New test - October 20, 2014
[389 marks]
1a. The first three terms of an arithmetic sequence are 5 , 6.7 , 8.4 .
[2 marks]
Find the common difference.
Markscheme
(M1)
valid method
e.g. subtracting terms, using sequence formula
d = 1.7
A1
N2
[2 marks]
Examiners report
Most candidates performed well on this question. A few were confused between the term number and the value of a term.
1b. The first three terms of an arithmetic sequence are 5 , 6.7 , 8.4 .
Find the
28th
[2 marks]
term of the sequence.
Markscheme
correct substitution into term formula
(A1)
e.g. 5 + 27(1.7)
28th term is 50.9 (exact)
A1
N2
[2 marks]
Examiners report
Most candidates performed well on this question. A few were confused between the term number and the value of a term.
1c. The first three terms of an arithmetic sequence are
5 , 6.7 , 8.4 .
Find the sum of the first 28 terms.
Markscheme
correct substitution into sum formula
e.g. S28 =
28
(2(5) + 27(1.7))
2
,
(A1)
28
(5 + 50.9)
2
S28 = 782.6 (exact) [782, 783] A1
N2
[2 marks]
Examiners report
Most candidates performed well on this question. A few were confused between the term number and the value of a term.
(2x + p
6
60
4
[2 marks]
2.
The third term in the expansion of
(2x + p)6
is 60x4 . Find the possible values of p .
[7 marks]
Markscheme
(M1)
attempt to expand binomial
6
n
e.g. (2x)6 p0 + ( ) (2x)5 (p)1 + … , ( ) (2x)r (p)n−r
1
r
one correct calculation for term in x4 in the expansion for power 6
(A1)
e.g. 15 , 16x4
correct expression for term in x4
(A1)
6
e.g. ( ) (2x)4 (p)2 , 15.24 p2
2
Notes: Accept sloppy notation e.g. omission of brackets around 2x .
Accept absence of x in middle factor.
correct term
e.g.
240p2 x4
(A1)
(accept absence of x4 )
setting up equation with their coefficient equal to 60
M1
6
60
e.g. ( ) (2)4 (p)2 = 60 , 240p2 x4 = 60x4 , p2 = 240
2
p = ± 12 (p = ±0.5)
A1A1
N3
[7 marks]
Examiners report
This question proved challenging for many students. Most candidates recognized the need to expand a binomial but many executed
this task incorrectly by selecting the wrong term, omitting brackets, or ignoring the binomial coefficient. Other candidates did not
recognize that there were two values for p when solving their quadratic equation.
3.
Given that (1 + 23 x) (3 + nx)2 = 9 + 84 + … , find the value of n .
n
Markscheme
attempt to expand (1 + 23 x)
n
(M1)
e.g. Pascal's triangle, (1 + 23 x) = 1 + 23 nx + …
n
correct first two terms of (1 + 23 x) (seen anywhere)
n
(A1)
e.g. 1 + 23 nx
correct first two terms of quadratic (seen anywhere)
e.g. 9 , 6nx , (9 + 6nx + n 2 x2 )
correct calculation for the x-term
e.g.
2
nx × 9 + 6nx
3
correct equation
, 6n + 6n , 12n
A1
e.g. 6n + 6n = 84 , 12nx = 84x
n=7
A1
[7 marks]
N1
A2
(A1)
[7 marks]
Examiners report
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.
4a. The first term of a geometric sequence is 200 and the sum of the first four terms is 324.8.
[4 marks]
Find the common ratio.
Markscheme
correct substitution into sum of a geometric sequence
(A1)
e.g. 200 ( 1−r
) , 200 + 200r + 200r2 + 200r3
1−r
4
attempt to set up an equation involving a sum and 324.8
M1
e.g. 200 ( 1−r
) = 324.8 , 200 + 200r + 200r2 + 200r3 = 324.8
1−r
4
r = 0.4 (exact)
A2
N3
[4 marks]
Examiners report
In part (a), although most candidates substituted correctly into the formula for the sum of a geometric series and set it equal to 324.8,
some used the formula for the sum to infinity and a few the formula for the sum of an arithmetic series. The overwhelming error made
was in attempting to solve the equation algebraically and getting nowhere, or getting a wrong answer. The great majority did not
recognize the need to use the GDC to find the value of r.
4b. The first term of a geometric sequence is 200 and the sum of the first four terms is 324.8.
[2 marks]
Find the tenth term.
Markscheme
correct substitution into formula
A1
9
e.g. u10 = 200 × 0.4
u10 = 0.0524288 (exact), 0.0524
A1
N1
[2 marks]
Examiners report
In part (b) many did not obtain any marks since they weren't able to find an answer to part (a). Those who were able to get a value for
r in part (a) generally went on to gain full marks in (b). However, this was one of the most common places for rounding errors to be
made.
The first three terms of an arithmetic sequence are 36, 40, 44,….
5a. (i)
Write down the value of d .
(ii) Find u8 .
[3 marks]
Markscheme
(i) d = 4
A1
N1
(M1)
(ii) evidence of valid approach
e.g. u8 = 36 + 7(4) , repeated addition of d from 36
u8 = 64
A1
N2
[3 marks]
Examiners report
The majority of candidates were successful with this question. Most had little difficulty with part (a).
5b. (i)
Show that Sn = 2n 2 + 34n .
[3 marks]
(ii) Hence, write down the value of S14 .
Markscheme
(i) correct substitution into sum formula
e.g. Sn =
n
2
{2 (36) + (n − 1)(4)} ,
n
2
A1
{72 + 4n − 4}
evidence of simplifying
{4n + 68}
A1
Sn = 2n 2 + 34n
AG
e.g.
n
2
(ii) 868
A1
N0
N1
[3 marks]
Examiners report
Some candidates were unable to show the required result in part (b), often substituting values for n rather than working with the
formula for the sum of an arithmetic series.
8
Consider the expansion of (2x3 + xb ) = 256x24 + 3072x20 + … + kx0 + … .
6a. Find b.
[3 marks]
Markscheme
valid attempt to find term in x20
(M1)
8
e.g. ( ) (27 )(b) , (2x3 )7 ( xb ) = 3072
1
correct equation
A1
8
e.g. ( ) (27 )(b) = 3072
1
b=3
A1
[3 marks]
N2
Examiners report
An unfamiliar presentation confused a number of candidates who attempted to set up an equation with the wrong term in part (a).
Time and again, candidates omitted the binomial coefficient in their set up leading to an incorrect result. In part (b) it was common to
see the constant term treated as the last term of the expansion rather than the 7th term.
6b. Find k.
[3 marks]
Markscheme
(M1)
evidence of choosing correct term
e.g. 7th term, r = 6
correct expression
A1
6
8
e.g. ( ) (2x3 )2 ( x3 )
6
k = 81648 (accept 81600 )
A1
N2
[3 marks]
Examiners report
An unfamiliar presentation confused a number of candidates who attempted to set up an equation with the wrong term in part (a).
Time and again, candidates omitted the binomial coefficient in their set up leading to an incorrect result. In part (b) it was common to
see the constant term treated as the last term of the expansion rather than the 7th term.
The following diagram shows two ships A and B. At noon, ship A was 15 km due north of ship B. Ship A was moving south at 15 km h–1
and ship B was moving east at 11 km h–1.
7a. Find the distance between the ships
(i)
at 13:00;
(ii) at 14:00.
[5 marks]
Markscheme
(M1)
(i) evidence of valid approach
e.g. ship A where B was, B 11 km away
distance = 11
A1
N2
(ii) evidence of valid approach
(M1)
e.g. new diagram, Pythagoras, vectors
−−−−−−−
s = √152 + 222 (A1)
−−−
√709 = 26.62705
s = 26.6
A1
N2
Note: Award M0A0A0 for using the formula given in part (b).
[5 marks]
Examiners report
Part (a) was generally well done although some candidates incorrectly used the function given in part (b) to find the required values.
There was evidence that some candidates are not comfortable with a 24-hour clock.
