Find \({\log _2}32\) .

Given that \({\log _2}\left( {\frac{{{{32}^x}}}{{{8^y}}}} \right)\) can be written as \(px + qy\) , find the value of p and of q.

5     A1     N1

[1 mark]

METHOD 1

\({\log _2}\left( {\frac{{{{32}^x}}}{{{8^y}}}} \right) = {\log _2}{32^x} - {\log _2}{8^y}\)     (A1)

\( = x{\log _2}32 - y{\log _2}8\)     (A1)

\({\log _2}8 = 3\)     (A1)

\(p = 5\) , \(q = - 3\) (accept \(5x - 3y\) )     A1      N3 

METHOD 2

\(\frac{{{{32}^x}}}{{{8^y}}} = \frac{{{{({2^5})}^x}}}{{{{({2^3})}^y}}}\)     (A1) 

\( = \frac{{{2^5}^x}}{{{2^3}^y}}\)     (A1)

\( = {2^{5x - 3y}}\)     (A1)

\({\log _2}({2^{5x - 3y}}) = 5x - 3y\)

\(p = 5\) , \(q = - 3\) (accept \(5x - 3y\) )     A1      N3

[4 marks]

Part (a) proved very accessible.

Although many found part (b) accessible as well, a good number of candidates could not complete their way to a final result. Many gave q as a positive value.

\({\log _6}36\)

\({\log _6}4 + {\log _6}9\)

\({\log _6}2 - {\log _6}12\)

correct approach     (A1)

eg     \({6^x} = 36,{\text{ }}{6^2}\)

\(2\)      A1     N2

[2 marks]

correct simplification     (A1)

eg     \({\log _6}36,{\text{ }}\log (4 \times 9)\)

\(2\)      A1      N2

[2 marks]

correct simplification     (A1)

eg     \({\log _6}\frac{2}{{12}},{\text{ }}\log (2 \div 12)\)

correct working     (A1)

eg     \({\log _6}\frac{1}{6},{\text{ }}{6^{ - 1}} = \frac{1}{6}{,6^x} = \frac{1}{6}\)

\(-1\)     A1     N2

[3 marks]

Expand \({(2 + x)^4}\) and simplify your result.

Hence, find the term in \({x^2}\) in \({(2 + x)^4}\left( {1 + \frac{1}{{{x^2}}}} \right)\) .

evidence of expanding     M1

e.g. \({2^4} + 4({2^3})x + 6({2^2}){x^2} + 4(2){x^3} + {x^4}\) , \((4 + 4x + {x^2})(4 + 4x + {x^2})\)

\({(2 + x)^4} = 16 + 32x + 24{x^2} + 8{x^3} + {x^4}\)     A2     N2

[3 marks]

finding coefficients 24 and 1     (A1)(A1)

term is \(25{x^2}\)     A1     N3

[3 marks]

Surprisingly few candidates employed the binomial theorem, choosing instead to expand by repeated use of the distributive property. This earned full marks if done correctly, but often proved prone to error.

Candidates often expanded the entire expression in part (b). Few recognized that only two distributions are required to answer the question. Some gave the coefficient as the final answer.

An arithmetic sequence has the first term \(\ln a\) and a common difference \(\ln 3\).

The 13th term in the sequence is \(8\ln 9\). Find the value of \(a\).

Note:     There are many approaches to this question, and the steps may be done in any order. There are 3 relationships they may need to apply at some stage, for the 3rd, 4th and 5th marks. These are

equating bases eg recognising 9 is \({{\text{3}}^2}\)

log rules: \(\ln b + \ln c = \ln (bc),{\text{ }}\ln b - \ln c = \ln \left( {\frac{b}{c}} \right)\),

exponent rule: \(\ln {b^n} = n\ln b\).

The exception to the FT rule applies here, so that if they demonstrate correct application of the 3 relationships, they may be awarded the A marks, even if they have made a previous error. However all applications of a relationship need to be correct. Once an error has been made, do not award A1FT for their final answer, even if it follows from their working.

Please check working and award marks in line with the markscheme.

correct substitution into \({u_{13}}\) formula     (A1)

eg\(\;\;\;\ln a + (13 - 1)\ln 3\)

set up equation for \({u_{13}}\) in any form (seen anywhere)     (M1)

eg\(\;\;\;\ln a + 12\ln 3 = 8\ln 9\)

correct application of relationships     (A1)(A1)(A1)

\(a = 81\)     A1     N3

[6 marks]

Examples of application of relationships

Example 1

correct application of exponent rule for logs     (A1)

eg\(\;\;\;\ln a + \ln {3^{12}} = \ln {9^8}\)

correct application of addition rule for logs     (A1)

eg\(\;\;\;\ln (a{3^{12}}) = \ln {9^8}\)

substituting for 9 or 3 in ln expression in equation     (A1)

eg\(\;\;\;\ln (a{3^{12}}) = \ln {3^{16}},{\text{ }}\ln (a{9^6}) = \ln {9^8}\)

Example 2

recognising \(9 = {3^2}\)      (A1)

eg\(\;\;\;\ln a + 12\ln 3 = 8\ln {3^2},{\text{ }}\ln a + 12\ln {9^{\frac{1}{2}}} = 8\ln 9\)

one correct application of exponent rule for logs relating \(\ln 9\) to \(\ln 3\)     (A1)

eg\(\;\;\;\ln a + 12\ln 3 = 16\ln 3,{\text{ }}\ln a + 6\ln 9 = 8\ln 9\)

another correct application of exponent rule for logs     (A1)

eg\(\;\;\;\ln a = \ln {3^4},{\text{ }}\ln a = \ln {9^2}\)

Given that Figure \(n\) contains 801 line segments, show that \(n = 200\).

Find the total number of line segments in the first 200 figures.

recognizing that it is an arithmetic sequence     (M1)

eg\(\,\,\,\,\,\)\(5,{\text{ }}5 + 4,{\text{ }}5 + 4 + 4,{\text{ }} \ldots ,{\text{ }}d = 4,{\text{ }}{u_n} = {u_1} + (n - 1)d,{\text{ }}4n + 1\)

correct equation     A1

eg\(\,\,\,\,\,\)\(5 + 4(n - 1) = 801\)

correct working (do not accept substituting \(n = 200\))     A1

eg\(\,\,\,\,\,\)\(4n - 4 = 796,{\text{ }}n - 1 = \frac{{796}}{4}\)

\(n = 200\)    AG     N0

[3 marks]

recognition of sum     (M1)

eg\(\,\,\,\,\,\)\({S_{200}},{\text{ }}{u_1} + {u_2} +  \ldots  + {u_{200}},{\text{ }}5 + 9 + 13 +  \ldots  + 801\)

correct working for AP     (A1)

eg\(\,\,\,\,\,\)\(\frac{{200}}{2}(5 + 801),{\text{ }}\frac{{200}}{2}{\text{ }}\left( {2(5) + 199(4)} \right)\)

\(80\,600\)     A1     N2

[3 marks]

Most candidates recognized that the series was arithmetic but many worked backwards using \(n = 200\) rather than creating and solving an equation of their own to produce the given answer.

Almost all students answered (b) correctly.

Given that \({f^{ - 1}}(1) = 8\) , find the value of \(k\) .

Find \({f^{ - 1}}\left( {\frac{2}{3}} \right)\) .

METHOD 1

recognizing that \(f(8) = 1\)     (M1)

e.g. \(1 = k{\log _2}8\)

recognizing that \({\log _2}8 = 3\)     (A1)

e.g. \(1 = 3k\)

\(k = \frac{1}{3}\)     A1     N2

METHOD 2

attempt to find the inverse of \(f(x) = k{\log _2}x\)     (M1)

e.g. \(x = k{\log _2}y\) , \(y = {2^{\frac{x}{k}}}\)

substituting 1 and 8     (M1)

e.g. \(1 = k{\log _2}8\) , \({2^{\frac{1}{k}}} = 8\)

\(k = \frac{1}{{{{\log }_2}8}}\) \(\left( {k = \frac{1}{3}} \right)\)     A1     N2

[3 marks]

METHOD 1

recognizing that \(f(x) = \frac{2}{3}\)     (M1)

e.g. \(\frac{2}{3} = \frac{1}{3}{\log _2}x\)

\({\log _2}x = 2\)     (A1)

\({f^{ - 1}}\left( {\frac{2}{3}} \right) = 4\) (accept \(x = 4\))     A2     N3

METHOD 2

attempt to find inverse of \(f(x) = \frac{1}{3}{\log _2}x\)     (M1)

e.g. interchanging x and y , substituting \(k = \frac{1}{3}\) into \(y = {2^{\frac{x}{k}}}\)

correct inverse     (A1)

e.g. \({f^{ - 1}}(x) = {2^{3x}}\) , \({2^{3x}}\)

\({f^{ - 1}}\left( {\frac{2}{3}} \right) = 4\)     A2    N3

[4 marks]

A very poorly done question. Most candidates attempted to find the inverse function for \(f\) and used that to answer parts (a) and (b). Few recognized that the explicit inverse function was not necessary to answer the question.

