correct.

$= 805.678 \ldots {\text{ (USD)}}$ (A1)

Note: Award (A3) for 805.678… (806) seen without working.

(Juan spends) 5.68 (USD) (5.67828… USD) (more than the original price) (A1)(ft) (C4)

[4 marks]

Examiners report

[N/A]

Last year a South American candy factory sold 4.8 × 10⁸ spherical sweets. Each sweet has a diameter of 2.5 cm.

The factory is producing an advertisement showing all of these sweets placed in a straight line.

The advertisement claims that the length of this line is x times the length of the Amazon River. The length of the Amazon River is 6400 km.

Find the length, in cm, of this line. Give your answer in the form a × 10^k , where 1 ≤ a < 10 and k ∈ $\mathbb{Z}$ .

[3]

Write down the length of the Amazon River in cm.

[1]

b.i.

Find the value of x.

[2]

b.ii.

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

4.8 × 10⁸ × 2.5 (M1)

Note: Award (M1) for multiplying by 2.5.

1.2 × 10⁹ (cm) (A1)(ft)(A1)(ft) (C3)

Note: Award (A0)(A0) for answers of the type 12 × 10⁸.

[3 marks]

640 000 000 (cm) (6.4 × 10⁸ (cm)) (A1)

[1 mark]

b.i.

$\frac{{1.2 \times {{10}^9}}}{{6.4 \times {{10}^8}}}$ (M1)

Note: Award (M1) for division by 640 000 000.

= 1.88 (1.875) (A1)(ft) (C3)

Note: Follow through from part (a) and part (b)(i).

[2 marks]

b.ii.

Examiners report

[N/A]

b.i.

[N/A]

b.ii.

A hydraulic hammer drives a metal post vertically into the ground by striking the top of the post. The distance that the post is driven into the ground, by the $n{\text{th}}$ strike of the hammer, is ${d_n}$ .

The distances ${d_1},{\text{ }}{d_2},{\text{ }}{d_3}{\text{ }} \ldots ,{\text{ }}{d_n}$ form a geometric sequence.

The distance that the post is driven into the ground by the first strike of the hammer, ${d_1}$ , is 64 cm.

The distance that the post is driven into the ground by the second strike of the hammer, ${d_2}$ , is 48 cm.

Find the value of the common ratio for this sequence.

[2]

Find the distance that the post is driven into the ground by the eighth strike of the hammer.

[2]

Find the total depth that the post has been driven into the ground after 10 strikes of the hammer.

[2]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$48 = 64r$ (M1)

Note: Award (M1) for correct substitution into geometric sequence formula.

$= 0.75\left( {\frac{3}{4},{\text{ }}\frac{{48}}{{64}}} \right)$ (A1) (C2)

[2 marks]

$64 \times {(0.75)^7}$ (M1)

Note: Award (M1) for correct substitution into geometric sequence formula or list of eight values using their $r$ . Follow through from part (a), only if answer is positive.

$= 8.54{\text{ }}({\text{cm}}){\text{ }}(8.54296 \ldots {\text{ cm}})$ (A1)(ft) (C2)

[2 marks]

${\text{depth}} = \frac{{64\left( {1 - {{(0.75)}^{10}}} \right)}}{{1 - 0.75}}$ (M1)

Note: Award (M1) for correct substitution into geometric series formula. Follow through from part (a), only if answer is positive.

$= 242({\text{cm}}){\text{ }}(241.583 \ldots )$ (A1)(ft) (C2)

[2 marks]

Examiners report

[N/A]

Consider the function $f (x) = a^{x}$ where $x, a \in ℝ$ and $x > 0, a > 1$ .

The graph of $f$ contains the point $(\frac{2}{3}, 4)$ .

Consider the arithmetic sequence $\log_{8} 27, \log_{8} p, \log_{8} q, \log_{8} 125,$ where $p > 1$ and $q > 1$ .

Show that $a = 8$ .

[2]

Write down an expression for $f^{- 1} (x)$ .

[1]

Find the value of $f^{- 1} (\sqrt{32})$ .

[3]

Show that $27, p, q$ and $125$ are four consecutive terms in a geometric sequence.

[4]

d.i.

Find the value of $p$ and the value of $q$ .

[5]

d.ii.

Markscheme

$f (\frac{2}{3}) = 4$ OR $a^{\frac{2}{3}} = 4$ (M1)

$a = 4^{\frac{3}{2}}$ OR $a = {(2^{2})}^{\frac{3}{2}}$ OR $a^{2} = 64$ OR $\sqrt[3]{a} = 2$ A1

$a = 8$ AG

[2 marks]

$f^{- 1} (x) = \log_{8} x$ A1

Note: Accept $f^{- 1} (x) = \log_{a} x$ .
Accept any equivalent expression for $f^{- 1}$ e.g. $f^{- 1} (x) = \frac{\ln x}{\ln 8}$ .

[1 mark]

correct substitution (A1)

$\log_{8} \sqrt{32}$ OR $8^{x} = 32^{\frac{1}{2}}$

correct working involving log/index law (A1)

$\frac{1}{2} \log_{8} 32$ OR $\frac{5}{2} \log_{8} 2$ OR $\log_{8} 2 = \frac{1}{3}$ OR $\log_{2} 2^{\frac{5}{2}}$ OR $\log_{2} 8 = 3$ OR $\frac{\ln 2^{\frac{5}{2}}}{\ln 2^{3}}$ OR $2^{3 x} = 2^{\frac{5}{2}}$

$f^{- 1} (\sqrt{32}) = \frac{5}{6}$ A1

[3 marks]

METHOD 1

equating a pair of differences (M1)

$u_{2} - u_{1} = u_{4} - u_{3} (= u_{3} - u_{2})$

$\log_{8} p - \log_{8} 27 = \log_{8} 125 - \log_{8} q$

$\log_{8} 125 - \log_{8} q = \log_{8} q - \log_{8} p$

$\log_{8} (\frac{p}{27}) = \log_{8} (\frac{125}{q}), \log_{8} (\frac{125}{q}) = \log_{8} (\frac{q}{p})$ A1A1

$\frac{p}{27} = \frac{125}{q}$ and $\frac{125}{q} = \frac{q}{p}$ A1

$27, p, q$ and $125$ are in geometric sequence AG

Note: If candidate assumes the sequence is geometric, award no marks for part (i). If $r = \frac{5}{3}$ has been found, this will be awarded marks in part (ii).

METHOD 2

expressing a pair of consecutive terms, in terms of $d$ (M1)

$p = 8^{d} \times 27$ and $q = 8^{2 d} \times 27$ OR $q = 8^{2 d} \times 27$ and $125 = 8^{3 d} \times 27$

two correct pairs of consecutive terms, in terms of $d$ A1

$\frac{8^{d} \times 27}{27} = \frac{8^{2 d} \times 27}{8^{d} \times 27} = \frac{8^{3 d} \times 27}{8^{2 d} \times 27}$ (must include 3 ratios) A1

all simplify to $8^{d}$ A1

$27, p, q$ and $125$ are in geometric sequence AG

[4 marks]

d.i.

METHOD 1 (geometric, finding $r$ )

$u_{4} = u_{1} r^{3}$ OR $125 = 27 {(r)}^{3}$ (M1)

$r = \frac{5}{3}$ (seen anywhere) A1

$p = 27 r$ OR $\frac{125}{q} = \frac{5}{3}$ (M1)

$p = 45, q = 75$ A1A1

METHOD 2 (arithmetic)

$u_{4} = u_{1} + 3 d$ OR $\log_{8} 125 = \log_{8} 27 + 3 d$ (M1)

$d = \log_{8} (\frac{5}{3})$ (seen anywhere) A1

$\log_{8} p = \log_{8} 27 + \log_{8} (\frac{5}{3})$ OR $\log_{8} q = \log_{8} 27 + 2 \log_{8} (\frac{5}{3})$ (M1)

$p = 45, q = 75$ A1A1

METHOD 3 (geometric using proportion)

recognizing proportion (M1)

$p q = 125 \times 27$ OR $q^{2} = 125 p$ OR $p^{2} = 27 q$

two correct proportion equations A1

attempt to eliminate either $p$ or $q$ (M1)

$q^{2} = 125 \times \frac{125 \times 27}{q}$ OR $p^{2} = 27 \times \frac{125 \times 27}{p}$

$p = 45, q = 75$ A1A1

[5 marks]

d.ii.

Examiners report

[N/A]

d.i.

[N/A]

d.ii.

Iron in the asteroid 16 Psyche is said to be valued at $8973$ quadrillion euros $(EUR)$ , where one quadrillion $= 10^{15}$ .

James believes the asteroid is approximately spherical with radius $113 km$ . He uses this information to estimate its volume.

Write down the value of the iron in the form $a \times 10^{k}$ where $1 \leq a < 10, k \in ℤ$ .

[2]

Calculate James’s estimate of its volume, in ${km}^{3}$ .

[2]

The actual volume of the asteroid is found to be $6.074 \times 10^{6} {km}^{3}$ .

Find the percentage error in James’s estimate of the volume.

[2]

Markscheme

$8.97 \times 10^{18} (EUR) (8.973 \times 10^{18})$ (A1)(A1) (C2)

Note: Award (A1) for $8.97 (8.973)$ , (A1) for $\times 10^{18}$ . Award (A1)(A0) for $8.97 E 18$ .
Award (A0)(A0) for answers of the type $8973 \times 10^{15}$ .

[2 marks]

$\frac{4 \times π \times 113^{3}}{3}$ (M1)

Note: Award (M1) for correct substitution in volume of sphere formula.