7b. Let s(t) be the distance between the ships t hours after noon, for 0 ≤ t ≤ 4 .
−−−−−−−−−−−−−−
Show that s(t) = √346t2 − 450t + 225 .
[6 marks]
Markscheme
evidence of valid approach
(M1)
e.g. a table, diagram, formula d = r × t
distance ship A travels t hours after noon is 15(t − 1)
distance ship B travels in t hours after noon is 11t
(A2)
(A1)
evidence of valid approach M1
−−−−−−−−−−−−−−−
e.g. s(t) = √[15(t − 1)]2 + (11t)2
correct simplification A1
−−−−−−−−−−−−−−−−−−
e.g. √225(t2 − 2t + 1) + 121t2
−−−−−−−−−−−−−−
s(t) = √346t2 − 450t + 225
AG
N0
[6 marks]
Examiners report
Candidates had difficulty generalizing the problem and therefore, were unable to show how the function s(t) was obtained in part (b).
7c. Sketch the graph of s(t) .
[3 marks]
Markscheme
A1A1A1
N3
Note: Award A1 for shape, A1 for minimum at approximately (0.7, 9), A1 for domain.
[3 marks]
Examiners report
Surprisingly, the graph in part (c) was not well done. Candidates often ignored the given domain, provided no indication of scale, and
drew "V" shapes or parabolas.
7d. Due to poor weather, the captain of ship A can only see another ship if they are less than 8 km apart. Explain why the captain
[3 marks]
cannot see ship B between noon and 16:00.
Markscheme
evidence of valid approach
(M1)
e.g. s ′ (t) = 0 , find minimum of s(t) , graph, reference to "more than 8 km"
min = 8.870455 … (accept 2 or more sf)
A1
since s min > 8 , captain cannot see ship B
R1
N0
[3 marks]
Examiners report
In part (d), candidates simply regurgitated the question without providing any significant evidence for their statements that the two
ships must have been more than 8 km apart.
Let f(x) = 14 x2 + 2 . The line L is the tangent to the curve of f at (4, 6) .
8a. Find the equation of L .
[4 marks]
Markscheme
finding f ′ (x) = 12 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]
Examiners report
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 .
Let g(x) =
90
3x+4
, for 2 ≤ x ≤ 12 . The following diagram shows the graph of g .
8b. 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 ∈ Z .
Markscheme
12
area = ∫2
90
dx
3x+4
A1A1
correct integral
e.g. 30 ln(3x + 4)
substituting limits and subtracting
(M1)
e.g. 30 ln(3 × 12 + 4) − 30 ln(3 × 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
e.g. 30 ln 40
10
area = 30 ln 4
[6 marks]
A1
N4
(A1)
[6 marks]
Examiners report
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.
Consider the triangle ABC, where AB =10 , BC = 7 and C ÂB = 30∘ .
9a. Find the two possible values of AĈ B .
[4 marks]
Markscheme
Note: accept answers given in degrees, and minutes.
(M1)
evidence of choosing sine rule
e.g.
sin A
a
=
sin B
b
correct substitution
e.g.
sin θ
10
=
sin 30∘
7
A1
, sin θ =
5
7
AĈ B = 45.6∘ ​ , AĈ B = 134∘
​ A1A1
N1N1
Note: If candidates only find the acute angle in part (a), award no marks for (b).
[4 marks]
Examiners report
Most candidates were comfortable applying the sine rule, although many were then unable to find the obtuse angle, demonstrating a
lack of understanding of the ambiguous case. This precluded them from earning marks in part (b). Those who found the obtuse angle
generally had no difficulty with part (b).
9b. Hence, find AB̂C , given that it is acute.
[2 marks]
Markscheme
attempt to substitute their larger value into angle sum of triangle
(M1)
e.g. 180∘ − (134.415…∘ + 30∘ )
AB̂C = 15.6∘
​ A1
N2
[2 marks]
Examiners report
Most candidates were comfortable applying the sine rule, although many were then unable to find the obtuse angle, demonstrating a
lack of understanding of the ambiguous case. This precluded them from earning marks in part (b). Those who found the obtuse angle
generally had no difficulty with part (b).
Consider the expansion of (3x2 + 2)9 .
10a. Write down the number of terms in the expansion.
Markscheme
10 terms
[1 mark]
A1
N1
[1 mark]
Examiners report
Many candidates were familiar with the binomial expansion, although some expanded entirely which at times led to careless errors.
Others attempted to use Pascal's Triangle. Common errors included misidentifying the binomial coefficient corresponding to this term
and not squaring the 3 in (3x2 ) .
10b. Find the term in x4 .
[5 marks]
Markscheme
evidence of binomial expansion
(M1)
9
9
9
e.g. a9 b0 + ( ) a8 b + ( ) a7 b2 + …, ( ) (a)n−r (b)r , Pascal’s triangle
1
2
r
(A1)
evidence of correct term
9
e.g. 8th term, r = 7 , ( ) , (3x2 )2 27
7
(A1)
correct expression of complete term
9
e.g. ( ) (3x2 )2 (2)7 , 92 C(3x2 )2 (2)7 , 36 × 9 × 128
7
41472x4 (accept 41500x4 )
A1
N2
[4 marks]
Examiners report
Many candidates were familiar with the binomial expansion, although some expanded entirely which at times led to careless errors.
Others attempted to use Pascal's triangle. Common errors included misidentifying the binomial coefficient corresponding to this term
and not squaring the 3 in (3x2 ) .
11a. Consider an infinite geometric sequence with u1 = 40 and r =
(i)
1
2
.
[4 marks]
Find u4 .
(ii) Find the sum of the infinite sequence.
Markscheme
(A1)
(i) correct approach
1 (4−1)
e.g. u4 = (40) 2
u4 = 5
A1
, listing terms
N2
(ii) correct substitution into formula for infinite sum
e.g. S∞ =
S∞ = 80
40
1−0.5
A1
, S∞ =
(A1)
40
0.5
N2
[4 marks]
Examiners report
Most candidates found part (a) straightforward, although a common error in (a)(ii) was to calculate 40 divided by
1
2
as 20.
11b. Consider an arithmetic sequence with n terms, with first term (−36) and eighth term (−8) .
(i)
[5 marks]
Find the common difference.
(ii) Show that Sn = 2n 2 − 38n .
Markscheme
(i) attempt to set up expression for u8
(M1)
e.g. −36 + (8 − 1)d
A1
correct working
e.g. −8 = −36 + (8 − 1)d ,
d=4
A1
−8−(−36)
7
N2
(A1)
(ii) correct substitution into formula for sum
e.g. Sn = n2 (2(−36) + (n − 1)4)
A1
correct working
e.g. Sn = n2 (4n − 76) , −36n + 2n2 − 2n
Sn = 2n 2 − 38n
AG
N0
[5 marks]
Examiners report
In part (b), some candidates had difficulty with the "show that" and worked backwards from the answer given.
11c. The sum of the infinite geometric sequence is equal to twice the sum of the arithmetic sequence. Find n .
[5 marks]
Markscheme
multiplying Sn (AP) by 2 or dividing S (infinite GP) by 2
e.g. 2Sn ,
S∞
2
(M1)
, 40
evidence of substituting into 2Sn = S∞
A1
e.g. 2n 2 − 38n = 40 , 4n 2 − 76n − 80 (= 0)
attempt to solve their quadratic (equation)
(M1)
e.g. intersection of graphs, formula
n = 20
A2
N3
[5 marks]
Examiners report
Most candidates obtained the correct equation in part (c), although some did not reject the negative value of n as impossible in this
context.
12a. Find the value of log2 40 − log2 5 .
[3 marks]
Markscheme
evidence of correct formula (M1)
eg log a − log b = log ab , log ( 405 ) , log 8 + log 5 − log 5
Note: Ignore missing or incorrect base.
correct working
(A1)
3
eg log2 8 , 2 = 8
log2 40 − log2 5 = 3
A1
N2
[3 marks]
Examiners report
Many candidates readily earned marks in part (a). Some interpreted log2 40 − log2 5 to mean
marks. Others left the answer as log2 5 where an integer answer is expected.