Although many candidates seem to know that they can find an inverse function by interchanging x and y, very few were able to actually get the correct inverse. Almost none recognized that if \({f^{ - 1}}(1) = 8\) , then \(f(8) = 1\) . Many thought that the letters "log" could be simply "cancelled out", leaving the \(2\) and the \(8\).

A very poorly done question. Most candidates attempted to find the inverse function for \(f\) and used that to answer parts (a) and (b). Few recognized that the explicit inverse function was not necessary to answer the question.

Although many candidates seem to know that they can find an inverse function by interchanging x and y, very few were able to actually get the correct inverse. Almost none recognized that if \({f^{ - 1}}(1) = 8\) , then \(f(8) = 1\) . Many thought that the letters "log" could be simply "cancelled out", leaving the \(2\) and the \(8\).

Find the common difference.

Find the tenth term.

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

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]

Write down an expression for the common ratio, \(r\).

Hence, show that \(m\) satisfies the equation \({m^2} + 3m - 40 = 0\).

Find the two possible values of \(m\).

Find the possible values of \(r\).

The sequence has a finite sum.

State which value of \(r\) leads to this sum and justify your answer.

The sequence has a finite sum.

Calculate the sum of the sequence.

correct expression for \(r\)     A1     N1

eg   \(r = \frac{6}{{m - 1}},{\text{ }}\frac{{m + 4}}{6}\)

[2 marks]

correct equation     A1

eg     \(\frac{6}{{m - 1}} = \frac{{m + 4}}{6},{\text{ }}\frac{6}{{m + 4}} = \frac{{m - 1}}{6}\)

correct working     (A1)

eg     \((m + 4)(m - 1) = 36\)

correct working     A1

eg     \({m^2} - m + 4m - 4 = 36,{\text{ }}{m^2} + 3m - 4 = 36\)

\({m^2} + 3m - 40 = 0\)     AG     N0

[2 marks] 

valid attempt to solve     (M1)

eg     \((m + 8)(m - 5) = 0,{\text{ }}m = \frac{{ - 3 \pm \sqrt {9 + 4 \times 40} }}{2}\)

\(m =  - 8,{\text{ }}m = 5\)     A1A1     N3

[3 marks]

attempt to substitute any value of \(m\) to find \(r\)     (M1)

eg     \(\frac{6}{{ - 8 - 1}},{\text{ }}\frac{{5 + 4}}{6}\)

\(r = \frac{3}{2},{\text{ }}r =  - \frac{2}{3}\)     A1A1     N3

[3 marks]

\(r =  - \frac{2}{3}\)   (may be seen in justification)     A1

valid reason     R1     N0

eg     \(\left| r \right| < 1,{\text{ }} - 1 < \frac{{ - 2}}{3} < 1\)

Notes: Award R1 for \(\left| r \right| < 1\) only if A1 awarded.

[2 marks]

Find the common difference.

Find the eighth term.

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

correct approach     (A1)

eg\(\;\;\;d = {u_2} - {u_1},{\text{ }}5 - 2\)

\(d = 3\)     A1     N2

[2 marks]

correct approach     (A1)

eg\(\;\;\;{u_8} = 2 + 7 \times 3\), listing terms

\({u_8} = 23\)     A1     N2

[2 marks]

correct approach     (A1)

eg\(\;\;\;{S_8} = \frac{8}{2}(2 + 23)\), listing terms, \(\frac{8}{2}\left( {2(2) + 7(3)} \right)\)

\({S_8} = 100\)     A1     N2

[2 marks]

Total [6 marks]

All three parts of this question were very well done by the candidates. The occasional mistakes that were seen tended to be arithmetic errors which happened after the candidates had substituted correctly into the formulas given in the formula booklet.

All three parts of this question were very well done by the candidates. The occasional mistakes that were seen tended to be arithmetic errors which happened after the candidates had substituted correctly into the formulas given in the formula booklet.

All three parts of this question were very well done by the candidates. The occasional mistakes that were seen tended to be arithmetic errors which happened after the candidates had substituted correctly into the formulas given in the formula booklet.

Write the expression \(3\ln 2 - \ln 4\) in the form \(\ln k\), where \(k \in \mathbb{Z}\).

Hence or otherwise, solve \(3\ln 2 - \ln 4 =  - \ln x\).

correct application of \(\ln {a^b} = b\ln a\) (seen anywhere)     (A1)

eg\(\;\;\;\ln 4 = 2\ln 2,{\text{ }}3\ln 2 = \ln {2^3},{\text{ }}3\log 2 = \log 8\)

correct working     (A1)

eg\(\;\;\;3\ln 2 - 2\ln 2,{\text{ }}\ln 8 - \ln 4\)

\(\ln 2\;\;\;{\text{(accept }}k = 2{\text{)}}\)     A1     N2

[3 marks]

METHOD 1

attempt to substitute their answer into the equation     (M1)

eg\(\;\;\;\ln 2 =  - \ln x\)

correct application of a log rule     (A1)

eg\(\;\;\;\ln \frac{1}{x},{\text{ }}\ln \frac{1}{2} = \ln x,{\text{ }}\ln 2 + \ln x = \ln 2x\;\;\;( = 0)\)

\(x = \frac{1}{2}\)     A1     N2

METHOD 2

attempt to rearrange equation, with  \(3\ln 2\) written as \(\ln {2^3}\) or \(\ln 8\)     (M1)

eg\(\;\;\;\ln x = \ln 4 - \ln {2^3},{\text{ }}\ln 8 + \ln x = \ln 4,{\text{ }}\ln {2^3} = \ln 4 - \ln x\)

correct working applying \(\ln a \pm \ln b\)     (A1)

eg\(\;\;\;\frac{4}{8},{\text{ }}8x = 4,{\text{ }}\ln {2^3} = \ln \frac{4}{x}\)

\(x = \frac{1}{2}\)     A1     N2

[3 marks]

Total [6 marks]

Part (a) was answered correctly by a large number of candidates, though there were quite a few who applied the rules of logarithms in the wrong order.

In part (b), many candidates knew to set their answer from part (a) equal to \( - \ln x\), but then a good number incorrectly said that \(\ln 2 =  - \ln x\) led to \(2 =  - x\).

Given that \({u_1} = 1 + k\), find \({u_2},{\text{ }}{u_3}\) and \({u_4}\).

Find a general expression for \({u_n}\).

valid method     (M1)

eg     \({u_2} = {S_2} - {S_1},{\text{ }}1 + k + {u_2} = 5 + 3k\)

\({u_2} = 4 + 2k,{\text{ }}{u_3} = 7 + 4k,{\text{ }}{u_4} = 10 + 8k\)     A1A1A1     N4

[4 marks]

correct AP or GP     (A1)

eg     finding common difference is \(3\), common ratio is \(2\)

valid approach using arithmetic and geometric formulas     (M1)

eg     \(1 + 3(n - 1)\)  and \({r^{n - 1}}k\)

\({u_n} = 3n - 2 + {2^{n - 1}}k\)     A1A1     N4

Note: Award A1 for \(3n - 2\), A1 for \({2^{n - 1}}k\).

[4 marks]

Consider the arithmetic sequence \(2{\text{, }}5{\text{, }}8{\text{, }}11{\text{,}} \ldots \) .

Find \({u_{101}}\) .

Consider the arithmetic sequence \(2{\text{, }}5{\text{, }}8{\text{, }}11{\text{,}} \ldots \) .

Find the value of n so that \({u_n} = 152\) .

\(d = 3\)     (A1)

evidence of substitution into \({u_n} = a + (n - 1)d\)     (M1)

e.g. \({u_{101}} = 2 + 100 \times 3\)

\({u_{101}} = 302\)     A1     N3

[3 marks]

correct approach     (M1)

e.g. \(152 = 2 + (n - 1) \times 3\)

correct simplification     (A1)

e.g. \(150 = (n - 1) \times 3\) , \(50 = n - 1\) , \(152 = - 1 + 3n\)

\(n = 51\)     A1     N2

[3 marks]

Candidates probably had the most success with this question with many good solutions which were written with the working clearly shown. Many used the alternate approach of \({u_n} = 3n - 1\) .

Candidates probably had the most success with this question with many good solutions which were written with the working clearly shown. Many used the alternate approach of \({u_n} = 3n - 1\) .

Find \(r\).

Show that the sum of the infinite sequence is \(4{\log _2}x\).

Find \(d\), giving your answer as an integer.

Show that \({S_{12}} = 12{\log _2}x - 66\).