$6 040 000 ({km}^{3}) (6.04 \times 10^{6}, \frac{5771588 π}{3}, 6 043 992.82)$ (A1) (C2)

[2 marks]

$|\frac{6 043 992.82 - 6.074 \times 10^{6}}{6.074 \times 10^{6}}| \times 100$ (M1)

Note: Award (M1) for their correct substitution into the percentage error formula (accept a consistent absence of “ $\times 10^{6}$ ” from all terms).

$0.494 (%) (0.494026 \dots (%))$ (A1)(ft) (C2)

Note: Follow through from their answer to part (b). If the final answer is negative, award at most (M1)(A0).

[2 marks]

Examiners report

[N/A]

Show that ${\left( {2n - 1} \right)^2} + {\left( {2n + 1} \right)^2} = 8{n^2} + 2$ , where $n \in \mathbb{Z}$ .

[2]

Hence, or otherwise, prove that the sum of the squares of any two consecutive odd integers is even.

[3]

Markscheme

attempting to expand the LHS (M1)

LHS $= \left( {4{n^2} - 4n + 1} \right) + \left( {4{n^2} + 4n + 1} \right)$ A1

$= 8{n^2} + 2$ (= RHS) AG

[2 marks]

METHOD 1

recognition that ${2n - 1}$ and ${2n + 1}$ represent two consecutive odd integers (for $n \in \mathbb{Z}$ ) R1

$8{n^2} + 2 = 2\left( {4{n^2} + 1} \right)$ A1

valid reason eg divisible by 2 (2 is a factor) R1

so the sum of the squares of any two consecutive odd integers is even AG

METHOD 2

recognition, eg that $_n$ and ${n + 2}$ represent two consecutive odd integers (for $n \in \mathbb{Z}$ ) R1

${n^2} + {\left( {n + 2} \right)^2} = 2\left( {{n^2} + 2n + 2} \right)$ A1

valid reason eg divisible by 2 (2 is a factor) R1

so the sum of the squares of any two consecutive odd integers is even AG

[3 marks]

Examiners report

[N/A]

Consider $f (x) = 4 \cos x (1 - 3 \cos 2 x + 3 \cos^{2} 2 x - \cos^{3} 2 x)$ .

Expand and simplify ${(1 - a)}^{3}$ in ascending powers of $a$ .

[2]

a.i.

By using a suitable substitution for $a$ , show that $1 - 3 \cos 2 x + 3 \cos^{2} 2 x - \cos^{3} 2 x = 8 \sin^{6} x$ .

[4]

a.ii.

Show that $\int_{0}^{m} f (x) d x = \frac{32}{7} \sin^{7} m$ , where $m$ is a positive real constant.

[4]

b.i.

It is given that $\int_{m}^{\frac{π}{2}} f (x) d x = \frac{127}{28}$ , where $0 \leq m \leq \frac{π}{2}$ . Find the value of $m$ .

[5]

b.ii.

Markscheme

EITHER

attempt to use binomial expansion (M1)

$1 + {}_{3}C_{1} \times 1 \times (- a) + {}_{3}C_{2} \times 1 \times {(- a)}^{2} + 1 \times {(- a)}^{3}$

$(1 - a) (1 - a) (1 - a)$

$= (1 - a) (1 - 2 a + a^{2})$ (M1)

THEN

$= 1 - 3 a + 3 a^{2} - a^{3}$ A1

[2 marks]

a.i.

$a = \cos 2 x$ (A1)

So, $1 - 3 \cos 2 x + 3 \cos^{2} 2 x - \cos^{3} 2 x =$

${(1 - \cos 2 x)}^{3}$ A1

attempt to substitute any double angle rule for $\cos 2 x$ into ${(1 - \cos 2 x)}^{3}$ (M1)

$= {(2 \sin^{2} x)}^{3}$ A1

$= 8 \sin^{6} x$ AG

Note: Allow working RHS to LHS.

[4 marks]

a.ii.

recognizing to integrate $\int (4 \cos x \times 8 \sin^{6} x) d x$ (M1)

EITHER

applies integration by inspection (M1)

$32 \int (\cos x \times {(\sin x)}^{6}) d x$

$= \frac{32}{7} \sin^{7} x (+ c)$ A1

${[\frac{32}{7} \sin^{7} x]}_{0}^{m} (= \frac{32}{7} \sin^{7} m - \frac{32}{7} \sin^{7} 0)$ A1

$u = \sin x \Rightarrow \frac{d u}{d x} = \cos x$ (M1)

$\int 32 \cos x (\sin^{6} x) d x = \int 32 u^{6} d u$

$= \frac{32}{7} u^{7} (+ c)$ A1

${[\frac{32}{7} \sin^{7} x]}_{0}^{m}$ OR ${[\frac{32}{7} u^{7}]}_{0}^{\sin m} (= \frac{32}{7} \sin^{7} m - \frac{32}{7} \sin^{7} 0)$ A1

THEN

$= \frac{32}{7} \sin^{7} m$ AG

[4 marks]

b.i.

EITHER

$\int_{m}^{\frac{π}{2}} f (x) d x (= {[\frac{32}{7} \sin^{7} x]}_{m}^{\frac{π}{2}}) = \frac{32}{7} \sin^{7} \frac{π}{2} - \frac{32}{7} \sin^{7} m$ M1

$\frac{32}{7} \sin^{7} \frac{π}{2} - \frac{32}{7} \sin^{7} m = \frac{127}{28}$ OR $\frac{32}{7} (1 - \sin^{7} m) = \frac{127}{28}$ (M1)

$\int_{0}^{\frac{π}{2}} f (x) d x = \int_{0}^{m} f (x) d x + \int_{m}^{\frac{π}{2}} f (x) d x$ M1

$\frac{32}{7} = \frac{32}{7} \sin^{7} m + \frac{127}{28}$ (M1)

THEN

$\sin^{7} m = \frac{1}{128} (= \frac{1}{2^{7}})$ (A1)

$\sin m = \frac{1}{2}$ (A1)

$m = \frac{π}{6}$ A1

[5 marks]

b.ii.

Examiners report

Many candidates successfully expanded the binomial, with the most common error being to omit the negative sign with a. The connection between (a)(i) and (ii) was often noted but not fully utilised with candidates embarking on unnecessary complex algebraic expansions of expressions involving double angle rules. Candidates often struggled to apply inspection or substitution when integrating. As a 'show that' question, b(i) provided a useful result to be utilised in (ii). So even without successfully completing (i) candidates could apply it in part (ii). Not many managed to do so.

a.i.

[N/A]

a.ii.

[N/A]

b.i.

[N/A]

b.ii.

Consider the binomial expansion ${(x + 1)}^{7} = x^{7} + a x^{6} + b x^{5} + 35 x^{4} + \dots + 1$ where $x \neq 0$ and $a, b \in ℤ^{+}$ .

Show that $b = 21$ .

[2]

The third term in the expansion is the mean of the second term and the fourth term in the expansion.

Find the possible values of $x$ .

[5]

Markscheme

EITHER

recognises the required term (or coefficient) in the expansion (M1)

$b x^{5} = {}_{7}C_{2} x^{5} 1^{2}$ OR $b = {}_{7}C_{2}$ OR ${}_{7}C_{5}$

$b = \frac{7!}{2! 5!} (= \frac{7!}{2! (7 - 2)!})$

correct working A1

$\frac{7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{2 \times 1 \times 5 \times 4 \times 3 \times 2 \times 1}$ OR $\frac{7 \times 6}{2!}$ OR $\frac{42}{2}$

lists terms from row $7$ of Pascal’s triangle (M1)

$1, 7, 21, \dots$ A1

THEN

$b = 21$ AG

[2 marks]

$a = 7$ (A1)

correct equation A1

$21 x^{5} = \frac{a x^{6} + 35 x^{4}}{2}$ OR $21 x^{5} = \frac{7 x^{6} + 35 x^{4}}{2}$

correct quadratic equation A1

$7 x^{2} - 42 x + 35 = 0$ OR $x^{2} - 6 x + 5 = 0$ (or equivalent)

valid attempt to solve their quadratic (M1)

$(x - 1) (x - 5) = 0$ OR $x = \frac{6 \pm \sqrt{{(- 6)}^{2} - 4 (1) (5)}}{2 (1)}$

$x = 1, x = 5$ A1

Note: Award final A0 for obtaining $x = 0, x = 1, x = 5$ .

[5 marks]

Examiners report

The majority of candidates answered part (a) correctly, either by using the ${}^{n}C_{r}$ formula or Pascal's Triangle. In part (b) of the question, most candidates were able to correctly find the value of $a = 7$ and set up a correct equation showing the mean of the second and fourth terms. While some struggled to complete the required algebra to solve the equation, the majority of candidates who found a correct quadratic equation were able to solve it correctly. A few candidates included $x = 0$ in their final answer, thus not earning the final mark.

[N/A]

The first three terms of a geometric sequence are ${u_1} = 486,{\text{ }}{u_2} = 162,{\text{ }}{u_3} = 54$ .

Find the value of $r$ , the common ratio of the sequence.

[2]

Find the value of $n$ for which ${u_n} = 2$ .

[2]

Find the sum of the first 30 terms of the sequence.

[2]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$\frac{{162}}{{486}}$ $\,\,\,$ OR $\,\,\,$ $\frac{{54}}{{162}}$ (M1)

Note: Award (M1) for dividing any ${u_{n + 1}}$ by ${u_n}$ .

$= \frac{1}{3}{\text{ }}(0.333,{\text{ }}0.333333 \ldots )$ (A1) (C2)

[2 marks]

$486{\left( {\frac{1}{3}} \right)^{n - 1}} = 2$ (M1)

Note: Award (M1) for their correct substitution into geometric sequence formula.