12b. Find the value of 8log25 .
log240
log25
, an error which led to no further
[4 marks]
Markscheme
attempt to write 8 as a power of 2 (seen anywhere)
3 log25
eg (2 )
3
,2 =8,2
multiplying powers
eg 2
3log25
(M1)
a
(M1)
, alog2 5
correct working
(A1)
eg 2log2125 , log2 53 , ( 2log25 )
8log25 = 125
A1
3
N3
[4 marks]
Examiners report
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 23log25 yet could not make that final leap to an integer.
An arithmetic sequence is given by 5, 8, 11, ….
13. (a)
(b)
(c)
Write down the value of d .
Find
(i)
u100 ;
(ii)
S100 .
Given that un = 1502 , find the value of n .
[7 marks]
Markscheme
(a)
d=3
A1
N1
[1 mark]
(b)
(i)
correct substitution into term formula
(A1)
e.g. u100 = 5 + 3(99) , 5 + 3(100 − 1)
u100 = 302
(ii)
A1
N2
correct substitution into sum formula
eg S100 =
100
2
(2(5) + 99(3)) , S100 =
S100 = 15350
A1
100
2
(A1)
(5 + 302)
N2
[4 marks]
(c)
correct substitution into term formula
eg
1502 = 5 + 3(n − 1) , 1502 = 3n + 2
n = 500
A1
(A1)
N2
[2 marks]
Total [7 marks]
Examiners report
The majority of candidates had little difficulty with this question. If errors were made, they were normally made out of carelessness. A
very few candidates mistakenly used the formulas for geometric sequences and series.
In the expansion of (3x − 2)12 , the term in x5 can be expressed as (
14. (a)
(b)
12
) × (3x)p × (−2)q .
r
Write down the value of p , of q and of r .
Find the coefficient of the term in
x5
[5 marks]
.
Markscheme
(a)
p=5,q=7,r=7
(accept r = 5)
A1A1A1
N3
[3 marks]
(b)
correct working
eg (
(A1)
12
) × (3x)5 × (−2)7 , 792 , 243 , −27 , 24634368
7
coefficient of term in x5 is −24634368
A1
N2
Note: Do not award the final A1 for an answer that contains x.
[2 marks]
Total [5 marks]
Examiners report
Candidates frequently made reasonable attempts at both parts of the question. Those who correctly stated the values in (a) were
generally successful in part (b). Many candidates offered the whole term rather than the coefficient in part (b) and lost the final mark.
Some candidates appeared to have misread the order of the variables, stating that p = 7 (instead of r = 7 ), q = 5 (instead of p = 5)
and r = 5 or 7 (instead of q = 5 or 7). A large number of candidates did not make the connection between parts (a) and (b).
Let log3 p = 6 and log3 q = 7 .
15. (a)
Find log3 p2 .
p
log3 ( q )
(b)
Find
(c)
Find log3 (9p) .
[7 marks]
.
Markscheme
(a)
METHOD 1
(M1)
evidence of correct formula
eg log un = n log u , 2log3 p
log3 (p2 ) = 12
A1
N2
METHOD 2
valid method using p = 36
6 2
eg log3 (3 ) , log 3
log3 (p2 ) = 12
12
(M1)
, 12log3 3
A1
N2
[2 marks]
(b)
METHOD 1
evidence of correct formula
(M1)
eg log ( q ) = log p − log q , 6 − 7
p
log3 ( q ) = −1
p
A1
N2
METHOD 2
valid method using p = 36 and q = 37
(M1)
eg log3 ( 37 ) , log 3−1 , −log3 3
6
3
p
log3 ( q )
= −1
A1
N2
[2 marks]
(c)
METHOD 1
evidence of correct formula
(M1)
eg log3 uv = log3 u + log3 v , log 9 + log p
log3 9 = 2 (may be seen in expression)
A1
eg 2 + log p
log3 (9p) = 8
A1
N2
METHOD 2
valid method using p = 36
(M1)
eg log3 (9 × 36 ) , log3 (32 × 36 )
A1
correct working
6
eg log3 9 + log3 3 , log3 38
log3 (9p) = 8
A1
N2
[3 marks]
Total [7 marks]
Examiners report
This question proved to be surprisingly challenging for many candidates. A common misunderstanding was to set p equal to 6 and q
equal to 7. A large number of candidates had trouble applying the rules of logarithms, and made multiple errors in each part of the
question. Common types of errors included incorrect working such as log3 p2 = 36 in part (a), log3 ( q ) =
p
p
log3 ( q )
= log3 6 − log3 7 in part (b), and log3 (9p) = 54 in part (c).
log36
log37
or
16. The sum of the first three terms of a geometric sequence is 62.755, and the sum of the infinite sequence is 440. Find the
[6 marks]
common ratio.
Markscheme
correct substitution into sum of a geometric sequence
A1
eg 62.755 = u1 ( 1−r
) , u1 + u1 r + u1 r2 = 62.755
1−r
3
correct substitution into sum to infinity
eg
u1
1−r
A1
= 440
attempt to eliminate one variable
(M1)
eg substituting u1 = 440(1 − r)
correct equation in one variable
(A1)
eg 62.755 = 440(1 − r) ( 1−r
) , 440(1 − r)(1 + r + r2 ) = 62.755
1−r
3
evidence of attempting to solve the equation in a single variable
(M1)
eg sketch, setting equation equal to zero, 62.755 = 440(1 − r3 )
r = −0.95 =
19
20
A1
N4
[6 marks]
Examiners report
Many candidates were able to successfully obtain two equations in two variables, but far fewer were able to correctly solve for the
value of r. Some candidates had misread errors for either 440 or 62.755, with some candidates taking the French and Spanish exams
mistaking the decimal comma for a thousands comma. Many candidates who attempted to solve algebraically did not cancel the 1 − r
from both sides and ended up with a 4th degree equation that they could not solve. Some of these candidates obtained the extraneous
answer of r − 1 as well. Some candidates used a minimum of algebra to eliminate the first term and then quickly solved the resulting
equation on their GDC.
17.
6
The constant term in the expansion of ( xa + ax ) , where a ∈ R is 1280. Find a .
2
Markscheme
evidence of binomial expansion
(M1)
0
1
2
2
6
6
5
eg selecting correct term,( xa ) ( ax ) + ( ) ( xa ) ( ax ) + …
1
evidence of identifying constant term in expansion for power 6
eg r = 3 , 4th term
evidence of correct term (may be seen in equation)
3
6
2
6
3
eg 20 a3 , ( ) ( xa ) ( ax )
a
3
attempt to set up their equation
(M1)
2
6
3
eg ( ) ( xa ) ( ax ) = 1280, a3 = 1280
3
3
correct equation in one variable a
eg 20a3 = 1280 , a3 = 64
a=4
A1
[7 marks]
N4
(A1)
A2
(A1)
[7 marks]
Examiners report
Many candidates struggled with this question. Some had difficulty with the binomial expansion, while others did not understand that
the constant term had no x , while still others were unable to simplify a ratio of exponentials with a common base. Some candidates
found r = 3 using algebraic methods while others found it by writing out the first several terms. In some cases, candidates just set the
entire expansion equal to 1280.
The following diagram shows a circle with centre O and radius r cm.
^ B = 1.4 radians .
Points A and B are on the circumference of the circle and AO
^ O = π radians .
The point C is on [OA] such that BC
2
18. Show that OC = r cos 1.4 .
[1 mark]
Markscheme
use right triangle trigonometry
eg cos 1.4 =
A1
OC
r
OC = r cos 1.4
AG
N0
[1 mark]
Examiners report
As to be expected, candidates found this problem challenging. In part (a), many were able to use right angle trigonometry to find the
length of OC.
The first three terms of an infinite geometric sequence are 32, 16 and 8.
19a. Write down the value of r .
[1 mark]
Markscheme
r=
16
32
(= 12 )
A1
N1
[1 mark]
Examiners report
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 .
19b. Find u6 .