Given that \({S_{12}}\) is equal to half the sum of the infinite geometric sequence, find \(x\), giving your answer in the form \({2^p}\), where \(p \in \mathbb{Q}\).

evidence of dividing terms (in any order)     (M1)

eg\(\,\,\,\,\,\)\(\frac{{{\mu _2}}}{{{\mu _1}}},{\text{ }}\frac{{2{{\log }_2}x}}{{{{\log }_2}x}}\)

\(r = \frac{1}{2}\)    A1     N2

[2 marks]

correct substitution     (A1)

eg\(\,\,\,\,\,\)\(\frac{{2{{\log }_2}x}}{{1 - \frac{1}{2}}}\)

correct working     A1

eg\(\,\,\,\,\,\)\(\frac{{2{{\log }_2}x}}{{\frac{1}{2}}}\)

\({S_\infty } = 4{\log _2}x\)     AG     N0

[2 marks]

evidence of subtracting two terms (in any order)     (M1)

eg\(\,\,\,\,\,\)\({u_3} - {u_2},{\text{ }}{\log _2}x - {\log _2}\frac{x}{2}\)

correct application of the properties of logs     (A1)

eg\(\,\,\,\,\,\)\({\log _2}\left( {\frac{{\frac{x}{2}}}{x}} \right),{\text{ }}{\log _2}\left( {\frac{x}{2} \times \frac{1}{x}} \right),{\text{ }}({\log _2}x - {\log _2}2) - {\log _2}x\)

correct working     (A1)

eg\(\,\,\,\,\,\)\({\log _2}\frac{1}{2},{\text{ }} - {\log _2}2\)

\(d =  - 1\)    A1     N3

[4 marks]

correct substitution into the formula for the sum of an arithmetic sequence     (A1)

eg\(\,\,\,\,\,\)\(\frac{{12}}{2}\left( {2{{\log }_2}x + (12 - 1)( - 1)} \right)\)

correct working     A1

eg\(\,\,\,\,\,\)\(6(2{\log _2}x - 11),{\text{ }}\frac{{12}}{2}(2{\log _2}x - 11)\)

\(12{\log _2}x - 66\)    AG     N0

[2 marks]

correct equation     (A1)

eg\(\,\,\,\,\,\)\(12{\log _2}x - 66 = 2{\log _2}x\)

correct working     (A1)

eg\(\,\,\,\,\,\)\(10{\log _2}x = 66,{\text{ }}{\log _2}x = 6.6,{\text{ }}{2^{66}} = {x^{10}},{\text{ }}{\log _2}\left( {\frac{{{x^{12}}}}{{{x^2}}}} \right) = 66\)

\(x = {2^{6.6}}\) (accept \(p = \frac{{66}}{{10}}\))     A1     N2

[3 marks]

Write down the value of \(n\).

Write down a and b, in terms of p and/or q.

Write down an expression for the sixth term in the expansion.

\(n = 10\)     A1     N1

[1 mark]

\(a = p\) , \(b = 2q\) (or \(a = 2q\) , \(b = p\) )     A1A1     N1N1

[2 marks]

\(\left( {\begin{array}{*{20}{c}}
{10}\\
5
\end{array}} \right){p^5}{(2q)^5}\)     A1A1A1     N3

[3 marks]

The most common error was in (c) where many candidates interpreted the "sixth" term as using \(\left( {\begin{array}{*{20}{c}}
{10}\\
6
\end{array}} \right)\) , with accompanying powers of 4 and 6 in the expression.

Show that \(d = {\text{lo}}{{\text{g}}_c}\left( q \right)\).

Let \(p = {c^2}\) and \(q = {c^3}\). Find the value of \(\sum\limits_{n = 1}^{20} {{u_n}} \).

valid approach involving addition or subtraction       M1
eg  \({u_2} = {\text{lo}}{{\text{g}}_c}\,p + d,\,\,{u_1} - {u_2}\)

correct application of log law      A1
eg  \({\text{lo}}{{\text{g}}_c}\left( {pq} \right) = {\text{lo}}{{\text{g}}_c}\,p + {\text{lo}}{{\text{g}}_c}\,q,\,\,{\text{lo}}{{\text{g}}_c}\left( {\frac{{pq}}{p}} \right)\)

\(d = {\text{lo}}{{\text{g}}_c}\,q\)    AG N0

[2 marks]

METHOD 1 (finding \({u_1}\) and d)

recognizing \(\sum { = {S_{20}}} \) (seen anywhere)      (A1)

attempt to find \({u_1}\) or d using \({\text{lo}}{{\text{g}}_c}\,{c^k} = k\)     (M1)
eg  \({\text{lo}}{{\text{g}}_c}\,c\), \({\text{3}}\,{\text{lo}}{{\text{g}}_c}\,c\), correct value of \({u_1}\) or d

\({u_1}\) = 2, d = 3 (seen anywhere)      (A1)(A1)

correct working     (A1)
eg  \({S_{20}} = \frac{{20}}{2}\left( {2 \times 2 + 19 \times 3} \right),\,\,{S_{20}} = \frac{{20}}{2}\left( {2 + 59} \right),\,\,10\left( {61} \right)\)

\(\sum\limits_{n = 1}^{20} {{u_n}} \) = 610     A1 N2

METHOD 2 (expressing S in terms of c)

recognizing \(\sum { = {S_{20}}} \) (seen anywhere)      (A1)

correct expression for S in terms of c      (A1)
eg  \(10\left( {2\,{\text{lo}}{{\text{g}}_c}\,{c^2} + 19\,{\text{lo}}{{\text{g}}_c}\,{c^3}} \right)\)

\({\text{lo}}{{\text{g}}_c}\,{c^2} = 2,\,\,\,{\text{lo}}{{\text{g}}_c}\,{c^3} = 3\)  (seen anywhere)     (A1)(A1)

correct working      (A1)

eg  \({S_{20}} = \frac{{20}}{2}\left( {2 \times 2 + 19 \times 3} \right),\,\,{S_{20}} = \frac{{20}}{2}\left( {2 + 59} \right),\,\,10\left( {61} \right)\)

\(\sum\limits_{n = 1}^{20} {{u_n}} \) = 610     A1 N2

METHOD 3 (expressing S in terms of c)

recognizing \(\sum { = {S_{20}}} \) (seen anywhere)      (A1)

correct expression for S in terms of c      (A1)
eg  \(10\left( {2\,{\text{lo}}{{\text{g}}_c}\,{c^2} + 19\,{\text{lo}}{{\text{g}}_c}\,{c^3}} \right)\)

correct application of log law     (A1)
eg  \(2\,{\text{lo}}{{\text{g}}_c}\,{c^2} = \,\,{\text{lo}}{{\text{g}}_c}\,{c^4},\,\,19\,{\text{lo}}{{\text{g}}_c}\,{c^3} = \,\,{\text{lo}}{{\text{g}}_c}\,{c^{57}},\,\,10\,\left( {{\text{lo}}{{\text{g}}_c}\,{{\left( {{c^2}} \right)}^2} + \,\,{\text{lo}}{{\text{g}}_c}\,{{\left( {{c^3}} \right)}^{19}}} \right),\,\,10\,\left( {{\text{lo}}{{\text{g}}_c}\,{c^4} + \,{\text{lo}}{{\text{g}}_c}\,{c^{57}}} \right),\,\,10\left( {{\text{lo}}{{\text{g}}_c}\,{c^{61}}} \right)\)

correct application of definition of log      (A1)
eg  \({\text{lo}}{{\text{g}}_c}\,{c^{61}} = 61,\,\,{\text{lo}}{{\text{g}}_c}\,{c^4} = 4,\,\,{\text{lo}}{{\text{g}}_c}\,{c^{57}} = 57\)

correct working     (A1)
eg  \({S_{20}} = \frac{{20}}{2}\left( {4 + 57} \right),\,\,10\left( {61} \right)\)

\(\sum\limits_{n = 1}^{20} {{u_n}} \) = 610     A1 N2

[6 marks]

Find the common difference.

Find the first term.

Find the sum of the first 20 terms of the sequence.

attempt to find \(d\)     (M1)

eg     \(\frac{{16 - 10}}{2}{\text{, }}10 - 2d = 16 - 4d{\text{, }}2d = 6{\text{, }}d = 6\)

\(d = 3\)     A1     N2

[2 marks]

correct approach     (A1)

eg     \(10 = {u_1} + 2 \times 3{\text{, }}10 - 3 - 3\)

\({u_1} = 4\)     A1     N2

[2 marks]

correct substitution into sum or term formula     (A1)

eg     \(\frac{{20}}{2}(2 \times 4 + 19 \times 3){\text{, }}{u_{20}} = 4 + 19 \times 3\)

correct simplification     (A1)

eg     \(8 + 57{\text{, }}4 + 61\)

\({S_{20}} = 650\)     A1     N2

[3 marks]

Consider the infinite geometric sequence \(3{\text{, }}3(0.9){\text{, }}3{(0.9)^2}{\text{, }}3{(0.9)^3}{\text{, }} \ldots \) .

Write down the 10th term of the sequence. Do not simplify your answer.

Consider the infinite geometric sequence \(3{\text{, }}3(0.9){\text{, }}3{(0.9)^2}{\text{, }}3{(0.9)^3}{\text{, }} \ldots \) .

Find the sum of the infinite sequence.