$n = 6$ (A1)(ft) (C2)

Note: Follow through from part (a).

Award (A1)(A0) for ${u_6} = 2$ or ${u_6}$ with or without working.

[2 marks]

${S_{30}} = \frac{{486\left( {1 - {{\frac{1}{3}}^{30}}} \right)}}{{1 - \frac{1}{3}}}$ (M1)

Note: Award (M1) for correct substitution into geometric series formula.

$= 729$ (A1)(ft) (C2)

[2 marks]

Examiners report

[N/A]

Show that $2 x - 3 - \frac{6}{x - 1} = \frac{2 x^{2} - 5 x - 3}{x - 1}, x \in ℝ, x \neq 1$ .

[2]

Hence or otherwise, solve the equation $2 \sin 2 θ - 3 - \frac{6}{\sin 2 θ - 1} = 0$ for $0 \leq θ \leq π, θ \neq \frac{π}{4}$ .

[5]

Markscheme

METHOD 1

attempt to write all LHS terms with a common denominator of $x - 1$ (M1)

$2 x - 3 - \frac{6}{x - 1} = \frac{2 x (x - 1) - 3 (x - 1) - 6}{x - 1}$ OR $\frac{(2 x - 3) (x - 1)}{x - 1} - \frac{6}{x - 1}$

$= \frac{2 x^{2} - 2 x - 3 x + 3 - 6}{x - 1}$ OR $\frac{2 x^{2} - 5 x + 3}{x - 1} - \frac{6}{x - 1}$ A1

$= \frac{2 x^{2} - 5 x - 3}{x - 1}$ AG

METHOD 2

attempt to use algebraic division on RHS (M1)

correctly obtains quotient of $2 x - 3$ and remainder $- 6$ A1

$= 2 x - 3 - \frac{6}{x - 1}$ as required. AG

[2 marks]

consider the equation $\frac{2 \sin^{2} 2 θ - 5 \sin 2 θ - 3}{\sin 2 θ - 1} = 0$ (M1)

$\Rightarrow 2 \sin^{2} 2 θ - 5 \sin 2 θ - 3 = 0$

EITHER

attempt to factorise in the form $(2 \sin 2 θ + a) (\sin 2 θ + b)$ (M1)

Note: Accept any variable in place of $\sin 2 θ$ .

$(2 \sin 2 θ + 1) (\sin 2 θ - 3) = 0$

attempt to substitute into quadratic formula (M1)

$\sin 2 θ = \frac{5 \pm \sqrt{49}}{4}$

THEN

$\sin 2 θ = - \frac{1}{2}$ or $\sin 2 θ = 3$ (A1)

Note: Award A1 for $\sin 2 θ = - \frac{1}{2}$ only.

one of $\frac{7 π}{6}$ OR $\frac{11 π}{6}$ (accept $210$ or $330$ ) (A1)

$θ = \frac{7 π}{12}, \frac{11 π}{12}$ (must be in radians) A1

Note: Award A0 if additional answers given.

[5 marks]

Examiners report

[N/A]

The volume of a hemisphere, V, is given by the formula

V = $\sqrt {\frac{{4\,{S^3}}}{{243\,\pi }}}$ ,

where S is the total surface area.

The total surface area of a given hemisphere is 350 cm².

Calculate the volume of this hemisphere in cm³.

Give your answer correct to one decimal place.

[3]

Write down your answer to part (a) correct to the nearest integer.

[1]

Write down your answer to part (b) in the form a × 10^k , where 1 ≤ a < 10 and k ∈ $\mathbb{Z}$ .

[2]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$\sqrt {\frac{{4\,{{\left( {350} \right)}^3}}}{{243\,\pi }}}$ OR $\sqrt {\frac{{171500\,000}}{{763.407\, \ldots }}}$ (M1)

Note: Award (M1) for substitution of 350 into volume formula.

= 473.973… (A1)

= 474 (cm³) (A1)(ft) (C3)

Note: The final (A1)(ft) is awarded for rounding their answer to 1 decimal place provided the unrounded answer is seen.

[3 marks]

474 (cm³) (A1)(ft) (C1)

Note: Follow through from part (a).

[1 mark]

4.74 × 10² (cm³) (A1)(ft)(A1)(ft) (C2)

Note: Follow through from part (b) only.

Award (A0)(A0) for answers of the type 0.474 × 10³.

[2 marks]

Examiners report

[N/A]

In an arithmetic sequence, the first term is 8 and the second term is 5.

Find the common difference.

[2]

Find the tenth term.

[2]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

subtracting terms (M1)

eg $\,\,\,\,\,$ $5 - 8,{\text{ }}{u_2} - {u_1}$

$d = - 3$ A1 N2

[2 marks]

correct substitution into formula (A1)

eg $\,\,\,\,\,$ ${u_{10}} = 8 + (10 - 1)( - 3),{\text{ }}8 - 27,{\text{ }} - 3(10) + 11$

${u_{10}} = - 19$ A1 N2

[2 marks]

Examiners report

[N/A]

Sergei is training to be a weightlifter. Each day he trains at the local gym by lifting a metal bar that has heavy weights attached. He carries out successive lifts. After each lift, the same amount of weight is added to the bar to increase the weight to be lifted.

The weights of each of Sergei’s lifts form an arithmetic sequence.

Sergei’s friend, Yuri, records the weight of each lift. Unfortunately, last Monday, Yuri misplaced all but two of the recordings of Sergei’s lifts.

On that day, Sergei lifted 21 kg on the third lift and 46 kg on the eighth lift.

For that day find how much weight was added after each lift.

[2]

a.i.

For that day find the weight of Sergei’s first lift.

[2]

a.ii.

On that day, Sergei made 12 successive lifts. Find the total combined weight of these lifts.

[2]

Markscheme

5d = 46 − 21 OR u₁ + 2d = 21 and u₁ + 7d = 46 (M1)

Note: Award (M1) for a correct equation in d or for two correct equations in u₁ and d.

(d =) 5 (kg) (A1) (C2)

[2 marks]

a.i.

u₁ + 2 × 5 = 21 (M1)

u₁ + 7 × 5 = 46 (M1)

Note: Award (M1) for substitution of their d into either of the two equations.

(u₁=) 11 (kg) (A1)(ft) (C2)

Note: Follow through from part (a)(i).

[2 marks]

a.ii.

$\frac{{12}}{2}\left( {2 \times 11 + \left( {12 - 1} \right) \times 5} \right)$ (M1)

Note: Award (M1) for correct substitution into arithmetic series formula.

= 462 (kg) (A1)(ft) (C2)

Note: Follow through from parts (a) and (b).

[2 marks]

Examiners report

[N/A]

a.i.

[N/A]

a.ii.

[N/A]

Consider any three consecutive integers, $n - 1$ , $n$ and $n + 1$ .

Prove that the sum of these three integers is always divisible by $3$ .

[2]

Prove that the sum of the squares of these three integers is never divisible by $3$ .

[4]

Markscheme

$(n - 1) + n + (n + 1)$ (A1)

$= 3 n$ A1

which is always divisible by $3$ AG

[2 marks]

${(n - 1)}^{2} + n^{2} + {(n + 1)}^{2} (= n^{2} - 2 n + 1 + n^{2} + n^{2} + 2 n + 1)$ A1

attempts to expand either ${(n - 1)}^{2}$ or ${(n + 1)}^{2}$ (do not accept $n^{2} - 1$ or $n^{2} + 1$ ) (M1)

$= 3 n^{2} + 2$ A1

demonstrating recognition that $2$ is not divisible by $3$ or $\frac{2}{3}$ seen after correct expression divided by $3$ R1

$3 n^{2}$ is divisible by $3$ and so $3 n^{2} + 2$ is never divisible by $3$

OR the first term is divisible by $3$ , the second is not

OR $3 (n^{2} + \frac{2}{3})$ OR $\frac{3 n^{2} + 2}{3} = n^{2} + \frac{2}{3}$

hence the sum of the squares is never divisible by $3$ AG

[4 marks]

Examiners report

Most candidates were able to earn full marks in part (a), though some were not able to provide the required reasoning to earn full marks in part (b). In many cases, candidates did not seem to understand the nature of a general deductive proof, and instead substituted different consecutive integers (such as 1, 2,3 ), showing the desired result for these specific values, rather than an algebraic generalization for any three consecutive integers.

[N/A]

In the Canadian city of Ottawa:

$\begin{array}{*{20}{l}} {{\text{97% of the population speak English,}}} \\ {{\text{38% of the population speak French,}}} \\ {{\text{36% of the population speak both English and French.}}} \end{array}$

The total population of Ottawa is $985\,000$ .

Calculate the percentage of the population of Ottawa that speak English but not French.

[2]

Calculate the number of people in Ottawa that speak both English and French.

[2]

Write down your answer to part (b) in the form $a \times {10^k}$ where $1 \leqslant a < 10$ and k $\in \mathbb{Z}$ .

[2]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$97 - 36$ (M1)

Note: Award (M1) for subtracting 36 from 97.

M17/5/MATSD/SP1/ENG/TZ1/02.a/M

(M1)

Note: Award (M1) for 61 and 36 seen in the correct places in the Venn diagram.

$= 61{\text{ }}(\% )$ (A1) (C2)

Note: Accept 61.0 (%).

[2 marks]

$\frac{{36}}{{100}} \times 985\,000$ (M1)

Note: Award (M1) for multiplying 0.36 (or equivalent) by $985\,000$ .