[2 marks]
Markscheme
correct calculation or listing terms
e.g. 32 × ( 12 )
u6 = 1
6−1
A1
(A1)
3
, 8 × ( 12 ) , 32, … 4, 2, 1
N2
[2 marks]
Examiners report
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 .
19c. Find the sum to infinity of this sequence.
[2 marks]
Markscheme
evidence of correct substitution in S∞
e.g.
32
1− 1
,
2
S∞ = 64
A1
32
1
2
A1
N1
[2 marks]
Examiners report
A handful of candidates struggled with the basic computation involved in part (c).
The diagram shows quadrilateral ABCD with vertices A(1, 0), B(1, 5), C(5, 2) and D(4, −1) .
−
−→
4
Show that AC = ( ) .
2
−→
(ii) Find BD .
−
−→
−→
(iii) Show that AC is perpendicular to BD .
20a.
(i)
[5 marks]
Markscheme
(i) correct approach
A1
−
−→ −
−→
5
1
e.g. OC − OA , ( ) − ( )
2
0
−
−→
4
AC = ( )
2
AG
N0
(M1)
(ii) appropriate approach
e.g. D − B , (
4
1
) − ( ) , move 3 to the right and 6 down
−1
5
−→
3
BD = (
)
−6
A1
N2
(iii) finding the scalar product
A1
e.g. 4(3) + 2(−6) , 12 − 12
valid reasoning
R1
e.g. 4(3) + 2(−6) = 0 , scalar product is zero
−
−→
−→
AC is perpendicular to BD AG N0
[5 marks]
Examiners report
The majority of candidates were successful on part (a), finding vectors between two points and using the scalar product to show two
vectors to be perpendicular.
20b. The line (AC) has equation r = u + sv .
(i)
[4 marks]
Write down vector u and vector v .
(ii) Find a vector equation for the line (BD).
Markscheme
(i) correct “position” vector for u; “direction” vector for v
A1A1
N2
5
1
4
−2
e.g. u = ( ) , u = ( ) ; v = ( ) , v = (
)
2
0
2
−1
5
−4
accept in equation e.g. ( ) + t (
)
2
−2
−→
(ii) any correct equation in the form r = a + tb , where b = BD
1
3
x
4
−1
r = ( ) +t(
),( )= (
) +t(
)
5
−6
y
−1
2
A2
N2
[4 marks]
Examiners report
Although a large number of candidates answered part (b) correctly, there were many who had trouble with the vector equation of a
line. Most notably, there were those who confused the position vector with the direction vector, and those who wrote their equation in
an incorrect form.
20c. The lines (AC) and (BD) intersect at the point P(3, k) .
Show that k = 1 .
Markscheme
METHOD 1
substitute (3, k) into equation for (AC) or (BD)
(M1)
e.g. 3 = 1 + 4s , 3 = 1 + 3t
A1
value of t or s
1
2
1
2
e.g. s = , − , t = 23 , − 13
A1
substituting
e.g. k = 0 +
k=1
1
(2)
2
AG
N0
METHOD 2
setting up two equations
(M1)
e.g. 1 + 4s = 4 + 3t , 2s = −1 − 6t ; setting vector equations of lines equal
value of t or s
A1
e.g. s = 12 , − 12 , t = 23 , − 13
A1
substituting
e.g. r = (
k=1
4
3
) − 13 (
)
−1
−6
AG
[3 marks]
N0
[3 marks]
Examiners report
In part (c), most candidates seemed to know what was required, though there were many who made algebraic errors when solving for
the parameters. A few candidates worked backward, using k = 1 , which is not allowed on a "show that" question.
20d. The lines (AC) and (BD) intersect at the point P(3, k) .
[5 marks]
Hence find the area of triangle ACD.
Markscheme
−→
1
PD = (
) (A1)
−2
−→
−−−−−
|PD | = √22 + 12 (= √5) (A1)
−
−→
−−−−−
−−
|AC | = √42 + 22 (= √20 ) (A1)
−
−→
−→
−−
area = 12 × |AC | × |PD | (= 12 × √20 × √5)
=5
A1
M1
N4
[5 marks]
Examiners report
In part (d), candidates attempted many different geometric and vector methods to find the area of the triangle. As the question said
"hence", it was required that candidates should use answers from their previous working - i.e. AC⊥BD and P(3, 1) . Some
geometric approaches, while leading to the correct answer, did not use "hence" or lacked the required justification.
f(x) =
3
Let f(x) = x3 . The following diagram shows part of the graph of f .
The point P(a,f(a)) , where a > 0 , lies on the graph of f . The tangent at P crosses the x-axis at the point Q ( 23 ,0) . This tangent intersects
the graph of f at the point R(−2, −8) .
21a. (i)
Show that the gradient of [PQ] is
a3
a− 2
3
(ii) Find f ′ (a) .
(iii) Hence show that a = 1 .
.
[7 marks]
Markscheme
(i) substitute into gradient =
e.g.
y1 − y2
x1− x2
(M1)
f(a)−0
a− 2
3
substituting f(a) = a3
e.g.
a3−0
A1
a− 2
3
gradient
a3
AG
a− 2
N0
3
A1
(ii) correct answer
e.g.
3a2
′
N1
′
, f (a) = 3 , f (a) =
a3
a− 2
3
(iii) METHOD 1
evidence of approach
(M1)
′
e.g. f (a) = gradient , 3a2 =
a3
a− 2
3
A1
simplify
e.g. 3a2 (a − 23 ) = a3
A1
rearrange
e.g.
3a3
− 2a2 = a3
A1
evidence of solving
e.g. 2a3 − 2a2 = 2a2 (a − 1) = 0
a=1
AG
N0
METHOD 2
gradient RQ =
e.g.
−8
A1
3
A1
simplify
−8
−8
−2− 2
,3
3
evidence of approach
(M1)
e.g. f ′ (a) = gradient , 3a2 =
e.g.
a=1
,
3
a3
a− 2
3
=3
A1
simplify
3a2
−8
−2− 2
= 3 , a2 = 1
AG
N0
[7 marks]
Examiners report
Part (a) seemed to be well-understood by many candidates, and most were able to earn at least partial marks here. Part (ai) was a
"show that" question, and unfortunately there were some candidates who did not show how they arrived at the given expression.
The equation of the tangent at P is y = 3x − 2 . Let T be the region enclosed by the graph of f , the tangent [PR] and the line x = k , between
x = −2 and x = k where −2 < k < 1 . This is shown in the diagram below.
21b. Given that the area of T is 2k + 4 , show that k satisfies the equation k4 − 6k2 + 8 = 0 .
[9 marks]
Markscheme
approach to find area of T involving subtraction and integrals (M1)
k
k
e.g. ∫ f − (3x − 2)dx , ∫−2 (3x − 2) − ∫−2 x3 , ∫ (x3 − 3x + 2)
correct integration with correct signs
e.g.
1 4
x
4
−
3 2
x
2
+ 2x ,
3 2
x
2
− 2x −
A1A1A1
1 4
x
4
correct limits −2 and k (seen anywhere)
A1
k
e.g. ∫−2
(x3 − 3x + 2)dx , [ 14 x4 − 32 x2 + 2x]
attempt to substitute k and −2
k
−2
(M1)
correct substitution into their integral if 2 or more terms
A1
e.g. ( 14 k4 − 32 k2 + 2k) − (4 − 6 − 4)
setting their integral expression equal to 2k + 4 (seen anywhere)
simplifying
e.g.
1 4
k
4
−
(M1)
A1
3 2
k
2
+2 = 0
k4 − 6k2 + 8 = 0
AG
N0
[9 marks]
Examiners report
In part (b), the concept seemed to be well-understood. Most candidates saw the necessity of using definite integrals and subtracting
the two functions, and the integration was generally done correctly. However, there were a number of algebraic and arithmetic errors
which prevented candidates from correctly showing the desired final result.
The nth term of an arithmetic sequence is given by un = 5 + 2n .
22a. Write down the common difference.
[1 mark]
Markscheme
d=2
A1
N1
[1 mark]
Examiners report
The majority of candidates could either recognize the common difference in the formula for the nth term or could find it by writing out
the first few terms of the sequence.