\({u_{10}} = 3{(0.9)^9}\)    A1     N1

[1 mark]

recognizing \(r = 0.9\)     (A1)

correct substitution     A1

e.g.  \(S = \frac{3}{{1 - 0.9}}\)

\(S = \frac{3}{{0.1}}\)    (A1)

\(S = 30\)    A1     N3

[4 marks]

This question was well done by most candidates. There were a surprising number of candidates who lost a mark for not simplifying \(\frac{3}{{0.1}}\) to 30 , and there were a few candidates who used the formula for the finite sum unsuccessfully.

This question was well done by most candidates. There were a surprising number of candidates who lost a mark for not simplifying \(\frac{3}{{0.1}}\) to 30, and there were a few candidates who used the formula for the finite sum unsuccessfully.

Show that \({f^{ - 1}}(x) = {3^{2x}}\) .

Write down the range of \({f^{ - 1}}\) .

Let \(g(x) = {\log _3}x\) , for \(x > 0\) .

Find the value of \(({f^{ - 1}} \circ g)(2)\) , giving your answer as an integer.

interchanging x and y (seen anywhere)     (M1)

e.g. \(x = \log \sqrt y \) (accept any base)

evidence of correct manipulation     A1

e.g. \(3^x = \sqrt y \) , \({3^y} = {x^{\frac{1}{2}}}\) , \(x = \frac{1}{2}{\log _3}y\) , \(2y = {\log _3}x\)

\({f^{ - 1}}(x) = {3^{2x}}\)     AG     N0 

[2 marks]

\(y > 0\) , \({f^{ - 1}}(x) > 0\)     A1     N1

[1 mark]

METHOD 1

finding \(g(2) = lo{g_3}2\) (seen anywhere)     A1

attempt to substitute     (M1)

e.g. \(({f^{ - 1}} \circ g)(2) = {3^{2\log {_3}2}}\)

evidence of using log or index rule     (A1)

e.g. \(({f^{ - 1}} \circ g)(2) = {3^{\log {_3}4}}\) , \({3^{{{\log }_3}2^2}}\)

\(({f^{ - 1}} \circ g)(2) = 4\)     A1     N1

METHOD 2

attempt to form composite (in any order)     (M1)

e.g. \(({f^{ - 1}} \circ g)(x) = {3^{2{{\log }_3}x}}\)

evidence of using log or index rule     (A1)

e.g. \(({f^{ - 1}} \circ g)(x) = {3^{{{\log }_3}{x^2}}}\) , \({3^{{{\log }_3}{x^{}}}}^2\)

\(({f^{ - 1}} \circ g)(x) = {x^2}\)     A1

\(({f^{ - 1}} \circ g)(2) = 4\)     A1     N1

[4 marks]

Candidates were generally skilled at finding the inverse of a logarithmic function.

Few correctly gave the range of this function, often stating “all real numbers” or “ \(y \ge 0\) ”, missing the idea that the range of an inverse is the domain of the original function.

Some candidates answered part (c) correctly, although many did not get beyond \({3^{2{{\log }_3}2}}\) . Some attempted to form the composite in the incorrect order. Others interpreted \(({f^{ - 1}} \circ g)(2)\) as multiplication by 2.

Write down the value of r .

Find \({u_6}\) .

Find the sum to infinity of this sequence.

\(r = \frac{{16}}{{32}}\left( { = \frac{1}{2}} \right)\)     A1     N1

[1 mark]

correct calculation or listing terms     (A1)

e.g. \(32 \times {\left( {\frac{1}{2}} \right)^{6 - 1}}\) , \(8 \times {\left( {\frac{1}{2}} \right)^3}\) , 32, \(\ldots \) 4, 2, 1

\({u_6} = 1\)     A1     N2

[2 marks]

evidence of correct substitution in \({S_\infty }\)      A1

e.g. \(\frac{{32}}{{1 - \frac{1}{2}}}\) , \(\frac{{32}}{{\frac{1}{2}}}\)

\({S_\infty } = 64\)     A1     N1

[2 marks]

This question was very well done by the majority of candidates. There were some who used a value of r greater than one, with the most common error being \(r = 2\) .

This question was very well done by the majority of candidates. There were some who used a value of r greater than one, with the most common error being \(r = 2\) . 

A handful of candidates struggled with the basic computation involved in part (c).

Find an expression for r in terms of θ.

Find the possible values of r.

Show that the sum of the infinite sequence is \(\frac{{54}}{{2 + {\text{cos}}\,\left( {2\theta } \right)}}\).

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

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]

recognizing that sinθ is bounded      (M1)

eg    0 ≤ sin² θ ≤ 1, −1 ≤ sinθ ≤ 1, −1 < sinθ < 1

0 < r ≤ \(\frac{2}{3}\)      A2 N3

Note: If working shown, award M1A1 for correct values with incorrect inequality sign(s).
If no working shown, award N1 for correct values with incorrect inequality sign(s).

[3 marks]

correct substitution into formula for infinite sum       A1

eg  \(\frac{{18}}{{1 - \frac{{2\,{\text{si}}{{\text{n}}^2}\,\theta }}{3}}}\)

evidence of choosing an appropriate rule for cos 2θ (seen anywhere)         (M1)

eg   cos 2θ = 1 − 2 sin² θ

correct substitution of identity/working (seen anywhere)      (A1)

eg   \(\frac{{18}}{{1 - \frac{2}{3}\left( {\frac{{1 - {\text{cos}}\,2\theta }}{2}} \right)}},\,\,\frac{{54}}{{3 - 2\left( {\frac{{1 - {\text{cos}}\,2\theta }}{2}} \right)}},\,\,\frac{{18}}{{\frac{{3 - 2\,{\text{si}}{{\text{n}}^2}\,\theta }}{3}}}\)

correct working that clearly leads to the given answer       A1

eg  \(\frac{{18 \times 3}}{{2 + \left( {1 - 2\,{\text{si}}{{\text{n}}^2}\,\theta } \right)}},\,\,\frac{{54}}{{3 - \left( {1 - {\text{cos}}\,2\theta } \right)}}\)

\(\frac{{54}}{{2 + {\text{cos}}\left( {2\theta } \right)}}\)    AG N0

[4 marks]

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]

Three consecutive terms of a geometric sequence are \(x - 3\), 6 and \(x + 2\).

Find the possible values of \(x\).

METHOD 1

valid approach     (M1)

eg\(\,\,\,\,\,\)\(r = \frac{6}{{x - 3}},{\text{ }}(x - 3) \times r = 6,{\text{ }}(x - 3){r^2} = x + 2\)

correct equation in terms of \(x\) only     A1

eg\(\,\,\,\,\,\)\(\frac{6}{{x - 3}} = \frac{{x + 2}}{6},{\text{ }}(x - 3)(x + 2) = {6^2},{\text{ }}36 = {x^2} - x - 6\)

correct working     (A1)

eg\(\,\,\,\,\,\)\({x^2} - x - 42,{\text{ }}{x^2} - x = 42\)

valid attempt to solve their quadratic equation     (M1)

eg\(\,\,\,\,\,\)factorizing, formula, completing the square

evidence of correct working     (A1)

eg\(\,\,\,\,\,\)\((x - 7)(x + 6),{\text{ }}\frac{{1 \pm \sqrt {169} }}{2}\)

\(x = 7,{\text{ }}x =  - 6\)     A1     N4

METHOD 2 (finding r first)

valid approach     (M1)

eg\(\,\,\,\,\,\)\(r = \frac{6}{{x - 3}},{\text{ }}6r = x + 2,{\text{ }}(x - 3){r^2} = x + 2\)

correct equation in terms of \(r\) only     A1

eg\(\,\,\,\,\,\)\(\frac{6}{r} + 3 = 6r - 2,{\text{ }}6 + 3r = 6{r^2} - 2r,{\text{ }}6{r^2} - 5r - 6 = 0\)

evidence of correct working     (A1)

eg\(\,\,\,\,\,\)\((3r + 2)(2r - 3),{\text{ }}\frac{{5 \pm \sqrt {25 + 144} }}{{12}}\)

\(r =  - \frac{2}{3},{\text{ }}r = \frac{3}{2}\)    A1

substituting their values of \(r\) to find \(x\)     (M1)

eg\(\,\,\,\,\,\)\((x - 3)\left( {\frac{2}{3}} \right) = 6,{\text{ }}x = 6\left( {\frac{3}{2}} \right) - 2\)

\(x = 7,{\text{ }}x =  - 6\)    A1     N4

[6 marks]

Nearly all candidates attempted to set up an expression, or pair of expressions, for the common ratio of the geometric sequence. When done correctly, these expressions led to a quadratic equation which was solved correctly by many candidates.

Write down the values in the fifth row of Pascal’s triangle.

Hence or otherwise, find the term in \({x^3}\) in the expansion of \({(2x + 3)^5}\).