$= 355\,000{\text{ }}(354\,600)$ (A1) (C2)

[2 marks]

$3.55 \times {10^5}{\text{ }}(3.546 \times {10^5})$ (A1)(ft)(A1)(ft) (C2)

Note: Award (A1)(ft) for 3.55 (3.546) must match part (b), and (A1)(ft) $\times {10^5}$ .

Award (A0)(A0) for answers of the type: $35.5 \times {10^4}$ . Follow through from part (b).

[2 marks]

Examiners report

[N/A]

Let $p = \frac{{\cos x + \sin y}}{{\sqrt {{w^2} - z} }}$ ,

where $x = 36^\circ ,{\text{ }}y = 18^\circ ,{\text{ }}w = 29$ and $z = 21.8$ .

Calculate the value of $p$ . Write down your full calculator display.

[2]

Write your answer to part (a)

(i) correct to two decimal places;

(ii) correct to three significant figures.

[2]

Write your answer to part (b)(ii) in the form $a \times {10^k}$ , where $1 \leqslant a < 10,{\text{ }}k \in \mathbb{Z}$ .

[2]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$\frac{{\cos 36^\circ + \sin 18^\circ }}{{\sqrt {{{29}^2} - 21.8} }}$ (M1)

Note: Award (M1) for correct substitution into formula.

$= 0.0390625$ (A1) (C2)

Note: Accept $\frac{5}{{128}}$ .

[2 marks]

(i) 0.04 (A1)(ft)

(ii) 0.0391 (A1)(ft) (C2)

Note: Follow through from part (a).

[2 marks]

$3.91 \times {10^{ - 2}}$ (A1)(ft)(A1)(ft) (C2)

Note: Answer should be consistent with their answer to part (b)(ii). Award (A1)(ft) for 3.91, and (A1)(ft) for ${10^{ - 2}}$ . Follow through from part (b)(ii).

[2 marks]

Examiners report

[N/A]

Consider the functions $f (x) = \frac{1}{x - 4} + 1$ , for $x \neq 4$ , and $g (x) = x - 3$ for $x \in ℝ$ .

The following diagram shows the graphs of $f$ and $g$ .

The graphs of $f$ and $g$ intersect at points $A$ and $B$ . The coordinates of $A$ are $(3, 0)$ .

In the following diagram, the shaded region is enclosed by the graph of $f$ , the graph of $g$ , the $x$ -axis, and the line $x = k$ , where $k \in ℤ$ .

The area of the shaded region can be written as $\ln (p) + 8$ , where $p \in ℤ$ .

Find the coordinates of $B$ .

[5]

Find the value of $k$ and the value of $p$ .

[10]

Markscheme

$\frac{1}{x - 4} + 1 = x - 3$ (M1)

$x^{2} - 8 x + 15 = 0$ OR ${(x - 4)}^{2} = 1$ (A1)

valid attempt to solve their quadratic (M1)

$(x - 3) (x - 5) = 0$ OR $x = \frac{8 \pm \sqrt{8^{2} - 4 (1) (15)}}{2 (1)}$ OR $(x - 4) = \pm 1$

$x = 5 (x = 3, x = 5)$ (may be seen in answer) A1

$B (5, 2)$ (accept $x = 5, y = 2$ ) A1

[5 marks]

recognizing two correct regions from $x = 3$ to $x = 5$ and from $x = 5$ to $x = k$ (R1)

triangle $+ \int_{5}^{k} f (x) d x$ OR $\int_{3}^{5} g (x) d x + \int_{5}^{k} f (x) d x$ OR $\int_{3}^{5} (x - 3) d x + \int_{5}^{k} (\frac{1}{x - 4} + 1) d x$

area of triangle is $2$ OR $\frac{2 \cdot 2}{2}$ OR $(\frac{5^{2}}{2} - 3 (5)) - (\frac{3^{2}}{2} - 3 (3))$ (A1)

correct integration (A1)(A1)

$\int (\frac{1}{x - 4} + 1) d x = \ln (x - 4) + x (+ C)$

Note: Award A1 for $\ln (x - 4)$ and A1 for $x$ .
Note: The first three A marks may be awarded independently of the R mark.

substitution of their limits (for $x$ ) into their integrated function (in terms of $x$ ) (M1)

$\ln (k - 4) + k - (\ln 1 + 5)$

${[\ln (x - 4) + x]}_{5}^{k} = \ln (k - 4) + k - 5$ A1

adding their two areas (in terms of $k$ ) and equating to $\ln p + 8$ (M1)

$2 + \ln (k - 4) + k - 5 = \ln p + 8$

equating their non-log terms to $8$ (equation must be in terms of $k$ ) (M1)

$k - 3 = 8$

$k = 11$ A1

$11 - 4 = p$

$p = 7$ A1

[10 marks]

Examiners report

Nearly all candidates knew to set up an equation with $f (x) = g (x)$ in order to find the intersection of the two graphs, and most were able to solve the resulting quadratic equation. Candidates were not as successful in part (b), however. While some candidates recognized that there were two regions to be added together, very few were able to determine the correct boundaries of these regions, with many candidates integrating one or both functions from $x = 3$ to $x = k$ . While a good number of candidates were able to correctly integrate the function(s), without the correct bounds the values of $k$ and $p$ were unattainable.

[N/A]

The expression $\frac{3 \sqrt{x} - 5}{\sqrt{x}}$ can be written as $3 - 5 x^{p}$ . Write down the value of $p$ .

[1]

Hence, find the value of $\int_{1}^{9} (\frac{3 \sqrt{x} - 5}{\sqrt{x}}) d x$ .

[4]

Markscheme

$\frac{3 \sqrt{x} - 5}{\sqrt{x}} = 3 - 5 x^{- \frac{1}{2}}$ A1

$p = - \frac{1}{2}$

[1 mark]

$\int \frac{3 \sqrt{x} - 5}{\sqrt{x}} d x = 3 x - 10 x^{\frac{1}{2}} (+ c)$ A1A1

substituting limits into their integrated function and subtracting (M1)

$3 (9) - 10 {(9)}^{\frac{1}{2}} - (3 (1) - 10 {(1)}^{\frac{1}{2}})$ OR $27 - 10 \times 3 - (3 - 10)$

$= 4$ A1

[4 marks]

Examiners report

Many candidates could give the value of p correctly. However, many did struggle with the integration, including substituting limits into the integrand, without integrating at all. An incorrect value of p often resulted in arithmetic of greater complexity.

[N/A]

In this question, give all answers to two decimal places.

Karl invests 1000 US dollars (USD) in an account that pays a nominal annual interest of 3.5%, compounded quarterly. He leaves the money in the account for 5 years.

Calculate the amount of money he has in the account after 5 years.

[3]

a.i.

Write down the amount of interest he earned after 5 years.

[1]

a.ii.

Karl decides to donate this interest to a charity in France. The charity receives 170 euros (EUR). The exchange rate is 1 USD = t EUR.

Calculate the value of t.

[2]

Markscheme

$1000{\left( {1 + \frac{{3.5}}{{4 \times 100}}} \right)^{4 \times 5}}$ (M1)(A1)

Note: Award (M1) for substitution in compound interest formula, (A1) for correct substitution.

N = 5

I = 3.5

PV = 1000

P/Y = 1

C/Y = 4

Note: Award (A1) for C/Y = 4 seen, (M1) for other correct entries.

N = 5 × 4

I = 3.5

PV = 1000

P/Y = 1

C/Y = 4

Note: Award (A1) for C/Y = 4 seen, (M1) for other correct entries.

= 1190.34 (USD) (A1)

Note: Award (M1) for substitution in compound interest formula, (A1) for correct substitution.

[3 marks]

a.i.

190.34 (USD) (A1)(ft) (C4)

Note: Award (A1)(ft) for subtraction of 1000 from their part (a)(i). Follow through from (a)(i).

[1 mark]

a.ii.

$\frac{{170}}{{190.34}}$ (M1)

Note: Award (M1) for division of 170 by their part (a)(ii).

= 0.89 (A1)(ft) (C2)

Note: Follow through from their part (a)(ii).

[2 marks]

Examiners report

[N/A]

a.i.

[N/A]

a.ii.

[N/A]

The first two terms of an infinite geometric sequence are u₁ = 18 and u₂ = 12sin² θ , where 0 < θ < 2 $\pi$ , and θ ≠ $\pi$ .

Find an expression for r in terms of θ.

[2]

a.i.

Find the values of θ which give the greatest value of the sum.

[6]

Markscheme

valid approach (M1)

eg $\frac{{{u_2}}}{{{u_1}}},\,\,\frac{{{u_1}}}{{{u_2}}}$

$r = \frac{{12\,{{\sin }^2}\,\theta }}{{18}}\left( { = \frac{{2\,{{\sin }^2}\,\theta }}{3}} \right)$ A1 N2

[2 marks]

a.i.

METHOD 1 (using differentiation)

recognizing $\frac{{{\text{d}}{S_\infty }}}{{{\text{d}}\theta }} = 0$ (seen anywhere) (M1)

finding any correct expression for $\frac{{{\text{d}}{S_\infty }}}{{{\text{d}}\theta }}$ (A1)

eg $\frac{{0 - 54 \times \left( { - 2\,{\text{sin}}\,2\,\theta } \right)}}{{{{\left( {2 + {\text{cos}}\,2\,\theta } \right)}^2}}},\,\, - 54{\left( {2 + {\text{cos}}\,2\,\theta } \right)^{ - 2}}\,\left( { - 2\,{\text{sin}}\,2\,\theta } \right)$

correct working (A1)

eg sin 2θ = 0

any correct value for sin⁻¹(0) (seen anywhere) (A1)

eg 0, $\pi$ , … , sketch of sine curve with x-intercept(s) marked both correct values for 2θ (ignore additional values) (A1)

2θ = $\pi$ , 3 $\pi$ (accept values in degrees)

both correct answers $\theta = \frac{\pi }{2},\,\frac{{3\pi }}{2}$ A1 N4

Note: Award A0 if either or both correct answers are given in degrees.
Award A0 if additional values are given.