22b. (i)
Given that the nth term of this sequence is 115, find the value of n .
[5 marks]
(ii) For this value of n , find the sum of the sequence.
Markscheme
(i) 5 + 2n = 115
n = 55
A1
(A1)
N2
(ii) u1 = 7 (may be seen in above)
(A1)
correct substitution into formula for sum of arithmetic series
e.g. S55 =
55
(7 + 115)
2
, S55 =
S55 = 3355 (accept 3360)
55
(2(7) + 54(2))
2
A1
(A1)
55
, ∑ (5 + 2k)
k=1
N3
[5 marks]
Examiners report
Part (b) demonstrated that candidates were not familiar with expression, "nth term". Many stated that the first term was 5 and then
decided to use their own version of the nth term formula leading to a great many errors in (b) (ii).
The following diagram represents a large Ferris wheel at an amusement park.
The points P, Q and R represent different positions of a seat on the wheel.
The wheel has a radius of 50 metres and rotates clockwise at a rate of one revolution every 30 minutes.
A seat starts at the lowest point P, when its height is one metre above the ground.
23a. Find the height of a seat above the ground after 15 minutes.
[2 marks]
Markscheme
valid approach
(M1)
e.g. 15 mins is half way, top of the wheel, d + 1
height = 101 (metres)
A1
N2
[2 marks]
Examiners report
Part (a) was well done with most candidates obtaining the correct answer.
23b. After six minutes, the seat is at point Q. Find its height above the ground at Q.
[5 marks]
Markscheme
evidence of identifying rotation angle after 6 minutes
e.g.
2π
5
,
1
5
of a rotation, 72
A1
∘
(M1)
evidence of appropriate approach
e.g. drawing a right triangle and using cosine ratio
correct working (seen anywhere)
e.g. cos
2π
5
=
x
50
A1
, 15.4(508 …)
evidence of appropriate method
M1
e.g. height = radius + 1 − 15.45 …
height = 35.5 (metres) (accept 35.6)
A1
N2
[5 marks]
Examiners report
Part (b) however was problematic with most errors resulting from incorrect, missing or poorly drawn diagrams. Many did not
recognize this as a triangle trigonometric problem while others used the law of cosines to find the chord length rather than the vertical
height, but this was only valid if they then used this to complete the problem. Many candidates misinterpreted the question as one that
was testing arc length and area of a sector and made little to no progress in part (b).
Still, others recognized that 6 minutes represented 15 of a rotation, but the majority then thought the height after 6 minutes should be
of the maximum height, treating the situation as linear. There were even a few candidates who used information given later in the
question to answer part (b). Full marks are not usually awarded for this approach.
23c. The height of the seat above ground after t minutes can be modelled by the function h(t) = 50 sin(b(t − c)) + 51.
Find the value of b and of c .
1
5
[6 marks]
Markscheme
METHOD 1
evidence of substituting into b =
2π
period
(M1)
correct substitution
e.g. period = 30 minutes, b =
π
b = 0.209 ( 15
)
A1
substituting into h(t)
2π
30
A1
N2
(M1)
e.g. h(0) = 1 , h(15) = 101
correct substitution
A1
π
1 = 50 sin (− 15
c) + 51
c = 7.5 A1
N2
METHOD 2
evidence of setting up a system of equations
(M1)
two correct equations
e.g. 1 = 50 sin b(0 − c) + 51 , 101 = 50 sin b(15 − c) + 51
attempt to solve simultaneously
A1A1
(M1)
e.g. evidence of combining two equations
π
b = 0.209 ( 15
) , c = 7.5
A1A1
N2N2
[6 marks]
Examiners report
Part (c) was not well done. It was expected that candidates simply use the formula
2π
period
to find the value of b and then substitute back
into the equation to find the value of c. However, candidates often preferred to set up a pair of equations and attempt to solve them
analytically, some successful, some not. No attempts were made to solve this system on the GDC indicating that candidates do not get
exposed to many “systems” that are not linear. Confusing radians and degrees here did nothing to improve the lack of success.
23d. The height of the seat above ground after t minutes can be modelled by the function h(t) = 50 sin(b(t − c)) + 51.
[3 marks]
Hence find the value of t the first time the seat is 96 m above the ground.
Markscheme
evidence of solving h(t) = 96
(M1)
e.g. equation, graph
t = 12.8 (minutes)
A2
N3
[3 marks]
Examiners report
In part (d), candidates were clear on what was required and set their equation equal to 96. Yet again however, solving this equation
graphically using a GDC proved too daunting a task for most.
In an arithmetic sequence u1 = 7 , u20 = 64 and un = 3709 .
24a. Find the value of the common difference.
[3 marks]
Markscheme
evidence of choosing the formula for 20th term
(M1)
e.g. u20 = u1 + 19d
A1
correct equation
e.g. 64 = 7 + 19d , d =
d=3
A1
64−7
19
N2
[3 marks]
Examiners report
The majority of candidates gained full marks in this question. However, the presentation of work was often disappointing, and there
were a few incorrect trial and error approaches in part (a) resulting in division by 20 rather than 19.
24b. Find the value of n .
[2 marks]
Markscheme
correct substitution into formula for un
A1
e.g. 3709 = 7 + 3(n − 1) , 3709 = 3n + 4
n = 1235
A1
N1
[2 marks]
Examiners report
The majority of candidates gained full marks in this question. However, the presentation of work was often disappointing, and there
were a few incorrect trial and error approaches in part (a) resulting in division by 20 rather than 19.
Let f(x) = log3 x2 + log3 16 − log3 4 , for x > 0 .
25a. Show that f(x) = log3 2x .
[2 marks]
Markscheme
(A1)
combining 2 terms
e.g. log3 8x − log3 4 , log3 12 x + log3 4
expression which clearly leads to answer given
e.g.
log3 8x
4
,
A1
log3 4x
2
f(x) = log3 2x
AG
N0
[2 marks]
Examiners report
Few candidates had difficulty with part (a) although it was often communicated using some very sloppy applications of the rules of
logarithm, writing
log 16
log 4
instead of log ( 164 ) .
25b. Find the value of f(0.5) and of f(4.5) .
[3 marks]
Markscheme
attempt to substitute either value into f
(M1)
e.g. log3 1 , log3 9
f(0.5) = 0 , f(4.5) = 2
A1A1
N3
[3 marks]
Examiners report
Part (b) was generally done well.
25c. The function f can also be written in the form f(x) =
(i)
ln ax
ln b
.
[6 marks]
Write down the value of a and of b .
(ii) Hence on graph paper, sketch the graph of f , for −5 ≤ x ≤ 5 , −5 ≤ y ≤ 5 , using a scale of 1 cm to 1 unit on each axis.
(iii) Write down the equation of the asymptote.
Markscheme
(i) a = 2 , b = 3
A1A1
N1N1
(ii)
A1A1A1
N3
Note: Award A1 for sketch approximately through (0.5 ± 0.1, 0 ± 0.1) , A1 for approximately correct shape, A1 for sketch
asymptotic to the y-axis.
(iii) x = 0 (must be an equation)
A1
N1
[6 marks]
Examiners report
Part (c) (i) was generally done well; candidates seemed quite comfortable changing bases. There were some very good sketches in (c)
(ii), but there were also some very poor ones with candidates only considering shape and not the location of the x-intercept or the
asymptote. A surprising number of candidates did not use the scale required by the question and/or did not use graph paper to sketch
the graph. In some cases, it was evident that students simply transposed their graphs from their GDC without any analytical
consideration.
−1
(0)
25d. Write down the value of f −1 (0) .
[1 mark]
Markscheme
f −1 (0) = 0.5
A1
N1
[1 mark]
Examiners report
Part (d) was poorly done as candidates did not consider the command term, “write down” and often proceeded to find the inverse
function before making the appropriate substitution.
25e. The point A lies on the graph of f . At A, x = 4.5 .
On your diagram, sketch the graph of f
−1
[4 marks]
, noting clearly the image of point A.