1, 5, 10, 10, 5, 1     A2     N2

[2 marks]

evidence of binomial expansion with binomial coefficient     (M1)

eg\(\,\,\,\,\,\)\(\left( {\begin{array}{*{20}{c}} n \\ r \end{array}} \right){a^{n - r}}{b^r}\), selecting correct term, \({(2x)^5}{(3)^0} + 5{(2x)^4}{(3)^1} + 10{(2x)^3}{(3)^2} +  \ldots \)

correct substitution into correct term     (A1)(A1)(A1)

eg\(\,\,\,\,\,\)\(10{(2)^3}{(3)^2},{\text{ }}\left( {\begin{array}{*{20}{c}} 5 \\ 3 \end{array}} \right){(2x)^3}{(3)^2}\)

Note: Award A1 for each factor.

\(720{x^3}\)     A1     N2

Notes: Do not award any marks if there is clear evidence of adding instead of multiplying.

Do not award final A1 for a final answer of 720, even if \(720{x^3}\) is seen previously.

[5 marks]

The following table gives the values of \({x_n}\) and \({A_n}\), for \(1 \leqslant n \leqslant 3\). Copy and complete the table. (Do not write on this page.)

The process described above is repeated. Find \({A_6}\).

Consider an initial square of side length \(k {\text{ cm}}\). The process described above is repeated indefinitely. The total area of the shaded regions is \(k {\text{ c}}{{\text{m}}^2}\). Find the value of \(k\).

valid method for finding side length     (M1)

eg   \({8^2} + {8^2} = {c^2},{\text{ }}45 - 45 - 90{\text{ side ratios, }}8\sqrt 2 ,{\text{ }}\frac{1}{2}{s^2} = 16,{\text{ }}{x^2} + {x^2} = {8^2}\)

correct working for area     (A1)

eg   \(\frac{1}{2} \times 4 \times 4\)

     A1A1     N2N2

[4 marks]

METHOD 1

recognize geometric progression for \({A_n}\)     (R1)

eg   \({u_n} = {u_1}{r^{n - 1}}\)

\(r = \frac{1}{2}\)     (A1)

correct working     (A1)

eg   \(32{\left( {\frac{1}{2}} \right)^5};{\text{ 4, 2, 1, }}\frac{1}{2},{\text{ }}\frac{1}{4},{\text{ }} \ldots \)

\({A_6} = 1\)     A1     N3

METHOD 2

attempt to find \({x_6}\)     (M1)

eg   \(8{\left( {\frac{1}{{\sqrt 2 }}} \right)^5},{\text{ }}2\sqrt 2 ,{\text{ 2, }}\sqrt 2 ,{\text{ 1, }} \ldots \)

\({x_6} = \sqrt 2 \)     (A1)

correct working     (A1)

eg   \(\frac{1}{2}{\left( {\sqrt 2 } \right)^2}\)

\({A_6} = 1\)     A1     N3

[4 marks]

METHOD 1

recognize infinite geometric series     (R1)

eg   \({S_n} = \frac{a}{{1 - r}},{\text{ }}\left| r \right| < 1\)

area of first triangle in terms of \(k\)     (A1)

eg   \(\frac{1}{2}{\left( {\frac{k}{2}} \right)^2}\)

attempt to substitute into sum of infinite geometric series (must have \(k\))     (M1)

eg   \(\frac{{\frac{1}{2}{{\left( {\frac{k}{2}} \right)}^2}}}{{1 - \frac{1}{2}}},{\text{ }}\frac{k}{{1 - \frac{1}{2}}}\)

correct equation     A1

eg   \(\frac{{\frac{1}{2}{{\left( {\frac{k}{2}} \right)}^2}}}{{1 - \frac{1}{2}}} = k,{\text{ }}k = \frac{{\frac{{{k^2}}}{8}}}{{\frac{1}{2}}}\)

correct working     (A1)

eg   \({k^2} = 4k\)

valid attempt to solve their quadratic     (M1)

eg   \(k(k - 4),{\text{ }}k = 4{\text{ or }}k = 0\)

\(k = 4\)     A1     N2

METHOD 2

recognizing that there are four sets of infinitely shaded regions with equal area     R1

area of original square is \({k^2}\)     (A1)

so total shaded area is \(\frac{{{k^2}}}{4}\)     (A1)

correct equation \(\frac{{{k^2}}}{4} = k\)     A1

\({k^2} = 4k\)    (A1)

valid attempt to solve their quadratic     (M1)

eg   \(k(k - 4),{\text{ }}k = 4{\text{ or }}k = 0\)

\(k = 4\)     A1     N2

[7 marks]

Find the value of \({\log _2}40 - {\log _2}5\) .

Find the value of \({8^{{{\log }_2}5}}\) .

evidence of correct formula    (M1)

eg   \(\log a - \log b = \log \frac{a}{b}\) , \(\log \left( {\frac{{40}}{5}} \right)\) , \(\log 8 + \log 5 - \log 5\)

Note: Ignore missing or incorrect base.

correct working     (A1)

eg   \({\log _2}8\) , \({2^3} = 8\)

\({\log _2}40 - {\log _2}5 = 3\)     A1     N2

[3 marks]

attempt to write \(8\) as a power of \(2\) (seen anywhere)     (M1)

eg   \({({2^3})^{{{\log }_2}5}}\) , \({2^3} = 8\) , \({2^a}\)

multiplying powers     (M1)

eg   \({2^{3{{\log }_2}5}}\) , \(a{\log _2}5\)

correct working     (A1)

eg   \({2^{{{\log }_2}125}}\) , \({\log _2}{5^3}\) , \({\left( {{2^{{{\log }_2}5}}} \right)^3}\)

\({8^{{{\log }_2}5}} = 125\)     A1     N3

[4 marks]

Many candidates readily earned marks in part (a). Some interpreted \({\log _2}40 - {\log _2}5\) to mean \(\frac{{{{\log }_2}40}}{{{{\log }_2}5}}\) , an error which led to no further marks. Others left the answer as \({\log _2}5\) where an integer answer is expected.

Part (b) proved challenging for most candidates, with few recognizing that changing \(8\) to base \(2\) is a helpful move. Some made it as far as \({2^{3{{\log }_2}5}}\) yet could not make that final leap to an integer.

Given that \({2^m} = 8\) and \({2^n} = 16\), write down the value of \(m\) and of \(n\).

Hence or otherwise solve \({8^{2x + 1}} = {16^{2x - 3}}\).

\(m = 3,{\text{ }}n = 4\)     A1A1     N2

[2 marks]

attempt to apply \({({2^a})^b} = {2^{ab}}\)     (M1)

eg\(\;\;\;6x + 3,{\text{ }}4(2x - 3)\)

equating their powers of \(2\) (seen anywhere)     M1

eg\(\;\;\;3(2x + 1) = 8x - 12\)

correct working     A1

eg\(\;\;\;8x - 12 = 6x + 3,{\text{ }}2x = 15\)

\(x = \frac{{15}}{2}\;\;\;(7.5)\)     A1     N2

[4 marks]

Total [6 marks]

Indices laws were well understood with many candidates solving the equation correctly. Some candidates used logs, which took longer, and errors crept in.

Indices laws were well understood with many candidates solving the equation correctly. Some candidates used logs, which took longer, and errors crept in.

Find the common ratio.

Solve \(\sum\limits_{k = 1}^\infty  {{2^{5 - k}}\ln x = 64} \).

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]

recognizing a sum (finite or infinite)     (M1)

eg\(\,\,\,\,\,\)\({2^4}\ln x + {2^3}\ln x,{\text{ }}\frac{a}{{1 - r}},{\text{ }}{S_\infty },{\text{ }}16\ln x +  \ldots \)

valid approach (seen anywhere)     (M1)

eg\(\,\,\,\,\,\)recognizing GP is the same as part (a), using their \(r\) value from part (a), \(r = \frac{1}{2}\)

correct substitution into infinite sum (only if \(\left| r \right|\) is a constant and less than 1)     A1

eg\(\,\,\,\,\,\)\(\frac{{{2^4}\ln x}}{{1 - \frac{1}{2}}},{\text{ }}\frac{{\ln {x^{16}}}}{{\frac{1}{2}}},{\text{ }}32\ln x\)

correct working     (A1)

eg\(\,\,\,\,\,\)\(\ln x = 2\)

\(x = {{\text{e}}^2}\)     A1     N3

[5 marks]

Find the probability that Ann wins on her first roll.

(i)     The probability that Ann wins on her third roll is \(\frac{5}{8} \times \frac{4}{8} \times p \times q\ \times \frac{3}{8}\).

Write down the value of \(p\) and of \(q\).

(ii)     The probability that Ann wins on her tenth roll is \(\frac{3}{8}{r^k}\) where \(r \in \mathbb{Q},{\text{ }}k \in \mathbb{Z}\).

Find the value of \(r\) and of \(k\).

