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Question
Each time a ball bounces, it reaches 95 % of the height reached on the previous bounce.
Initially, it is dropped from a height of 4 metres.
a. What height does the ball reach after its fourth bounce?[2]
Answer/Explanation
Markscheme
a.
height \( = 4 \times {0.95^4}\) (A1)
= 3.26 (metres) A1
[2 marks]
\(4 \times {0.95^n} < 1\) (M1)
\({0.95^n} < 0.25\)
\( \Rightarrow n > \frac{{\ln 0.25}}{{\ln 0.95}}\) (A1)
\( \Rightarrow n > 27.0\)
Note: Do not penalize improper use of inequalities.
\( \Rightarrow n = 28\) A1
Note: If candidates have used n – 1 rather than n throughout penalise in part (a) and treat as follow through in parts (b) and (c).
[3 marks]
METHOD 1
recognition of geometric series with sum to infinity, first term of \(4 \times 0.95\) and common ratio 0.95 M1
recognition of the need to double this series and to add 4 M1
total distance travelled is \(2\left( {\frac{{4 \times 0.95}}{{1 – 0.95}}} \right) + 4 = 156{\text{ (metres)}}\) A1
[3 marks]
Note: If candidates have used n – 1 rather than n throughout penalise in part (a) and treat as follow through in parts (b) and (c).
METHOD 2
recognition of a geometric series with sum to infinity, first term of 4 and common ratio 0.95 M1
recognition of the need to double this series and to subtract 4 M1
total distance travelled is \({\text{2}}\left( {\frac{4}{{1 – 0.95}}} \right) – 4 = 156{\text{ (metres)}}\) A1
[3 marks]
Examiners report
The majority of candidates were able to start this question and gain some marks, but only better candidates gained full marks. In part (a) the common error was to assume the wrong number of bounces and in part (b) many candidates lost marks due to rounding the inequality in the wrong direction. Part (c) was found difficult with only a limited number recognising the need for the sum to infinity of a geometric sequence and many of those did not recognise how to link the sum to infinity to the total distance travelled.
The majority of candidates were able to start this question and gain some marks, but only better candidates gained full marks. In part (a) the common error was to assume the wrong number of bounces and in part (b) many candidates lost marks due to rounding the inequality in the wrong direction. Part (c) was found difficult with only a limited number recognising the need for the sum to infinity of a geometric sequence and many of those did not recognise how to link the sum to infinity to the total distance travelled.
The majority of candidates were able to start this question and gain some marks, but only better candidates gained full marks. In part (a) the common error was to assume the wrong number of bounces and in part (b) many candidates lost marks due to rounding the inequality in the wrong direction. Part (c) was found difficult with only a limited number recognising the need for the sum to infinity of a geometric sequence and many of those did not recognise how to link the sum to infinity to the total distance travelled.
Question
Find the sum of all the multiples of 3 between 100 and 500.
Answer/Explanation
Markscheme
METHOD 1
102 + 105 + … + 498 (M1)
so number of terms = 133 (A1)
EITHER
\( = \frac{{133}}{2}(2 \times 102 + 132 \times 3)\) (M1)
= 39900 A1
OR
\( = (102 + 498) \times \frac{{133}}{2}\) (M1)
= 39900 A1
OR
\(\sum\limits_{n = 34}^{166} {3n} \) (M1)
= 39900 A1
METHOD 2
\(500 \div 3 = 166.666…{\text{ and }}100 \div 3 = 33.333…\)
\(102 + 105 + … + 498 = \sum\limits_{n = 1}^{166} {3n} – \sum\limits_{n = 1}^{33} {3n} \) (M1)
\(\sum\limits_{n = 1}^{166} {3n} = 41583\) (A1)
\(\sum\limits_{n = 1}^{33} {3n} = 1683\) (A1)
the sum is 39900 A1
[4 marks]
Examiners report
Most candidates got full marks in this question. Some mistakes were detected when trying to find the number of terms of the arithmetic sequence, namely the use of the incorrect value n = 132 ; a few interpreted the question as the sum of multiples between the 100th and 500th terms. Occasional application of geometric series was attempted.
Question
A metal rod 1 metre long is cut into 10 pieces, the lengths of which form a geometric sequence. The length of the longest piece is 8 times the length of the shortest piece. Find, to the nearest millimetre, the length of the shortest piece.
Answer/Explanation
Markscheme
the pieces have lengths \(a,{\text{ }}ar,{\text{ …, }}a{r^9}\) (M1)
\(8a = a{r^9}{\text{ }}({\text{or }}8 = {r^9})\) A1
\(r = \sqrt[9]{8} = 1.259922…\) A1
\(a\frac{{{r^{10}} – 1}}{{r – 1}} = 1\,\,\,\,\,\left( {{\text{or }}a\frac{{{r^{10}} – 1}}{{r – 1}} = 1000} \right)\) M1
\(a = \frac{{r – 1}}{{{r^{10}} – 1}} = 0.0286…\,\,\,\,\,\left( {{\text{or }}a = \frac{{r – 1}}{{{r^{10}} – 1}} = 28.6…} \right)\) (A1)
a = 29 mm (accept 0.029 m or any correct answer regardless the units) A1
[6 marks]
Examiners report
This question was generally well done by most candidates. Some candidates recurred to a diagram to comprehend the nature of the problem but a few thought it was an arithmetic sequence.
A surprising number of candidates missed earning the final A1 mark because they did not read the question instructions fully and missed the accuracy instruction to give the answer correct to the nearest mm.
Question
The integral \({I_n}\) is defined by \({I_n} = \int_{n\pi }^{(n + 1)\pi } {{{\text{e}}^{ – x}}|\sin x|{\text{d}}x,{\text{ for }}n \in \mathbb{N}} \) .