Markscheme
A1A1A1A1
N4
Note: Award A1 for sketch approximately through (0 ± 0.1, 0.5 ± 0.1) , A1 for approximately correct shape of the graph reflected
over y = x , A1 for sketch asymptotic to x-axis, A1 for point (2 ± 0.1, 4.5 ± 0.1) clearly marked and on curve.
[4 marks]
Examiners report
Part (e) eluded a great many candidates as most preferred to attempt to find the inverse analytically rather than simply reflecting the
graph of f in the line y = x . This graph also suffered from the same sort of problems as the graph in (c) (ii). Some students did not
have their curve passing through (2, 4.5) nor did they clearly indicate its position as instructed. This point was often mislabelled on
the graph of f. The efforts in this question demonstrated that students often work tenuously from one question to the next, without
considering the "big picture", thereby failing to make important links with earlier parts of the question.
In an arithmetic sequence, u1 = 2 and u3 = 8 .
26a. Find d .
[2 marks]
Markscheme
(M1)
attempt to find d
e.g.
u3− u1
2
d=3
, 8 = 2 + 2d
A1
N2
[2 marks]
Examiners report
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.
26b. Find u20 .
[2 marks]
Markscheme
(A1)
correct substitution
e.g. u20 = 2 + (20 − 1)3 , u20 = 3 × 20 − 1
u20 = 59
A1
N2
[2 marks]
Examiners report
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.
26c. Find S20 .
[2 marks]
Markscheme
(A1)
correct substitution
e.g. S20 =
S20 = 610
20
(2 + 59)
2
A1
, S20 =
20
(2 × 2 + 19 × 3)
2
N2
[2 marks]
Examiners report
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.
Let f(x) = 3 ln x and g(x) = ln 5x3 .
27a. Express g(x) in the form f(x) + ln a , where a ∈ Z+ .
[4 marks]
Markscheme
attempt to apply rules of logarithms
e.g.
ln ab
(M1)
= b ln a , ln ab = ln a + ln b
correct application of ln ab = b ln a (seen anywhere)
A1
e.g. 3 ln x = ln x3
correct application of ln ab = ln a + ln b (seen anywhere)
A1
e.g. ln 5x3 = ln 5 + ln x3
so ln 5x3 = ln 5 + 3 ln x
g(x) = f(x) + ln 5 (accept g(x) = 3 ln x + ln 5 )
A1
N1
[4 marks]
Examiners report
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 5x3 = 3 ln 5x . This error was often followed by other errors.
27b. The graph of g is a transformation of the graph of f . Give a full geometric description of this transformation.
[3 marks]
Markscheme
transformation with correct name, direction, and value
e.g. translation by (
A3
0
) , shift up by ln 5 , vertical translation of ln 5
ln 5
[3 marks]
Examiners report
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.
The diagram below shows a plan for a window in the shape of a trapezium.
Three sides of the window are 2 m long. The angle between the sloping sides of the window and the base is θ , where 0 < θ < π2 .
28a. Show that the area of the window is given by y = 4 sin θ + 2 sin 2θ .
[5 marks]
Markscheme
evidence of finding height, h
e.g. sin θ =
h
2
(A1)
, 2 sin θ
evidence of finding base of triangle, b
(A1)
e.g. cos θ = , 2 cos θ
b
2
attempt to substitute valid values into a formula for the area of the window
(M1)
e.g. two triangles plus rectangle, trapezium area formula
correct expression (must be in terms of θ )
A1
e.g. 2 ( 12 × 2 cos θ × 2 sin θ) + 2 × 2 sin θ , 12 (2 sin θ)(2 + 2 + 4 cos θ)
attempt to replace 2 sin θ cos θ by sin 2θ
M1
e.g. 4 sin θ + 2(2 sin θ cos θ)
y = 4 sin θ + 2 sin 2θ
AG
N0
[5 marks]
Examiners report
As the final question of the paper, this question was understandably challenging for the majority of the candidates. Part (a) was
generally attempted, but often with a lack of method or correct reasoning. Many candidates had difficulty presenting their ideas in a
clear and organized manner. Some tried a "working backwards" approach, earning no marks.
28b. Zoe wants a window to have an area of 5 m2 . Find the two possible values of θ .
[4 marks]
Markscheme
correct equation
A1
e.g. y = 5 , 4 sin θ + 2 sin 2θ = 5
evidence of attempt to solve
(M1)
e.g. a sketch, 4 sin θ + 2 sin θ − 5 = 0
θ = 0.856 (49.0∘ ) , θ = 1.25 (71.4∘ )
A1A1
N3
[4 marks]
Examiners report
In part (b), most candidates understood what was required and set up an equation, but many did not make use of the GDC and instead
attempted to solve this equation algebraically which did not result in the correct solution. A common error was finding a second
solution outside the domain.
28c. John wants two windows which have the same area A but different values of θ .
Find all possible values for A .
[7 marks]
Markscheme
recognition that lower area value occurs at θ = π2
finding value of area at θ =
π
2
(M1)
(M1)
e.g. 4 sin ( π2 ) + 2 sin (2 × π2 ) , draw square
A=4
(A1)
(M1)
recognition that maximum value of y is needed
A = 5.19615 …
(A1)
4 < A < 5.20 (accept 4 < A < 5.19 )
A2
N5
[7 marks]
Examiners report
A pleasing number of stronger candidates made progress on part (c), recognizing the need for the end point of the domain and/or the
maximum value of the area function (found graphically, analytically, or on occasion, geometrically). However, it was evident from
candidate work and teacher comments that some candidates did not understand the wording of the question. This has been taken into
consideration for future paper writing.
29a. Expand (2 + x)4 and simplify your result.
[3 marks]
Markscheme
evidence of expanding
M1
e.g. 24 + 4(23 )x + 6(22 )x2 + 4(2)x3 + x4 , (4 + 4x + x2 )(4 + 4x + x2 )
(2 + x)4 = 16 + 32x + 24x2 + 8x3 + x4
A2
N2
[3 marks]
Examiners report
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.
29b. Hence, find the term in x2 in (2 + x)4 (1 +
1
)
x2
.
[3 marks]
Markscheme
finding coefficients 24 and 1
term is
25x2
A1
(A1)(A1)
N3
[3 marks]
Examiners report
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.
The straight line with equation y = 34 x makes an acute angle θ with the x-axis.
30a. Write down the value of tan θ .
[1 mark]
Markscheme
tan θ = 34 (do not accept 34 x )
A1
N1
[1 mark]
Examiners report
Many candidates drew a diagram to correctly find tan θ , although few recognized that a line through the origin can be expressed as
y = x tan θ , with gradient tan θ , which is explicit in the syllabus.
30b. Find the value of
(i)
[6 marks]
sin 2θ ;
(ii) cos 2θ .
Markscheme
(i) sin θ = 35 , cos θ =
4
5
correct substitution
A1
(A1)(A1)
e.g. sin 2θ = 2 ( 35 ) ( 45 )
sin 2θ =
24
25
A1
N3
A1
(ii) correct substitution
2
2
e.g. cos 2θ = 1 − 2( 35 ) , ( 45 ) − ( 35 )
cos 2θ =
7
25
A1
2
N1
[6 marks]
Examiners report
A surprising number were unable to find the ratios for sin θ and cos θ from tan θ . It was not uncommon for candidates to use
unreasonable values, such as sin θ = 3 and cos θ = 4 , or to write nonsense such as 2 sin 35 cos 45 .
Let f(x) = log3 √x , for x > 0 .
31a. Show that f −1 (x) = 32x .
[2 marks]
Markscheme
interchanging x and y (seen anywhere)
(M1)
e.g. x = log √y (accept any base)
evidence of correct manipulation
A1
1
2
e.g. 3x = √y , 3y = x , x = 12 log3 y , 2y = log3 x
f −1 (x) = 32x
AG
N0
[2 marks]
Examiners report
Candidates were generally skilled at finding the inverse of a logarithmic function.
31b. Write down the range of f −1 .