METHOD 2 (using denominator)

recognizing when S_∞ is greatest (M1)

eg 2 + cos 2θ is a minimum, 1−r is smallest
correct working (A1)

eg minimum value of 2 + cos 2θ is 1, minimum r = $\frac{2}{3}$

correct working (A1)

eg ${\text{cos}}\,2\,\theta = - 1,\,\,\frac{2}{3}\,{\text{si}}{{\text{n}}^2}\,\theta = \frac{2}{3},\,\,{\text{si}}{{\text{n}}^2}\theta = 1$

EITHER (using cos 2θ)

any correct value for cos⁻¹(−1) (seen anywhere) (A1)

eg $\pi$ , 3 $\pi$ , … (accept values in degrees), sketch of cosine curve with x-intercept(s) marked

both correct values for 2θ (ignore additional values) (A1)

2θ = $\pi$ , 3 $\pi$ (accept values in degrees)

OR (using sinθ)

sinθ = ±1 (A1)

sin⁻¹(1) = $\frac{\pi }{2}$ (accept values in degrees) (seen anywhere) A1

THEN

both correct answers $\theta = \frac{\pi }{2},\,\frac{{3\pi }}{2}$ A1 N4

Note: Award A0 if either or both correct answers are given in degrees.
Award A0 if additional values are given.

[6 marks]

Examiners report

[N/A]

a.i.

[N/A]

The company Snakezen’s Ladders makes ladders of different lengths. All the ladders that the company makes have the same design such that:

the first rung is 30 cm from the base of the ladder,

the second rung is 57 cm from the base of the ladder,

the distance between the first and second rung is equal to the distance between all adjacent rungs on the ladder.

The ladder in the diagram was made by this company and has eleven equally spaced rungs.

M17/5/MATSD/SP1/ENG/TZ1/05

Find the distance from the base of this ladder to the top rung.

[3]

The company also makes a ladder that is 1050 cm long.

Find the maximum number of rungs in this 1050 cm long ladder.

[3]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$30 + (11 - 1) \times 27$ (M1)(A1)

Note: Award (M1) for substituted arithmetic sequence formula, (A1) for correct substitutions.

$= 300{\text{ (cm)}}$ (A1) (C3)

Note: Units are not required.

[3 marks]

$1050 \geqslant 30 + (n - 1) \times 27$ (M1)(A1)(ft)

Note: Award (M1) for substituted arithmetic sequence formula $\leqslant 1050$ , accept an equation, (A1) for correct substitutions.

$n = 38$ (A1)(ft) (C3)

Note: Follow through from their 27 in part (a). The answer must be an integer and rounded down.

[3 marks]

Examiners report

[N/A]

Consider the geometric sequence ${u_1} = 18,{\text{ }}{u_2} = 9,{\text{ }}{u_3} = 4.5,{\text{ }} \ldots$ .

Write down the common ratio of the sequence.

[1]

Find the value of ${u_5}$ .

[2]

Find the smallest value of $n$ for which ${u_n}$ is less than ${10^{ - 3}}$ .

[3]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$\frac{1}{2}{\text{ }}(0.5)$ (A1) (C1)

[1 mark]

$18 \times {\left( {\frac{1}{2}} \right)^4}$ (M1)

Note: Award (M1) for their correct substitution into the geometric sequence formula. Accept a list of their five correct terms.

$1.125{\text{ }}\left( {1.13,{\text{ }}\frac{9}{8}} \right)$ (A1)(ft) (C2)

Note: Follow through from their common ratio from part (a).

[2 marks]

$18 \times {\left( {\frac{1}{2}} \right)^{n - 1}} < {10^{ - 3}}$ (M1)(M1)

Notes: Award (M1) for their correct substitution into the geometric sequence formula with a variable in the exponent, (M1) for comparing their expression with ${10^{ - 3}}{\text{ }}\left( {\frac{1}{{1000}}} \right)$ .

Accept an equation.

$n = 16$ (A1)(ft) (C3)

Note: Follow through from their common ratio from part (a). “ $n$ ” must be a positive integer for the (A1) to be awarded.

[3 marks]

Examiners report

[N/A]

Let $g\left( x \right) = {p^x} + q$ , for $x{\text{, }}p{\text{, }}q \in \mathbb{R}{\text{, }}p > 1$ . The point ${\text{A}}\left( {0{\text{, }}a} \right)$ lies on the graph of $g$ .

Let $f\left( x \right) = {g^{ - 1}}\left( x \right)$ . The point ${\text{B}}$ lies on the graph of $f$ and is the reflection of point ${\text{A}}$ in the line $y = x$ .

The line ${L_1}$ is tangent to the graph of $f$ at ${\text{B}}$ .

Write down the coordinates of ${\text{B}}$ .

[2]

Given that $f'\left( a \right) = \frac{1}{{{\text{ln}}\,p}}$ , find the equation of ${L_1}$ in terms of $x$ , $p$ and $q$ .

[5]

The line ${L_2}$ is tangent to the graph of $g$ at ${\text{A}}$ and has equation $y = \left( {{\text{ln}}\,p} \right)x + q + 1$ .

The line ${L_2}$ passes through the point $\left( { - 2{\text{, }} - 2} \right)$ .

The gradient of the normal to $g$ at ${\text{A}}$ is $\frac{1}{{{\text{ln}}\left( {\frac{1}{3}} \right)}}$ .

Find the equation of ${L_1}$ in terms of $x$ .

[7]

Markscheme

${\text{B}}\left( {a{\text{, }}0} \right)$ (accept ${\text{B}}\left( {q + 1{\text{, }}0} \right)$ ) A2 N2

[2 marks]

Note: There are many approaches to this part, and the steps may be done in any order. Please check working and award marks in line with the markscheme, noting that candidates may work with the equation of the line before finding $a$ .

FINDING $a$

valid attempt to find an expression for $a$ in terms of $q$ (M1)

$g\left( 0 \right) = a{\text{, }}\,{p^0} + q = a$

$a = q + 1$ (A1)

FINDING THE EQUATION OF ${L_1}$

EITHER

attempt to substitute tangent gradient and coordinates into equation of straight line (M1)

eg $y - 0 = f'\left( a \right)\left( {x - a} \right){\text{, }}\,y = f'\left( a \right)\left( {x - \left( {q + 1} \right)} \right)$

correct equation in terms of $a$ and $p$ (A1)

eg $y - 0 = \frac{1}{{{\text{ln}}\left( p \right)}}\left( {x - a} \right)$

attempt to substitute tangent gradient and coordinates to find $b$

eg $0 = \frac{1}{{{\text{ln}}\left( p \right)}}\left( a \right) + b$

$b = \frac{{ - a}}{{{\text{ln}}\left( p \right)}}$ (A1)

THEN (must be in terms of both $p$ and $q$ )

$y = \frac{1}{{{\text{ln}}\,p}}\left( {x - q - 1} \right){\text{, }}\,y = \frac{1}{{{\text{ln}}\,p}}x - \frac{{q + 1}}{{{\text{ln}}\,p}}$ A1 N3

Note: Award A0 for final answers in the form ${L_1} = \frac{1}{{{\text{ln}}\,p}}\left( {x - q - 1} \right)$

[5 marks]

Note: There are many approaches to this part, and the steps may be done in any order. Please check working and award marks in line with the markscheme, noting that candidates may find $q$ in terms of $p$ before finding a value for $p$ .

FINDING $p$

valid approach to find the gradient of the tangent (M1)

eg ${m_1}{m_2} = - 1{\text{, }}\,\, - \frac{1}{{\frac{1}{{{\text{ln}}\left( {\frac{1}{3}} \right)}}}}{\text{, }}\,\, - {\text{ln}}\left( {\frac{1}{3}} \right){\text{, }}\,\, - \frac{1}{{{\text{ln}}\,p}} = \frac{1}{{{\text{ln}}\left( {\frac{1}{3}} \right)}}$

correct application of log rule (seen anywhere) (A1)

eg ${\text{ln}}{\left( {\frac{1}{3}} \right)^{ - 1}}{\text{, }}\,\, - \left( {{\text{ln}}\left( 1 \right) - {\text{ln}}\left( 3 \right)} \right)$

correct equation (seen anywhere) A1

eg ${\text{ln}}\,p = {\text{ln}}\,3{\text{, }}\,\,p = 3$

FINDING $q$

correct substitution of $\left( { - 2{\text{, }} - 2} \right)$ into ${L_2}$ equation (A1)

eg $- 2 = \left( {{\text{ln}}\,p} \right)\left( { - 2} \right) + q + 1$

$q = 2\,{\text{ln}}\,p - 3{\text{, }}\,\,q = 2\,{\text{ln}}\,3 - 3$ (seen anywhere) A1

FINDING ${L_1}$

correct substitution of their $p$ and $q$ into their ${L_1}$ (A1)

eg $y = \frac{1}{{{\text{ln}}\,3}}\left( {x - \left( {2\,{\text{ln}}\,3 - 3} \right) - 1} \right)$

$y = \frac{1}{{{\text{ln}}\,3}}\left( {x - 2\,{\text{ln}}\,3 + 2} \right){\text{, }}\,\,y = \frac{1}{{{\text{ln}}\,3}}x - \frac{{2\,{\text{ln}}\,3 - 2}}{{{\text{ln}}\,3}}$ A1 N2

Note: Award A0 for final answers in the form ${L_1} = \frac{1}{{{\text{ln}}\,3}}\left( {x - 2\,{\text{ln}}\,3 + 2} \right)$ .