Show that \({I_0} = \frac{1}{2}(1 + {{\text{e}}^{ – \pi }})\) .[6]
By letting \(y = x – n\pi \) , show that \({I_n} = {{\text{e}}^{ – n\pi }}{I_0}\) .[4]
Hence determine the exact value of \(\int_0^\infty {{{\text{e}}^{ – x}}|\sin x|{\text{d}}x} \) .[5]
Answer/Explanation
Markscheme
\({I_0} = \int_0^\pi {{{\text{e}}^{ – x}}\sin x{\text{d}}x} \) M1
Note: Award M1 for \({I_0} = \int_0^\pi {{{\text{e}}^{ – x}}|\sin x|{\text{d}}x} \)
Attempt at integration by parts, even if inappropriate modulus signs are present. M1
\( = – \left[ {{{\text{e}}^{ – x}}\cos x} \right]_0^\pi – \int_0^\pi {{{\text{e}}^{ – x}}\cos x{\text{d}}x} \) or \( = – \left[ {{{\text{e}}^{ – x}}\sin x} \right]_0^\pi – \int_0^\pi {{{\text{e}}^{ – x}}\cos x{\text{d}}x} \) A1
\( = – \left[ {{{\text{e}}^{ – x}}\cos x} \right]_0^\pi – \left[ {{{\text{e}}^{ – x}}\sin x} \right]_0^\pi – \int_0^\pi {{{\text{e}}^{ – x}}\sin x{\text{d}}x} \) or \( = – \left[ {{{\text{e}}^{ – x}}\sin x + {{\text{e}}^{ – x}}\cos x} \right]_0^\pi – \int_0^\pi {{{\text{e}}^{ – x}}\sin x{\text{d}}x} \) A1
\( = – \left[ {{{\text{e}}^{ – x}}\cos x} \right]_0^\pi – \left[ {{{\text{e}}^{ – x}}\sin x} \right]_0^\pi – {I_0}\) or \( – \left[ {{{\text{e}}^{ – x}}\sin x + {{\text{e}}^{ – x}}\cos x} \right]_0^\pi – {I_0}\) M1
Note: Do not penalise absence of limits at this stage
\({I_0} = {{\text{e}}^{ – \pi }} + 1 – {I_0}\) A1
\({I_0} = \frac{1}{2}(1 + {{\text{e}}^{ – \pi }})\) AG
Note: If modulus signs are used around cos x , award no accuracy marks but do not penalise modulus signs around sin x .
[6 marks]
\({I_n} = \int_{n\pi }^{(n + 1)\pi } {{{\text{e}}^{ – x}}|\sin x|{\text{d}}x} \)
Attempt to use the substitution \(y = x – n\pi \) M1
(putting \(y = x – n\pi \) , \({\text{d}}y = {\text{d}}x\) and \(\left[ {n\pi ,{\text{ }}(n + 1)\pi } \right] \to [0,{\text{ }}\pi ]\))
so \({I_n} = \int_0^\pi {{{\text{e}}^{ – (y + n\pi )}}|\sin (y + n\pi )|{\text{d}}y} \) A1
\( = {{\text{e}}^{ – n\pi }}\int_0^\pi {{{\text{e}}^{ – y}}|\sin (y + n\pi )|{\text{d}}y} \) A1
\( = {{\text{e}}^{ – n\pi }}\int_0^\pi {{{\text{e}}^{ – y}}\sin y{\text{d}}y} \) A1
\( = {{\text{e}}^{ – n\pi }}{I_0}\) AG
[4 marks]
\(\int_0^\infty {{{\text{e}}^{ – x}}|\sin x|{\text{d}}x} = \sum\limits_{n = 0}^\infty {{I_n}} \) M1
\( = \sum\limits_{n = 0}^\infty {{{\text{e}}^{ – n\pi }}{I_0}} \) (A1)
the \(\sum \) term is an infinite geometric series with common ratio \({{\text{e}}^{ – \pi }}\) (M1)
therefore
\(\int_0^\infty {{{\text{e}}^{ – x}}|\sin x|{\text{d}}x} = \frac{{{I_0}}}{{1 – {{\text{e}}^{ – \pi }}}}\) (A1)
\( = \frac{{1 + {{\text{e}}^{ – \pi }}}}{{2(1 – {{\text{e}}^{ – \pi }})}}{\text{ }}\left( { = \frac{{{{\text{e}}^\pi } + 1}}{{2({{\text{e}}^\pi } – 1)}}} \right)\) A1
[5 marks]
Examiners report
Part (a) is essentially core work requiring repeated integration by parts and many candidates realised that. However, some candidates left the modulus signs in \({I_0}\) which invalidated their work. In parts (b) and (c) it was clear that very few candidates had a complete understanding of the significance of the modulus sign and what conditions were necessary for it to be dropped. Overall, attempts at (b) and (c) were disappointing with few correct solutions seen.
Part (a) is essentially core work requiring repeated integration by parts and many candidates realised that. However, some candidates left the modulus signs in \({I_0}\) which invalidated their work. In parts (b) and (c) it was clear that very few candidates had a complete understanding of the significance of the modulus sign and what conditions were necessary for it to be dropped. Overall, attempts at (b) and (c) were disappointing with few correct solutions seen.
Part (a) is essentially core work requiring repeated integration by parts and many candidates realised that. However, some candidates left the modulus signs in \({I_0}\) which invalidated their work. In parts (b) and (c) it was clear that very few candidates had a complete understanding of the significance of the modulus sign and what conditions were necessary for it to be dropped. Overall, attempts at (b) and (c) were disappointing with few correct solutions seen.