[1 mark]
Markscheme
y > 0 , f −1 (x) > 0
A1
N1
[1 mark]
Examiners report
Few correctly gave the range of this function, often stating “all real numbers” or “ y ≥ 0 ”, missing the idea that the range of an
inverse is the domain of the original function.
31c. Let g(x) = log3 x , for x > 0 .
[4 marks]
Find the value of (f −1 ∘ g)(2) , giving your answer as an integer.
Markscheme
METHOD 1
finding g(2) = log3 2 (seen anywhere)
(M1)
attempt to substitute
e.g. (f
−1
∘ g)(2) = 3
2 log 32
evidence of using log or index rule
e.g. (f
−1
∘ g)(2) = 3
(f −1 ∘ g)(2) = 4
A1
log 34
,3
A1
(A1)
log322
N1
METHOD 2
attempt to form composite (in any order)
(M1)
e.g. (f −1 ∘ g)(x) = 32log3x
evidence of using log or index rule
2
e.g. (f −1 ∘ g)(x) = 3log3x , 3log3x
(f −1 ∘ g)(x) = x2
(f −1 ∘ g)(2) = 4
(A1)
2
A1
A1
N1
[4 marks]
Examiners report
Some candidates answered part (c) correctly, although many did not get beyond 32log32 . Some attempted to form the composite in the
incorrect order. Others interpreted (f −1 ∘ g)(2) as multiplication by 2.
32. Let f(x) = kx4 . The point P(1, k) lies on the curve of f . At P, the normal to the curve is parallel to y = − 1 x . Find the value [6 marks]
of k.
8
Markscheme
gradient of tangent = 8 (seen anywhere)
′
f (x) =
4kx3
(seen anywhere)
(A1)
A1
recognizing the gradient of the tangent is the derivative
setting the derivative equal to 8
e.g.
4kx3
=8,
kx3
(A1)
=2
substituting x = 1 (seen anywhere)
k=2
A1
(M1)
(M1)
N4
[6 marks]
Examiners report
Candidates‟ success with this question was mixed. Those who understood the relationship between the derivative and the gradient of
the normal line were not bothered by the lack of structure in the question, solving clearly with only a few steps, earning full marks.
Those who were unclear often either gained a few marks for finding the derivative and substituting x = 1 , or no marks for working
that did not employ the derivative. Misunderstandings included simply finding the equation of the tangent or normal line, setting the
derivative equal to the gradient of the normal, and equating the function with the normal or tangent line equation. Among the
candidates who demonstrated greater understanding, more used the gradient of the normal (the equation − 14 k = − 18 ) than the
gradient of the tangent (4k = 8 ) ; this led to more algebraic errors in obtaining the final answer of k = 2 . A number of unsuccessful
candidates wrote down a lot of irrelevant mathematics with no plan in mind and earned no marks.
33. Solve log2 x + log2 (x − 2) = 3 , for x > 2 .
[7 marks]
Markscheme
recognizing log a + log b = log ab (seen anywhere)
e.g. log2 (x(x − 2)) ,
x2
(A1)
− 2x
recognizing loga b = x ⇔ ax = b
(A1)
3
e.g. 2 = 8
A1
correct simplification
e.g. x(x − 2) = 23 , x2 − 2x − 8
evidence of correct approach to solve
(M1)
e.g. factorizing, quadratic formula
correct working
A1
e.g. (x − 4)(x + 2) ,
x=4
A2
2± √36
2
N3
[7 marks]
Examiners report
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.
Let f(x) = klog2 x .
34a. Given that f −1 (1) = 8 , find the value of k .
[3 marks]
Markscheme
METHOD 1
recognizing that f(8) = 1
(M1)
e.g. 1 = klog2 8
recognizing that log2 8 = 3
(A1)
e.g. 1 = 3k
k=
1
3
A1
N2
METHOD 2
attempt to find the inverse of f(x) = klog2 x
e.g. x = klog2 y , y = 2
(M1)
x
k
(M1)
substituting 1 and 8
1
k
e.g. 1 = klog2 8 , 2 = 8
k=
1
log28
(k = 13 )
A1
N2
[3 marks]
Examiners report
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.
34b. Find f −1 ( 2 ) .
3
[4 marks]
Markscheme
METHOD 1
recognizing that f(x) =
e.g.
2
3
2
3
(M1)
= 13 log2 x
log2 x = 2
(A1)
f −1 ( 23 ) = 4 (accept x = 4)
A2
N3
METHOD 2
attempt to find inverse of f(x) = 13 log2 x
(M1)
x
e.g. interchanging x and y , substituting k = 13 into y = 2 k
(A1)
correct inverse
e.g. f
−1
(x) = 2
f −1 ( 23 ) = 4
[4 marks]
3x
, 23x
A2
N3
Examiners report
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.
In a class of 100 boys, 55 boys play football and 75 boys play rugby. Each boy must play at least one sport from football and rugby.
35. (i)
Find the number of boys who play both sports.
[3 marks]
(ii) Write down the number of boys who play only rugby.
Markscheme
(i) evidence of substituting into n(A ∪ B) = n(A) + n(B) − n(A ∩ B)
(M1)
e.g. 75 + 55 − 100 , Venn diagram
30
A1
(ii) 45
N2
A1
N1
[3 marks]
Examiners report
Overall, this question was very well done. There were some problems with the calculation of conditional probability, where a
considerable amount of candidates tried to use a formula instead of using its concept and analysing the problem. It is the kind of
question where it can be seen if the concept is not clear to candidates.
36. In an arithmetic sequence, S40 = 1900 and u40 = 106 . Find the value of u1 and of d .
[6 marks]
Markscheme
METHOD 1
substituting into formula for S40
(M1)
A1
correct substitution
e.g. 1900 =
40(u1+106)
2
u1 = −11
A1
N2
substituting into formula for u40 or S40
correct substitution
(M1)
A1
e.g. 106 = −11 + 39d , 1900 = 20(−22 + 39d)
d=3
A1
N2
METHOD 2
substituting into formula for S40
correct substitution
(M1)
A1
e.g. 20(2u1 + 39d) = 1900
substituting into formula for u40
correct substitution
(M1)
A1
e.g. 106 = u1 + 39d
u1 = −11 , d = 3
A1A1
N2N2
[6 marks]
Examiners report
Most candidates answered this question correctly. Those who chose to solve with a system of equations often did so algebraically,
using a fair bit of time doing so and sometimes making a careless error in the process. Few candidates took advantage of the system
solving features of the GDC.
Let f(x) = ex+3 .
37. Solve the equation f −1 (x) = ln 1 .
x
[4 marks]
Markscheme
collecting like terms; using laws of logs
(A1)(A1)
e.g. ln x − ln ( x1 ) = 3 , ln x + ln x = 3 , ln ( x1 ) = 3 , ln x2 = 3
x
simplify
(A1)
e.g. ln x = 32 , x2 = e3
3
−
−
x = e 2 (= √e3 )
A1
N2
[4 marks]
Examiners report
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.
A rectangle is inscribed in a circle of radius 3 cm and centre O, as shown below.
The point P(x , y) is a vertex of the rectangle and also lies on the circle. The angle between (OP) and the x-axis is θ radians, where 0 ≤ θ ≤ π2 .
38. Let the area of the rectangle be A.
[3 marks]
Show that A = 18 sin 2θ .
Markscheme
finding area
(M1)
e.g. A = 2x × 2y , A = 8 × 12 bh
substituting
A1
e.g. A = 4 × 3 sin θ × 3 cos θ , 8 × 12 × 3 cos θ × 3 sin θ
A = 18(2 sin θ cos θ) A1
A = 18 sin 2θ
AG
N0
[3 marks]
Examiners report
Those with answers from (a) could begin part (b), but many worked backwards and thus earned no marks. In a "show that" question,
a solution cannot begin with the answer given. The area of the rectangle could be found by using 2x × 2y , or by using the eight
small triangles, but it was essential that the substitution of the double-angle formula was shown before writing the given answer.
In an arithmetic series, the first term is –7 and the sum of the first 20 terms is 620.