[7 marks]

Examiners report

[N/A]

The speed of light is ${\text{300}}\,{\text{000}}$ kilometres per second. The average distance from the Sun to the Earth is 149.6 million km.

A light-year is the distance light travels in one year and is equal to ${\text{9}}\,{\text{467}}\,{\text{280}}$ million km. Polaris is a bright star, visible from the Northern Hemisphere. The distance from the Earth to Polaris is 323 light-years.

Calculate the time, in minutes, it takes for light from the Sun to reach the Earth.

[3]

Find the distance from the Earth to Polaris in millions of km. Give your answer in the form $a \times {10^k}$ with $1 \leqslant a < 10$ and $k \in \mathbb{Z}$ .

[3]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$\frac{{149600000}}{{300000 \times 60}}$ (M1)(M1)

Note: Award (M1) for dividing the correct numerator (which can be presented in a different form such as $149.6 \times {10^6}$ or $1.496 \times {10^8}$ ) by ${\text{300}}\,{\text{000}}$ and (M1) for dividing by 60.

$= 8.31{\text{ }}({\text{minutes}}){\text{ }}(8.31111 \ldots {\text{, 8 minutes 19 seconds}})$ (A1) (C3)

[3 marks]

$323 \times 9467\,280$ (M1)

Note: Award (M1) for multiplying 323 by $9\,467\,280$ , seen with any power of 10; therefore only penalizing incorrect power of 10 once.

$= 3.06 \times {10^9}{\text{ ( = }}3.05793 \ldots \times {10^9})$ (A1)(A1) (C3)

Note: Award (A1) for 3.06.

Award (A1) for $\times {10^9}$

Award (A0)(A0) for answers of the type: $30.6 \times {10^8}$

[3 marks]

Examiners report

[N/A]

A sphere with diameter 3 474 000 metres can model the shape of the Moon.

Use this model to calculate the circumference of the Moon in kilometres. Give your full calculator display.

[3]

Give your answer to part (a) correct to three significant figures.

[1]

Write your answer to part (b) in the form $a \times {10^k}$ , where 1 ≤ $a$ < 10 , $k \in \mathbb{Z}$ .

[2]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$\frac{{3\,474\,000 \times \pi }}{{1000}}$ (M1)(M1)

Note: Award (M1) for correct numerator and (M1) for dividing by 1000 OR equivalent, such as $\frac{{3\,474\,000 \times 2 \times \pi }}{{2000}}$ ie diameter.
Do not accept use of area formula ie $\pi {r^2}$ .

10 913.89287… (km) (A1) (C3)

[3 marks]

10 900 (km) (A1)(ft) (C1)

Note: Follow through from part (a).

[1 mark]

1.09 × 10⁴ (A1)(ft)(A1)(ft) (C2)

Note: Follow through from part (b) only. Award (A1)(ft) for 1.09, and (A1)(ft) × 10⁴. Award (A0)(A0) for answers of the type: 10.9 × 10³.

[2 marks]

Examiners report

[N/A]

The first three terms of an arithmetic sequence are $u_{1}, 5 u_{1} - 8$ and $3 u_{1} + 8$ .

Show that $u_{1} = 4$ .

[2]

Prove that the sum of the first $n$ terms of this arithmetic sequence is a square number.

[4]

Markscheme

EITHER

uses $u_{2} - u_{1} = u_{3} - u_{2}$ (M1)

$(5 u_{1} - 8) - u_{1} = (3 u_{1} + 8) - (5 u_{1} - 8)$

$6 u_{1} = 24$ A1

uses $u_{2} = \frac{u_{1} + u_{3}}{2}$ (M1)

$5 u_{1} - 8 = \frac{u_{1} + (3 u_{1} + 8)}{2}$

$3 u_{1} = 12$ A1

THEN

so $u_{1} = 4$ AG

[2 marks]

$d = 8$ (A1)

uses $S_{n} = \frac{n}{2} (2 u_{1} + (n - 1) d)$ M1

$S_{n} = \frac{n}{2} (8 + 8 (n - 1))$ A1

$= 4 n^{2}$

$= {(2 n)}^{2}$ A1

Note: The final A1 can be awarded for clearly explaining that $4 n^{2}$ is a square number.

so sum of the first $n$ terms is a square number AG

[4 marks]

Examiners report

[N/A]

Consider an arithmetic sequence where $u_{8} = S_{8} = 8$ . Find the value of the first term, $u_{1}$ , and the value of the common difference, $d$ .

Markscheme

METHOD 1 (finding $u_{1}$ first, from $S_{8}$ )

$4 (u_{1} + 8) = 8$ (A1)

$u_{1} = - 6$ A1

$u_{1} + 7 d = 8$ OR $4 (2 u_{1} + 7 d) = 8$ (may be seen with their value of $u_{1}$ ) (A1)

attempt to substitute their $u_{1}$ (M1)

$d = 2$ A1

METHOD 2 (solving simultaneously)

$u_{1} + 7 d = 8$ (A1)

$4 (u_{1} + 8) = 8$ OR $4 (2 u_{1} + 7 d) = 8$ OR $u_{1} = - 3 d$ (A1)

attempt to solve linear or simultaneous equations (M1)

$u_{1} = - 6, d = 2$ A1A1

[5 marks]

Examiners report

[N/A]

Consider the numbers $p = 2.78 \times {10^{11}}$ and $q = 3.12 \times {10^{ - 3}}$ .

Calculate $\sqrt[3]{{\frac{p}{q}}}$ . Give your full calculator display.

[2]

Write down your answer to part (a) correct to two decimal places;

[1]

b.i.

Write down your answer to part (a) correct to three significant figures.

[1]

b.ii.

Write your answer to part (b)(ii) in the form $a \times {10^k}$ , where $1 \leqslant a < 10,{\text{ }}k \in \mathbb{Z}$ .

[2]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$\sqrt[3]{{\frac{{2.78 \times {{10}^{11}}}}{{3.12 \times {{10}^{ - 3}}}}}}$ $\,\,\,$ OR $\,\,\,$ $\sqrt[3]{{8.91025 \ldots \times {{10}^{13}}}}$ (M1)

Note: Award (M1) for correct substitution into given expression.

44664.59503 (A1) (C2)

Note: Award (A1) for a correct answer with at least 8 digits.

Accept 44664.5950301.

[2 marks]

44664.60 (A1)(ft) (C1)

Note: For a follow through mark, the answer to part (a) must be to at least 3 decimal places.

[1 mark]

b.i.

44700 (A1)(ft) (C1)

Notes: Answer to part (a) must be to at least 4 significant figures.

Accept any equivalent notation which is correct to 3 significant figures.

For example $447 \times {10^2}$ or $44.7 \times {10^3}$ .

Follow through from part (a).

[1 mark]

b.ii.

$4.47 \times {10^4}$ (A1)(ft)(A1)(ft) (C2)

Notes: Award (A1)(ft) for 4.47 and (A1)(ft) for ${10^4}$ .

Award (A0)(A0) for answers such as $44.7 \times {10^3}$ .

Follow through from part (b)(ii) only.

[2 marks]

Examiners report

[N/A]

b.i.

[N/A]

b.ii.

[N/A]

Give your answers in this question correct to the nearest whole number.

Imon invested $25 000$ Singapore dollars ( $SGD$ ) in a fixed deposit account with a nominal annual interest rate of $3.6 %$ , compounded monthly.

Calculate the value of Imon’s investment after $5$ years.

[3]

At the end of the $5$ years, Imon withdrew $x SGD$ from the fixed deposit account and reinvested this into a super-savings account with a nominal annual interest rate of $5.7 %$ , compounded half-yearly.

The value of the super-savings account increased to $20 000 SGD$ after $18$ months.

Find the value of $x$ .

[3]

Markscheme

$(F V =) 25 000 \times {(1 + \frac{3.6}{100 \times 12})}^{12 \times 5}$ (M1)(A1)

Note: Award (M1) for substituted compound interest formula, (A1) for correct substitutions.

OR

$N = 5$
$I % = 3.6$
$P V = \mp 25 000$
$P / Y = 1$
$C / Y = 12$ (A1)(M1)

Note: Award (A1) for $C / Y = 12$ seen, (M1) for all other correct entries.

OR

$N = 60$
$I % = 3.6$
$P V = \mp 25 000$
$P / Y = 12$
$C / Y = 12$ (A1)(M1)

Note: Award (A1) for $C / Y = 12$ seen, (M1) for all other correct entries.

$(F V =) 29 922 (SGD)$ (A1) (C3)

Note: Do not award the final (A1) if answer is not given correct to the nearest integer.

[3 marks]

$20 000 = P V \times {(1 + \frac{5.7}{100 \times 2})}^{2 \times 1.5}$ (M1)(A1)

Note: Award (M1) for substituted compound interest equated to $20 000$ . Award (A1) for correct substitutions.

OR

$N = 1.5$
$I % = 5.7$
$F V = \pm 20 000$
$P / Y = 1$
$C / Y = 2$ (A1)(M1)

Note: Award (A1) for $C / Y = 2$ seen, (M1) for all other correct entries.

OR

$N = 3$
$I % = 5.7$
$F V = \pm 20 000$
$P / Y = 2$
$C / Y = 2$ (A1)(M1)

Note: Award (A1) for $C / Y = 2$ seen, (M1) for all other correct entries.