Find the probability that Ann wins the game.

recognizing Ann rolls green     (M1)

eg\(\;\;\;{\text{P(G)}}\)

\(\frac{3}{8}\)     A1     N2

[2 marks]

(i)     \(p = \frac{4}{8},{\text{ }}q = \frac{5}{8}\) or \(q = \frac{4}{8},{\text{ }}p = \frac{5}{8}\)     A1A1     N2

(ii)     recognizes Ann and Bob lose \(9\) times     (M1)

eg\(\;\;\;\(\overbrace {{A_L}{B_\(\overbrace {{A_L}{B_ \ldots \(\overbrace {{A_L}{B_{\text{ 9 times, }}\underbrace {\left( {\frac{5}{8} \times \frac{4}{8}} \right) \times  \ldots  \times \left( {\frac{5}{8} \times \frac{4}{8}} \right)}_{{\text{9 times}}}\)

\(k = 9\;\;\;\)(seen anywhere)     A1     N2

correct working     (A1)

eg\(\;\;\;{\left( {\frac{5}{8} \times \frac{4}{8}} \right)^9} \times \frac{3}{8},{\text{ }}\left( {\frac{5}{8} \times \frac{4}{8}} \right) \times  \ldots  \times \left( {\frac{5}{8} \times \frac{4}{8}} \right) \times \frac{3}{8}\)

\(r = \frac{{20}}{{64}}\;\;\;\left( { = \frac{5}{{16}}} \right)\)     A1     N2

[6 marks]

recognize the probability is an infinite sum     (M1)

eg\(\;\;\;\)Ann wins on her \({{\text{1}}^{{\text{st}}}}\) roll or \({{\text{2}}^{{\text{nd}}}}\) roll or \({{\text{3}}^{{\text{rd}}}}\) roll…, \({S_\infty }\)

recognizing GP     (M1)

\({u_1} = \frac{3}{8}\;\;\;\)(seen anywhere)     A1

\(r = \frac{{20}}{{64}}\;\;\;\)(seen anywhere)     A1

correct substitution into infinite sum of GP     A1

eg\(\;\;\;\frac{{\frac{3}{8}}}{{1 - \frac{5}{{16}}}},{\text{ }}\frac{3}{8}\left( {\frac{1}{{1 - \left( {\frac{5}{8} \times \frac{4}{8}} \right)}}} \right),{\text{ }}\frac{1}{{1 - \frac{5}{{16}}}}\)

correct working     (A1)

eg\(\;\;\;\frac{{\frac{3}{8}}}{{\frac{{11}}{{16}}}},{\text{ }}\frac{3}{8} \times \frac{{16}}{{11}}\)

\({\text{P (Ann wins)}} = \frac{{48}}{{88}}\;\;\;\left( { = \frac{6}{{11}}} \right)\)     A1     N1

[7 marks]

Total [15 marks]

Some teachers’ comments suggested that the word ‘loses’ in the diagram was misleading. But candidate scripts did not indicate any adverse effect.

a)     Very well answered.

b)     i) Probabilities \(p\) and \(q\) were typically found correctly. ii) Fewer candidates identified the common ratio and number of rolls correctly.

Few candidates recognized that this was an infinite geometric sum although some did see that a geometric progression was involved.

Some teachers’ comments suggested that the word ‘loses’ in the diagram was misleading, But candidate scripts did not indicate any adverse effect.

a)     Very well answered.

b)     i) Probabilities \(p\) and \(q\) were typically found correctly. ii) Fewer candidates identified the common ratio and number of rolls correctly.

Few candidates recognized that this was an infinite geometric sum although some did see that a geometric progression was involved.

Some teachers’ comments suggested that the word ‘loses’ in the diagram was misleading, But candidate scripts did not indicate any adverse effect.

a)     Very well answered.

b)     i) Probabilities \(p\) and \(q\) were typically found correctly. ii) Fewer candidates identified the common ratio and number of rolls correctly.

Few candidates recognized that this was an infinite geometric sum although some did see that a geometric progression was involved.

Solve \({\log _2}(2\sin x) + {\log _2}(\cos x) =  - 1\), for \(2\pi  < x < \frac{{5\pi }}{2}\).

correct application of \(\log a + \log b = \log ab\)     (A1)

eg\(\,\,\,\,\,\)\({\log _2}(2\sin x\cos x),{\text{ }}\log 2 + \log (\sin x) + \log (\cos x)\)

correct equation without logs     A1

eg\(\,\,\,\,\,\)\(2\sin x\cos x = {2^{ - 1}},{\text{ }}\sin x\cos x = \frac{1}{4},{\text{ }}\sin 2x = \frac{1}{2}\)

recognizing double-angle identity (seen anywhere)     A1

eg\(\,\,\,\,\,\)\(\log (\sin 2x),{\text{ }}2\sin x\cos x = \sin 2x,{\text{ }}\sin 2x = \frac{1}{2}\)

evaluating \({\sin ^{ - 1}}\left( {\frac{1}{2}} \right) = \frac{\pi }{6}{\text{ }}(30^\circ )\)     (A1)

correct working     A1

eg\(\,\,\,\,\,\)\(x = \frac{\pi }{{12}} + 2\pi ,{\text{ }}2x = \frac{{25\pi }}{6},{\text{ }}\frac{{29\pi }}{6},{\text{ }}750^\circ ,{\text{ }}870^\circ ,{\text{ }}x = \frac{\pi }{{12}}\)and \(x = \frac{{5\pi }}{{12}}\), one correct final answer

\(x = \frac{{25\pi }}{{12}},{\text{ }}\frac{{29\pi }}{{12}}\) (do not accept additional values)     A2     N0

[7 marks]

(i)     \({\log _3}27\);

(ii)     \({\log _8}\frac{1}{8}\);

(iii)     \({\log _{16}}4\).

Hence, solve \({\log _3}27 + {\log _8}\frac{1}{8} - {\log _{16}}4 = {\log _4}x\).

(i)     \({\log _3}27 = 3\)     A1     N1

[1 mark]

(ii)     \({\log _8}\frac{1}{8} =  - 1\)     A1     N1

[1 mark]

(iii)     \({\log _{16}}4 = \frac{1}{2}\)     A1     N1

[1 mark]

correct equation with their three values     (A1)

eg     \(\frac{3}{2} = {\log _4}x{\text{, }}3 + ( - 1) - \frac{1}{2} = {\log _4}x\)

correct working involving powers     (A1)

eg     \(x = {4^{\frac{3}{2}}}{\text{, }}{4^{\frac{3}{2}}} = {4^{{{\log }_4}x}}\)

\(x = 8\)     A1     N2

[3 marks]

Solve \({\log _2}x + {\log _2}(x - 2) = 3\) , for \(x > 2\) .

recognizing \(\log a + \log b = \log ab\) (seen anywhere)     (A1)

e.g. \({\log _2}(x(x - 2))\) , \({x^2} - 2x\)

recognizing \({\log _a}b = x \Leftrightarrow {a^x} = b\)     (A1)

e.g. \({2^3} = 8\)

correct simplification     A1

e.g. \(x(x - 2) = {2^3}\) , \({x^2} - 2x - 8\)

evidence of correct approach to solve     (M1)

e.g. factorizing, quadratic formula

correct working     A1

e.g. \((x - 4)(x + 2)\) , \(\frac{{2 \pm \sqrt {36} }}{2}\)

\(x = 4\)     A2     N3

[7 marks]

Candidates secure in their understanding of logarithm properties usually had success with this problem, solving the resulting quadratic either by factoring or using the quadratic formula. The majority of successful candidates correctly rejected the solution that was not in the domain. A number of candidates, however, were unclear on logarithm properties. Some unsuccessful candidates were able to demonstrate understanding of one property but without both were not able to make much progress. A few candidates employed a “guess and check” strategy, but this did not earn full marks.

In the expansion of \({(3x + 1)^n}\), the coefficient of the term in \({x^2}\) is \(135n\), where \(n \in {\mathbb{Z}^ + }\). Find \(n\).

Note:     Accept sloppy notation (such as missing brackets, or binomial coefficient which includes \({x^2}\)).

evidence of valid binomial expansion with binomial coefficients     (M1)

eg\(\;\;\;\left( {\begin{array}{*{20}{c}} n \\ r \end{array}} \right){(3x)^r}{(1)^{n - r}},{\text{ }}{(3x)^n} + n{(3x)^{n - 1}} + \left( {\begin{array}{*{20}{c}} n \\ 2 \end{array}} \right){(3x)^{n - 2}} +  \ldots ,{\text{ }}\left( {\begin{array}{*{20}{c}} n \\ r \end{array}} \right){(1)^{n - r}}{(3x)^r}\)

attempt to identify correct term     (M1)

eg\(\;\;\;\left( {\begin{array}{*{20}{c}} n \\ {n - 2} \end{array}} \right),{\text{ }}{(3x)^2},{\text{ }}n - r = 2\)

setting correct coefficient or term equal to \(135n\) (may be seen later)     A1

eg\(\;\;\;9\left( {\begin{array}{*{20}{c}} n \\ 2 \end{array}} \right) = 135n,{\text{ }}\left( {\begin{array}{*{20}{c}} n \\ {n - 2} \end{array}} \right){(3x)^2} = 135n,{\text{ }}\frac{{9n(n - 1)}}{2} = 135n{x^2}\)

correct working for binomial coefficient (using \(_n{C_r}\) formula)     (A1)

eg\(\;\;\;\frac{{n(n - 1)(n - 2)(n - 3) \ldots }}{{2 \times 1 \times (n - 2)(n - 3)(n - 4) \ldots }},{\text{ }}\frac{{n(n - 1)}}{2}\)

EITHER

evidence of correct working (with linear equation in \(n\))     (A1)

eg\(\;\;\;\frac{{9(n - 1)}}{2} = 135,{\text{ }}\frac{{9(n - 1)}}{2}{x^2} = 135{x^2}\)

correct simplification     (A1)

eg\(\;\;\;n - 1 = \frac{{135 \times 2}}{9},{\text{ }}\frac{{(n - 1)}}{2} = 15\)

\(n = 31\)     A1     N2

OR

evidence of correct working (with quadratic equation in \(n\))     (A1)

eg\(\;\;\;9{n^2} - 279n = 0,{\text{ }}{n^2} - n = 30n,{\text{ (9}}{{\text{n}}^2} - 9n){x^2} = 270n{x^2}\)

evidence of solving     (A1)

eg\(\;\;\;9n(n - 31) = 0,{\text{ }}9{n^2} = 279n\)

\(n = 31\)     A1     N2

Note:     Award A0 for additional answers.

[7 marks]

Express \(g(x)\) in the form \(f(x) + \ln a\) , where \(a \in {{\mathbb{Z}}^ + }\) .

The graph of g is a transformation of the graph of f . Give a full geometric description of this transformation.

attempt to apply rules of logarithms     (M1)

e.g. \(\ln {a^b} = b\ln a\) , \(\ln ab = \ln a + \ln b\)

correct application of \(\ln {a^b} = b\ln a\) (seen anywhere)     A1

e.g. \(3\ln x = \ln {x^3}\)

correct application of \(\ln ab = \ln a + \ln b\) (seen anywhere)     A1

e.g. \(\ln 5{x^3} = \ln 5 + \ln {x^3}\)

so \(\ln 5{x^3} = \ln 5 + 3\ln x\)

\(g(x) = f(x) + \ln 5\) (accept \(g(x) = 3\ln x + \ln 5\) )     A1     N1

[4 marks]

transformation with correct name, direction, and value     A3

e.g. translation by \(\left( {\begin{array}{*{20}{c}}
0\\
{\ln 5}
\end{array}} \right)\) , shift up by \(\ln 5\) , vertical translation of \(\ln 5\)

[3 marks]

This question was very poorly done by the majority of candidates. While candidates seemed to have a vague idea of how to apply the rules of logarithms in part (a), very few did so successfully. The most common error in part (a) was to begin incorrectly with \(\ln 5{x^3} = 3\ln 5x\) . This error was often followed by other errors.

In part (b), very few candidates were able to describe the transformation as a vertical translation (or shift). Many candidates attempted to describe numerous incorrect transformations, and some left part (b) entirely blank.

Find d .

Find \({u_{20}}\) .

Find \({S_{20}}\) .

attempt to find d     (M1)

e.g. \(\frac{{{u_3} - {u_1}}}{2}\) , \(8 = 2 + 2d\)

\(d = 3\)     A1     N2

[2 marks]

correct substitution     (A1)

e.g. \({u_{20}} = 2 + (20 - 1)3\) , \({u_{20}} = 3 \times 20 - 1\)

\({u_{20}} = 59\)     A1     N2

[2 marks]

correct substitution     (A1)

e.g. \({S_{20}} = \frac{{20}}{2}(2 + 59)\) , \({S_{20}} = \frac{{20}}{2}(2 \times 2 + 19 \times 3)\)

\({S_{20}} = 610\)    A1     N2

[2 marks]

This question was answered correctly by the large majority of candidates. The few mistakes seen were due to either incorrect substitution into the formula or simple arithmetic errors. Even where candidates made mistakes, they were usually able to earn full follow-through marks in the subsequent parts of the question.

This question was answered correctly by the large majority of candidates. The few mistakes seen were due to either incorrect substitution into the formula or simple arithmetic errors. Even where candidates made mistakes, they were usually able to earn full follow-through marks in the subsequent parts of the question.

This question was answered correctly by the large majority of candidates. The few mistakes seen were due to either incorrect substitution into the formula or simple arithmetic errors. Even where candidates made mistakes, they were usually able to earn full follow-through marks in the subsequent parts of the question.

Given that \({\left( {1 + \frac{2}{3}x} \right)^n}{(3 + nx)^2} = 9 + 84x + \ldots \) , find the value of n .

attempt to expand \({\left( {1 + \frac{2}{3}x} \right)^n}\)     (M1)

e.g. Pascal's triangle, \({\left( {1 + \frac{2}{3}x} \right)^n} = 1 + \frac{2}{3}nx + \ldots \)

correct first two terms of \({\left( {1 + \frac{2}{3}x} \right)^n}\) (seen anywhere)     (A1)

e.g. \(1 + \frac{2}{3}nx\)

correct first two terms of quadratic (seen anywhere)     (A1)

e.g. 9 , \(6nx\) , \((9 + 6nx + {n^2}{x^2})\)

correct calculation for the x-term     A2

e.g. \(\frac{2}{3}nx \times 9 + 6nx\) , \(6n + 6n\) , \(12n\)

correct equation     A1

e.g. \(6n + 6n = 84\) , \(12nx = 84x\)

\(n = 7\)     A1     N1

[7 marks]

This question proved quite challenging for the majority of candidates, although there were a small number who were able to find the correct value of n using algebraic and investigative methods. While most candidates recognized the need to apply the binomial theorem, the majority seemed to have no idea how to do so when the exponent was a variable, n, rather than a known integer. Most candidates who attempted this question did expand the quadratic correctly, but many went no further, or simply set the x-term of the quadratic equal to \(84x\), ignoring the expansion of the first binomial altogether.

The following diagram shows [AB], with length 2 cm. The line is divided into an infinite number of line segments. The diagram shows the first three segments.

The length of the line segments are \(p{\text{ cm}},{\text{ }}{p^2}{\text{ cm}},{\text{ }}{p^3}{\text{ cm}},{\text{ }} \ldots \), where \(0 < p < 1\).

Show that \(p = \frac{2}{3}\).

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.

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

infinite sum of segments is 2 (seen anywhere)     (A1)

eg\(\,\,\,\,\,\)\(p + {p^2} + {p^3} +  \ldots  = 2,{\text{ }}\frac{{{u_1}}}{{1 - r}} = 2\)

recognizing GP     (M1)

eg\(\,\,\,\,\,\)ratio is \(p,{\text{ }}\frac{{{u_1}}}{{1 - r}},{\text{ }}{u_n} = {u_1} \times {r^{n - 1}},{\text{ }}\frac{{{u_1}({r^n} - 1)}}{{r - 1}}\)

correct substitution into \({S_\infty }\) formula (may be seen in equation)     A1

eg\(\,\,\,\,\,\)\(\frac{p}{{1 - p}}\)

correct equation     (A1)

eg\(\,\,\,\,\,\)\(\frac{p}{{1 - p}} = 2,{\text{ }}p = 2 - 2p\)

correct working leading to answer     A1

eg\(\,\,\,\,\,\)\(3p = 2,{\text{ }}2 - 3p = 0\)

\(p = \frac{2}{3}{\text{ (cm)}}\)     AG     N0

[5 marks]

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]

Find the equation of L .

Find the area of the region enclosed by the curve of g , the x-axis, and the lines \(x = 2\) and \(x = 12\) . Give your answer in the form \(a\ln b\) , where \(a,b \in \mathbb{Z}\) .

The graph of g is reflected in the x-axis to give the graph of h . The area of the region enclosed by the lines L , \(x = 2\) , \(x = 12\) and the x-axis is 120 \(120{\text{ c}}{{\text{m}}^2}\) .

Find the area enclosed by the lines L , \(x = 2\) , \(x = 12\) and the graph of h .

finding \(f'(x) = \frac{1}{2}x\)     A1

attempt to find \(f'(4)\)     (M1)

correct value \(f'(4) = 2\)     A1

correct equation in any form     A1     N2

e.g. \(y - 6 = 2(x - 4)\) , \(y = 2x - 2\)

[4 marks]

\({\rm{area}} = \int_2^{12} {\frac{{90}}{{3x + 4}}} {\rm{d}}x\)

correct integral     A1A1

e.g. \(30\ln (3x + 4)\)

substituting limits and subtracting     (M1)

e.g. \(30\ln (3 \times 12 + 4) - 30\ln (3 \times 2 + 4)\) , \(30\ln 40 - 30\ln 10\)

correct working     (A1)

e.g. \(30(\ln 40 - \ln 10)\)

correct application of \(\ln b - \ln a\)     (A1)

e.g. \(30\ln \frac{{40}}{{10}}\)

\({\rm{area}} = 30\ln 4\)     A1     N4

[6 marks]

valid approach     (M1)

e.g. sketch, area h = area g , 120 + their answer from (b)

\({\rm{area}} = 120 + 30\ln 4\)     A2     N3

[3 marks]

While most candidates answered part (a) correctly, finding the equation of the tangent, there were some who did not consider the value of their derivative when \(x = 4\) .