$(x =) 18 383 (SGD)$ (A1) (C3)

Note: Do not award the final (A1) if answer is not given correct to the nearest integer (unless already penalized in part(a)).

[3 marks]

Examiners report

[N/A]

Consider $f(x) = \sqrt x \,{\text{sin}}\left( {\frac{\pi }{4}x} \right)$ and $g(x) = \sqrt x$ for $x$ ≥ 0. The first time the graphs of $f$ and $g$ intersect is at $x = 0$ .

The set of all non-zero values that satisfy $f(x) = g(x)$ can be described as an arithmetic sequence, ${u_n} = a + bn$ where $n$ ≥ 1.

Find the two smallest non-zero values of $x$ for which $f(x) = g(x)$ .

[5]

At point P, the graphs of $f$ and $g$ intersect for the 21st time. Find the coordinates of P.

[4]

The following diagram shows part of the graph of $g$ reflected in the $x$ -axis. It also shows part of the graph of $f$ and the point P.

Find an expression for the area of the shaded region. Do not calculate the value of the expression.

[4]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

correct working (A1)

eg ${\text{sin}}\left( {\frac{\pi }{4}x} \right) = 1$ , $\sqrt x \left( {1 - {\text{sin}}\left( {\frac{\pi }{4}x} \right)} \right) = 0$

${\text{sin}}\left( {\frac{\pi }{2}} \right) = 1$ (seen anywhere) (A1)

correct working (ignore additional values) (A1)

eg $\frac{\pi }{4}x = \frac{\pi }{2}$ , $\frac{\pi }{4}x = \frac{\pi }{2} + 2\pi$

$x$ = 2, 10 A1A1 N1N1

[5 marks]

valid approach (M1)

eg first intersection at $x = 0$ , $n = 20$

correct working A1

eg $- 6 + 8 \times 20$ , $2 + \left( {20 - 1} \right) \times 8$ , ${u_{20}} = 154$

P(154, $\sqrt {154}$ ) (accept $x = 154$ and $y = \sqrt {154}$ ) A1A1 N3

[4 marks]

valid attempt to find upper boundary (M1)

eg half way between ${u_{20}}$ and ${u_{21}}$ , ${u_{20}} + \frac{d}{2}$ , 154 + 4, $- 2 + 8n$ , at least two values of new sequence {6, 14, ...}

upper boundary at $x = 158$ (seen anywhere) (A1)

correct integral expression (accept missing ${{\text{d}}x}$ ) A1A1 N4

eg $\int_0^{158} {\left( {\sqrt x \,{\text{sin}}\left( {\frac{\pi }{4}x} \right) + \sqrt x } \right)} {\text{d}}x$ , $\int_0^{158} {\left( {g + f} \right)} \left. {{\text{d}}x} \right)$ , $\int_0^{158} {\sqrt x \,{\text{sin}}\left( {\frac{\pi }{4}x} \right)} {\text{d}}x - \int_0^{158} { - \sqrt x \,{\text{d}}x}$

Note: Award A1 for two correct limits and A1 for correct integrand. The A1 for correct integrand may be awarded independently of all the other marks.

[4 marks]

Examiners report

[N/A]

In an arithmetic sequence, ${u_2} = 5$ and ${u_3} = 11$ .

Find the common difference.

[2]

Find the first term.

[2]

Find the sum of the first $20$ terms.

[2]

Markscheme

valid approach (M1)

eg $11 - 5$ , $11 = 5 + d$

$d = 6$ A1 N2

[2 marks]

valid approach (M1)

eg ${u_2} - d$ , $5 - 6$ , ${u_1} + \left( {3 - 1} \right)\left( 6 \right) = 11$

${u_1} = - 1$ A1 N2

[2 marks]

correct substitution into sum formula

eg $\frac{{20}}{2}\left( {2\left( { - 1} \right) + 19\left( 6 \right)} \right)$ , $\frac{{20}}{2}\left( { - 1 + 113} \right)$ (A1)

${S_{20}} = 1120$ A1 N2

[2 marks]

Examiners report

[N/A]

In an arithmetic sequence, u₁ = −5 and d = 3.

Find u₈.

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

correct working (A1)

eg −5 + (8 − 1)(3)

u₈ = 16 A1 N2

[2 marks]

Examiners report

[N/A]

Explain why any integer can be written in the form $4k$ or $4k + 1$ or $4k + 2$ or $4k + 3$ , where $k \in \mathbb{Z}$ .

[2]

Hence prove that the square of any integer can be written in the form $4t$ or $4t + 1$ , where $t \in {\mathbb{Z}^ + }$ .

[6]

Markscheme

Upon division by 4 M1

any integer leaves a remainder of 0, 1, 2 or 3. R1

Hence, any integer can be written in the form $4k$ or $4k + 1$ or $4k + 2$ or $4k + 3$ , where $k \in \mathbb{Z}$ AG

[2 marks]

${\left( {4k} \right)^2} = 16{k^2} = 4t$ M1A1

${\left( {4k + 1} \right)^2} = 16{k^2} + 8k + 1 = 4t + 1$ M1A1

${\left( {4k + 2} \right)^2} = 16{k^2} + 16k + 4 = 4t$ A1

${\left( {4k + 3} \right)^2} = 16{k^2} + 24k + 9 = 4t + 1$ A1

Hence, the square of any integer can be written in the form $4t$ or $4t + 1$ , where $t \in {\mathbb{Z}^ + }$ . AG

[6 marks]

Examiners report

[N/A]

The following diagram shows [CD], with length $b{\text{ cm}}$ , where $b > 1$ . Squares with side lengths $k{\text{ cm}},{\text{ }}{k^2}{\text{ cm}},{\text{ }}{k^3}{\text{ cm}},{\text{ }} \ldots$ , where $0 < k < 1$ , are drawn along [CD]. This process is carried on indefinitely. The diagram shows the first three squares.

N17/5/MATME/SP1/ENG/TZ0/10.b

The total sum of the areas of all the squares is $\frac{9}{{16}}$ . Find the value of $b$ .

Markscheme

recognizing infinite geometric series with squares (M1)

eg $\,\,\,\,\,$ ${k^2} + {k^4} + {k^6} + \ldots ,{\text{ }}\frac{{{k^2}}}{{1 - {k^2}}}$

correct substitution into ${S_\infty } = \frac{9}{{16}}$ (must substitute into formula) (A2)

eg $\,\,\,\,\,$ $\frac{{{k^2}}}{{1 - {k^2}}} = \frac{9}{{16}}$

correct working (A1)

eg $\,\,\,\,\,$ $16{k^2} = 9 - 9{k^2},{\text{ }}25{k^2} = 9,{\text{ }}{k^2} = \frac{9}{{25}}$

$k = \frac{3}{5}$ (seen anywhere) A1

valid approach with segments and CD (may be seen earlier) (M1)

eg $\,\,\,\,\,$ $r = k,{\text{ }}{S_\infty } = b$

correct expression for $b$ in terms of $k$ (may be seen earlier) (A1)

eg $\,\,\,\,\,$ $b = \frac{k}{{1 - k}},{\text{ }}b = \sum\limits_{n = 1}^\infty {{k^n},{\text{ }}b = k + {k^2} + {k^3} + \ldots }$

substituting their value of $k$ into their formula for $b$ (M1)

eg $\,\,\,\,\,$ $\frac{{\frac{3}{5}}}{{1 - \frac{3}{5}}},{\text{ }}\frac{{\left( {\frac{3}{5}} \right)}}{{\left( {\frac{2}{5}} \right)}}$

$b = \frac{3}{2}$ A1 N3

[9 marks]

Examiners report

[N/A]

A comet orbits the Sun and is seen from Earth every 37 years. The comet was first seen from Earth in the year 1064.

Find the year in which the comet was seen from Earth for the fifth time.

[3]

Determine how many times the comet has been seen from Earth up to the year 2014.

[3]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$1064 + (5 - 1) \times 37$ (M1)(A1)

Note: Award (M1) for substituted arithmetic sequence formula, (A1) for correct substitution.

$= 1212$ (A1) (C3)

[3 marks]

$2014 > 1064 + (n - 1) \times 37$ (M1)

Note: Award (M1) for a correct substitution into arithmetic sequence formula.

Accept an equation.

$(n < ){\text{ }}26.6756 \ldots$ (A1)

26 (times) (A1) (C3)

Note: Award the final (A1) for the correct rounding down of their unrounded answer.

$2014 > 1064 + 37t$ (M1)

Note: Award (M1) for a correct substitution into a linear model (where $t = n - 1$ ).

Accept an equation or weak inequality.

Accept $\frac{{2014 - 1064}}{{37}}$ for (M1).

$(t < ){\text{ }}25.6756 \ldots$ (A1)

26 (times) (A1) (C3)

Note: Award the final (A1) for adding 1 to the correct rounding down of their unrounded answer.

[3 marks]

Examiners report

[N/A]

In the expansion of $(x + k)^{7}$ , where $k \in ℝ$ , the coefficient of the term in $x^{5}$ is $63$ .

Find the possible values of $k$ .

Markscheme

EITHER

attempt to use the binomial expansion of ${(x + k)}^{7}$ (M1)

${}^{7}C_{0} x^{7} k^{0} + {}^{7}C_{1} x^{6} k^{1} + {}^{7}C_{2} x^{5} k^{2} + \dots$ (or ${}^{7}C_{0} k^{7} x^{0} + {}^{7}C_{1} k^{5} x^{1} + {}^{7}C_{2} k^{5} x^{2} + \dots$ )

identifying the correct term ${}^{7}C_{2} x^{5} k^{2}$ (or ${}^{7}C_{5} k^{2} x^{5}$ ) (A1)

attempt to use the general term ${}^{7}C_{r} x^{r} k^{7 - r}$ (or ${}^{7}C_{r} k^{r} x^{7 - r}$ ) (M1)

$r = 2$ (or $r = 5$ ) (A1)

THEN

${}^{7}C_{2} = 21$ (or ${}^{7}C_{5} = 21$ (seen anywhere) (A1)

$21 x^{5} k^{2} = 63 x^{5} (21 k^{2} = 63, k^{2} = 3)$ A1

$k = \pm \sqrt{3}$ A1

Note: If working shown, award M1A1A1A1A0 for $k = \sqrt{3}$ .

[5 marks]

Examiners report

[N/A]

The first three terms of a geometric sequence are $\ln {x^{16}}$ , $\ln {x^8}$ , $\ln {x^4}$ , for $x > 0$ .

Find the common ratio.

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

correct use $\log {x^n} = n\log x$ A1

eg $\,\,\,\,\,$ $16\ln x$

valid approach to find $r$ (M1)

eg $\,\,\,\,\,$ $\frac{{{u_{n + 1}}}}{{{u_n}}},{\text{ }}\frac{{\ln {x^8}}}{{\ln {x^{16}}}},{\text{ }}\frac{{4\ln x}}{{8\ln x}},{\text{ }}\ln {x^4} = \ln {x^{16}} \times {r^2}$

$r = \frac{1}{2}$ A1 N2

[3 marks]

Examiners report

[N/A]

Show that ${\text{lo}}{{\text{g}}_9}\left( {{\text{cos}}\,2x + 2} \right) = {\text{lo}}{{\text{g}}_3}\sqrt {{\text{cos}}\,2x + 2}$ .

[3]

Hence or otherwise solve ${\text{lo}}{{\text{g}}_3}\left( {{\text{2}}\,{\text{sin}}\,x} \right) = {\text{lo}}{{\text{g}}_9}\left( {{\text{cos}}\,2x + 2} \right)$ for $0 < x < \frac{\pi }{2}$ .

[5]

Markscheme

attempting to use the change of base rule M1

${\text{lo}}{{\text{g}}_9}\left( {{\text{cos}}\,2x + 2} \right) = \frac{{{\text{lo}}{{\text{g}}_3}\left( {{\text{cos}}\,2x + 2} \right)}}{{{\text{lo}}{{\text{g}}_3}9}}$ A1

$= \frac{1}{2}{\text{lo}}{{\text{g}}_3}\left( {{\text{cos}}\,2x + 2} \right)$ A1

$= {\text{lo}}{{\text{g}}_3}\sqrt {{\text{cos}}\,2x + 2}$ AG

[3 marks]

${\text{lo}}{{\text{g}}_3}\left( {{\text{2}}\,{\text{sin}}\,x} \right) = {\text{lo}}{{\text{g}}_3}\sqrt {{\text{cos}}\,2x + 2}$

${\text{2}}\,{\text{sin}}\,x = \sqrt {{\text{cos}}\,2x + 2}$ M1

${\text{4}}\,{\text{si}}{{\text{n}}^2}\,x = {\text{cos}}\,2x + 2$ (or equivalent) A1

use of ${\text{cos}}\,2x = 1 - 2\,{\text{si}}{{\text{n}}^2}\,x$ (M1)

$6\,{\text{si}}{{\text{n}}^2}\,x = 3$

${\text{sin}}\,x = \left( \pm \right)\frac{1}{{\sqrt 2 }}$ A1

$x = \frac{\pi }{4}$ A1

Note: Award A0 if solutions other than $x = \frac{\pi }{4}$ are included.

[5 marks]

Examiners report

[N/A]

Consider the curve with equation $y = (2 x - 1) e^{k x}$ , where $x \in ℝ$ and $k \in ℚ$ .

The tangent to the curve at the point where $x = 1$ is parallel to the line $y = 5 e^{k} x$ .

Find the value of $k$ .

Markscheme

evidence of using product rule (M1)

$\frac{d y}{d x} = (2 x - 1) \times (k e^{k x}) + 2 \times e^{k x} (= e^{k x} (2 k x - k + 2))$ A1

correct working for one of (seen anywhere) A1

$\frac{d y}{d x}$ at $x = 1 \Rightarrow k e^{k} + 2 e^{k}$

slope of tangent is $5 e^{k}$

their $\frac{d y}{d x}$ at $x = 1$ equals the slope of $y = 5 e^{k} x (= 5 e^{k})$ (seen anywhere) (M1)

$k e^{k} + 2 e^{k} = 5 e^{k}$

$k = 3$ A1

[5 marks]

Examiners report

The product rule was well recognised and used with 𝑥=1 properly substituted into this expression. Although the majority of the candidates tried equating the derivative to the slope of the tangent line, the slope of the tangent line was not correctly identified; many candidates incorrectly substituted 𝑥=1 into the tangent equation, thus finding the y-coordinate instead of the slope.

The diameter of a spherical planet is $6 \times 10^{4} km$ .

Write down the radius of the planet.

[1]

The volume of the planet can be expressed in the form $π (a \times 10^{k}) {km}^{3}$ where $1 \leq a < 10$ and $k \in ℤ$ .

Find the value of $a$ and the value of $k$ .

[3]

Markscheme

$3 \times 10^{4}$ OR $30000 (km)$ (accept $3 ∙ 10^{4}$ ) A1

[1 mark]

$\frac{4}{3} π {(3 \times 10^{4})}^{3}$ OR $\frac{4}{3} π {(30000)}^{3}$ (A1)

$= \frac{4}{3} π \times 27 \times 10^{12} (= π (36 \times 10^{12}))$ OR $= \frac{4}{3} π \times 27000000000000$ (A1)

$= π (36 \times 10^{13}) ({km}^{3})$ OR $a = 3.6, k = 13$ A1

[3 marks]

Examiners report

[N/A]

Consider two consecutive positive integers, $n$ and $n + 1$ .

Show that the difference of their squares is equal to the sum of the two integers.

Markscheme

attempt to subtract squares of integers (M1)

${(n + 1)}^{2} - n^{2}$

EITHER

correct order of subtraction and correct expansion of ${(n + 1)}^{2}$ , seen anywhere A1A1

$= n^{2} + 2 n + 1 - n^{2} (= 2 n + 1)$

correct order of subtraction and correct factorization of difference of squares A1A1

$= (n + 1 - n) (n + 1 + n) (= 2 n + 1)$

THEN

$= n + n + 1 = RHS$ A1

Note: Do not award final A1 unless all previous working is correct.

which is the sum of $n$ and $n + 1$ AG

Note: If expansion and order of subtraction are correct, award full marks for candidates who find the sum of the integers as $2 n + 1$ and then show that the difference of the squares (subtracted in the correct order) is $2 n + 1$ .

[4 marks]

Examiners report

[N/A]

Tomás is playing with sticks and he forms the first three diagrams of a pattern. These diagrams are shown below.

M17/5/MATSD/SP1/ENG/TZ2/05

Tomás continues forming diagrams following this pattern.

Tomás forms a total of 24 diagrams.

Diagram $n$ is formed with 52 sticks. Find the value of $n$ .

[3]

Find the total number of sticks used by Tomás for all 24 diagrams.

[3]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$4 + 3(n - 1) = 52$ (M1)(A1)

Note: Award (M1) for substitution into the formula of the $n$ th term of an arithmetic sequence, (A1) for correct substitution.

$n = 17$ (A1) (C3)

[3 marks]

$\frac{{24}}{2}(2 \times 4 + 23 \times 3)$ $\,\,\,$ OR $\,\,\,$ $\frac{{24}}{2}(4 + 73)$ (M1)(A1)(ft)

Notes: Award (M1) for substitution into the sum of the first $n$ terms of an arithmetic sequence formula, (A1)(ft) for their correct substitution, consistent with part (a).

924 (A1)(ft) (C3)

Note: Follow through from part (a).

[3 marks]

Examiners report

[N/A]

Jashanti is saving money to buy a car. The price of the car, in US Dollars (USD), can be modelled by the equation

$P = 8500{\text{ }}{(0.95)^t}.$

Jashanti’s savings, in USD, can be modelled by the equation

$S = 400t + 2000.$

In both equations $t$ is the time in months since Jashanti started saving for the car.

Jashanti does not want to wait too long and wants to buy the car two months after she started saving. She decides to ask her parents for the extra money that she needs.

Write down the amount of money Jashanti saves per month.

[1]

Use your graphic display calculator to find how long it will take for Jashanti to have saved enough money to buy the car.

[2]

Calculate how much extra money Jashanti needs.

[3]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

400 (USD) (A1) (C1)

[1 mark]

$8500{\text{ }}{(0.95)^t} = 400 \times t + 2000$ (M1)

Note: Award (M1) for equating $8500{(0.95)^t}$ to $400 \times t + 2000$ or for comparing the difference between the two expressions to zero or for showing a sketch of both functions.

$(t = ){\text{ }}8.64{\text{ (months) }}\left( {8.6414 \ldots {\text{ (months)}}} \right)$ (A1) (C2)

Note: Accept 9 months.

[2 marks]

$8500{(0.95)^2} - (400 \times 2 + 2000)$ (M1)(M1)

Note: Award (M1) for correct substitution of $t = 2$ into equation for $P$ , (M1) for finding the difference between a value/expression for $P$ and a value/expression for $S$ . The first (M1) is implied if 7671.25 seen.

4870 (USD) (4871.25) (A1) (C3)

Note: Accept 4871.3.

[3 marks]

Examiners report

[N/A]