In part (b), most candidates knew that they needed to integrate to find the area, but errors in integration, and misapplication of the rules of logarithms kept many from finding the correct area.

In part (c), it was clear that a significant number of candidates understood the idea of the reflected function, and some recognized that the integral was the negative of the integral from part (b), but only a few recognized the relationship between the areas. Many thought the area between h and the x-axis was 120.

\(\ln \left( {\frac{5}{3}} \right)\).

\(\ln 45\).

correct approach     (A1)

eg\(\,\,\,\,\,\)\(\ln 5 - \ln 3\)

\(\ln \left( {\frac{5}{3}} \right) = y - x\)    A1     N2

[2 marks]

recognizing factors of 45 (may be seen in log expansion)     (M1)

eg\(\,\,\,\,\,\)\(\ln (9 \times 5),{\text{ }}3 \times 3 \times 5,{\text{ }}\log {3^2} \times \log 5\)

correct application of \(\log (ab) = \log a + \log b\)     (A1)

eg\(\,\,\,\,\,\)\(\ln 9 + \ln 5,{\text{ }}\ln 3 + \ln 3 + \ln 5,{\text{ }}\ln {3^2} + \ln 5\)

correct working     (A1)

eg\(\,\,\,\,\,\)\(2\ln 3 + \ln 5,{\text{ }}x + x + y\)

\(\ln 45 = 2x + y\)    A1     N3

[4 marks]

Most candidates were able to earn some or all the marks on this question. Part (a) was answered correctly by nearly all candidates.

Most candidates were able to earn some or all the marks on this question. In part (b), the majority of candidates knew they needed to factor 45, though some did not apply the log rules correctly to earn all the available marks here.

Find \({\log _3}{p^2}\) .

Find \({\log _3}\left( {\frac{p}{q}} \right)\) .

Find \({\log _3}(9p)\) .

METHOD 1

evidence of correct formula     (M1)

eg   \(\log {u^n} = n\log u\) , \(2{\log _3}p\)

\({\log _3}({p^2}) = 12\)     A1     N2

METHOD 2

valid method using \(p = {3^6}\)     (M1)

eg \({\log _3}{({3^6})^2}\) , \(\log {3^{12}}\) , \(12{\log _3}3\)

\({\log _3}({p^2}) = 12\)     A1     N2

[2 marks]

METHOD 1

evidence of correct formula     (M1)

eg   \(\log \left( {\frac{p}{q}} \right) = \log p - \log q\) , \(6 - 7\)

\({\log _3}\left( {\frac{p}{q}} \right) =  - 1\)     A1     N2

METHOD 2

valid method using \(p = {3^6}\) and \(q = {3^7}\)     (M1)

eg   \({\log _3}\left( {\frac{{{3^6}}}{{{3^7}}}} \right)\) , \(\log {3^{ - 1}}\) , \( - {\log _3}3\)

\({\log _3}\left( {\frac{p}{q}} \right) =  - 1\)     A1     N2

[2 marks]

METHOD 1

evidence of correct formula     (M1)

eg   \({\log _3}uv = {\log _3}u + {\log _3}v\) , \(\log 9 + \log p\)

\({\log _3}9 = 2\) (may be seen in expression)     A1

eg   \(2 + \log p\)

\({\log _3}(9p) = 8\)     A1     N2

METHOD 2

valid method using \(p = {3^6}\)     (M1)

eg   \({\log _3}(9 \times {3^6})\) , \({\log _3}({3^2} \times {3^6})\)

correct working     A1

eg   \({\log _3}9 + {\log _3}{3^6}\) , \({\log _3}{3^8}\)

\({\log _3}(9p) = 8\)     A1     N2

[3 marks]

Total [7 marks]

(i)     Show that \({f^{ - 1}}(x) = \ln x - 3\) .

(ii)    Write down the domain of \({f^{ - 1}}\) .

Solve the equation \({f^{ - 1}}(x) = \ln \frac{1}{x}\) .

(i) interchanging x and y (seen anywhere)     M1

e.g. \(x = {{\rm{e}}^{y + 3}}\)

correct manipulation     A1

e.g. \(\ln x = y + 3\) , \(\ln y = x + 3\)

\({f^{ - 1}}(x) = \ln x - 3\)     AG     N0

(ii) \(x > 0\)     A1     N1 

[3 marks]

collecting like terms; using laws of logs     (A1)(A1)

e.g. \(\ln x - \ln \left( {\frac{1}{x}} \right) = 3\) , \(\ln x + \ln x = 3\) , \(\ln \left( {\frac{x}{{\frac{1}{x}}}} \right) = 3\) , \(\ln {x^2} = 3\)

simplify     (A1)

e.g. \(\ln x = \frac{3}{2}\) ,  \({x^2} = {{\rm{e}}^3}\)

\(x = {{\rm{e}}^{\frac{3}{2}}}\left( { = \sqrt {{{\rm{e}}^3}} } \right)\)     A1     N2

[4 marks]

Many candidates interchanged the \(x\) and \(y\) to find the inverse function, but very few could write down the correct domain of the inverse, often giving \(x \ge 0\) , \(x > 3\) and "all real numbers" as responses.

Where students attempted to solve the equation in (b), most treated \(\ln x - 3\) as \(\ln (x - 3)\) and created an incorrect equation from the outset. The few who applied laws of logarithms often carried the algebra through to completion.

Find the common difference.

Find the tenth term.

Find the sum of the first ten terms.

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]

correct substitution into formula for sum     (A1)

eg\(\,\,\,\,\,\)\({S_{10}} = \frac{{10}}{2}(8 - 19),{\text{ 5}}\left( {2(8) + (10 - 1)( - 3)} \right)\)

\({S_{10}} =  - 55\)     A1     N2

[2 marks]

Find the \(x\)-coordinate of P.

Find \(f(x)\), expressing your answer as a single logarithm.

The graph of \(f\) is transformed by a vertical stretch with scale factor \(\frac{1}{{\ln 3}}\). The image of P under this transformation has coordinates \((a,{\text{ }}b)\).

Find the value of \(a\) and of \(b\), where \(a,{\text{ }}b \in \mathbb{N}\).

recognizing \(f'(x) = 0\)     (M1)

correct working     (A1)

eg\(\,\,\,\,\,\)\(6 - 2x = 0\)

\(x = 3\)    A1     N2

[3 marks]

evidence of integration     (M1)

eg\(\,\,\,\,\,\)\(\int {f',{\text{ }}\int {\frac{{6 - 2x}}{{6x - {x^2}}}{\text{d}}x} } \)

using substitution     (A1)

eg\(\,\,\,\,\,\)\(\int {\frac{1}{u}{\text{d}}u} \) where \(u = 6x - {x^2}\)

correct integral     A1

eg\(\,\,\,\,\,\)\(\ln (u) + c,{\text{ }}\ln (6x - {x^2})\)

substituting \((3,{\text{ }}\ln 27)\) into their integrated expression (must have \(c\))     (M1)

eg\(\,\,\,\,\,\)\(\ln (6 \times 3 - {3^2}) + c = \ln 27,{\text{ }}\ln (18 - 9) + \ln k = \ln 27\)

correct working     (A1)

eg\(\,\,\,\,\,\)\(c = \ln 27 - \ln 9\)

EITHER

\(c = \ln 3\)    (A1)

attempt to substitute their value of \(c\) into \(f(x)\)     (M1)

eg\(\,\,\,\,\,\)\(f(x) = \ln (6x - {x^2}) + \ln 3\)     A1     N4

OR

attempt to substitute their value of \(c\) into \(f(x)\)     (M1)

eg\(\,\,\,\,\,\)\(f(x) = \ln (6x - {x^2}) + \ln 27 - \ln 9\)

correct use of a log law     (A1)

eg\(\,\,\,\,\,\)\(f(x) = \ln (6x - {x^2}) + \ln \left( {\frac{{27}}{9}} \right),{\text{ }}f(x) = \ln \left( {27(6x - {x^2})} \right) - \ln 9\)

\(f(x) = \ln \left( {3(6x - {x^2})} \right)\)    A1     N4

[8 marks]

\(a = 3\)    A1     N1

correct working     A1

eg\(\,\,\,\,\,\)\(\frac{{\ln 27}}{{\ln 3}}\)

correct use of log law     (A1)

eg\(\,\,\,\,\,\)\(\frac{{3\ln 3}}{{\ln 3}},{\text{ }}{\log _3}27\)

\(b = 3\)    A1     N2

[4 marks]

Part a) was well answered.

In part b) most candidates realised that integration was required but fewer recognised the need to use integration by substitution. Quite a number of candidates who integrated correctly omitted finding the constant of integration.

In part c) many candidates showed good understanding of transformations and could apply them correctly, however, correct use of the laws of logarithms was challenging for many. In particular, a common error was \(\frac{{\ln 27}}{{\ln 3}} = \ln 9\).