39a. Find the common difference.
[3 marks]
Markscheme
attempt to substitute into sum formula for AP (accept term formula)
e.g. S20 =
20
2
{2(−7) + 19 d} , (or 202 (−7 + u20 ))
setting up correct equation using sum formula
e.g.
20
2
(M1)
A1
{2(−7) + 19d} = 620
d=4
A1
N2
[3 marks]
Examiners report
This question was generally well answered. Many candidates who got part (a) wrong, recovered and received full follow through
marks in part (b). There were a few candidates who confused the term and sum formula in part (a).
39b. Find the value of the 78th term.
[2 marks]
Markscheme
correct substitution u78 = −7 + 77(4)
= 301
A1
(A1)
N2
[2 marks]
Examiners report
This question was generally well answered. Many candidates who got part (a) wrong, recovered and received full follow through
marks in part (b). There were a few candidates who confused the term and sum formula in part (a).
The fifth term in the expansion of the binomial (a + b)n is given by (
40a. Write down the value of n .
10
) p6 (2q)4 .
4
[1 mark]
Markscheme
n = 10
A1
N1
[1 mark]
Examiners report
[N/A]
40b. Write down a and b, in terms of p and/or q.
[2 marks]
Markscheme
a = p , b = 2q (or a = 2q , b = p )
[2 marks]
A1A1
N1N1
Examiners report
[N/A]
40c. Write down an expression for the sixth term in the expansion.
[3 marks]
Markscheme
(
10
) p5 (2q)5
5
A1A1A1
N3
[3 marks]
Examiners report
The most common error was in (c) where many candidates interpreted the "sixth" term as using (
of 4 and 6 in the expression.
41a. Find log2 32 .
10
) , with accompanying powers
6
[1 mark]
Markscheme
5
A1
N1
[1 mark]
Examiners report
Part (a) proved very accessible.
41b. Given that log ( 32x ) can be written as px + qy , find the value of p and of q.
2
8y
Markscheme
METHOD 1
log2 ( 328y ) = log2 32x − log2 8y
x
= xlog2 32 − ylog2 8
log2 8 = 3
(A1)
(A1)
(A1)
p = 5 , q = −3 (accept 5x − 3y )
A1
N3
A1
N3
METHOD 2
32x
8y
=
=
x
(25)
(A1)
3 y
(2 )
x
25
y
23
(A1)
= 25x−3y
(A1)
log2 (25x−3y ) = 5x − 3y
p = 5 , q = −3 (accept 5x − 3y )
[4 marks]
[4 marks]
Examiners report
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.
Consider the arithmetic sequence 3, 9, 15, … , 1353 .
42a. Write down the common difference.
[1 mark]
Markscheme
A1
common difference is 6
N1
[1 mark]
Examiners report
Most candidates did well on this question. Any errors were usually arithmetic in nature but candidates were able to obtain followthrough marks on errors made in earlier parts.
42b. Find the number of terms in the sequence.
[3 marks]
Markscheme
evidence of appropriate approach
(M1)
e.g. un = 1353
correct working
A1
1353+3
6
e.g. 1353 = 3 + (n − 1)6 ,
n = 226
A1
N2
[3 marks]
Examiners report
Most candidates did well on this question. Any errors were usually arithmetic in nature but candidates were able to obtain followthrough marks on errors made in earlier parts.
42c. Find the sum of the sequence.
[2 marks]
Markscheme
evidence of correct substitution
e.g. S226 =
226(3+1353)
2
,
226
2
A1
(2 × 3 + 225 × 6)
S226 = 153228 (accept 153000)
A1
N1
[2 marks]
Examiners report
Most candidates did well on this question. Any errors were usually arithmetic in nature but candidates were able to obtain followthrough marks on errors made in earlier parts.
An arithmetic sequence, u1 , u2 , u3 … , has d = 11 and u27 = 263 .
43a. Find u1 .
[2 marks]
Markscheme
evidence of equation for u27
M1
e.g. 263 = u1 + 26 × 11 , u27 = u1 + (n − 1) × 11 , 263 − (11 × 26)
u1 = −23
A1
N1
[2 marks]
Examiners report
This problem was done well by the vast majority of candidates. Most students set out their working very neatly and logically and
gained full marks.
43b. (i)
Given that un = 516 , find the value of n .
[4 marks]
(ii) For this value of n , find Sn .
Markscheme
(i) correct equation
A1
e.g. 516 = −23 + (n − 1) × 11 , 539 = (n − 1) × 11
n = 50
A1
N1
(ii) correct substitution into sum formula
e.g. S50 =
50(−23+516)
2
, S50 =
S50 = 12325 (accept 12300)
A1
50(2×(−23)+49×11)
2
A1
N1
[4 marks]
Examiners report
This problem was done well by the vast majority of candidates. Most students set out their working very neatly and logically and
gained full marks.
44.
5
Find the term in x4 in the expansion of (3x2 − x2 ) .
[6 marks]
Markscheme
evidence of substituting into binomial expansion
(M1)
5
5
e.g. a5 + ( ) a4 b + ( ) a3 b2 + …
1
2
identifying correct term for x4
(M1)
evidence of calculating the factors, in any order
5
e.g. ( ) ,27x6 , 42 ; 10(3x2 )3 ( −2
)
x
x
2
A1A1A1
2
Note: Award A1 for each correct factor.
term = 1080x4
A1
N2
Note: Award M1M1A1A1A1A0 for 1080 with working shown.
[6 marks]
Examiners report
Although a great number of students recognized they could use the binomial theorem, fewer were successful in finding the term in x4
.
Candidates showed various difficulties when trying to solve this problem:
choosing the incorrect term
5
attempting to expand (3x2 − x2 ) by hand
finding only the coefficient of the term
not being able to determine which term would yield an x4
errors in the calculations of the coefficient.
45a. Consider the infinite geometric sequence 3, 3(0.9), 3(0.9)2 , 3(0.9)3 , … .
[1 mark]
Write down the 10th term of the sequence. Do not simplify your answer.
Markscheme
u10 = 3(0.9)9
A1
N1
[1 mark]
Examiners report
This question was well done by most candidates. There were a surprising number of candidates who lost a mark for not simplifying
3
to 30 , and there were a few candidates who used the formula for the finite sum unsuccessfully.
0.1
45b. Consider the infinite geometric sequence 3, 3(0.9), 3(0.9)2 , 3(0.9)3 , … .
Find the sum of the infinite sequence.
[4 marks]
Markscheme
recognizing r = 0.9
correct substitution
e.g. S =
3
1−0.9
3
S = 0.1
(A1)
S = 30
A1
(A1)
A1
N3
[4 marks]
Examiners report
This question was well done by most candidates. There were a surprising number of candidates who lost a mark for not simplifying
3
to 30, and there were a few candidates who used the formula for the finite sum unsuccessfully.
0.1
46a. Expand (x − 2)4 and simplify your result.
[3 marks]
Markscheme
evidence of expanding
4
e.g. (x − 2) =
x4
M1
+ 4x3 (−2) + 6x2 (−2)2
+ 4x(−2)3 + (−2)4
A2
N2
(x − 2)4 = x4 − 8x3 + 24x2 − 32x + 16
[3 marks]
Examiners report
Where candidates recognized the binomial nature of the expression, many completed the expansion successfully, although some
omitted the negative signs.
46b. Find the term in x3 in (3x + 4)(x − 2)4 .
[3 marks]
Markscheme
finding coefficients, 3 × 24(= 72) , 4 × (−8)(= −32)
term is 40x3
A1
(A1)(A1)
N3
[3 marks]
Examiners report
Where candidates recognized the binomial nature of the expression, many completed the expansion successfully, although some
omitted the negative signs. Few recognized that only the multiplications that achieve an index of 3 are required in part (b) and
distributed over the entire expression. Others did not recognize that two terms in the expansion must be combined.
47a. Consider the arithmetic sequence 2, 5, 8, 11,… .
Find u101 .
[3 marks]
Markscheme
d=3
(A1)
evidence of substitution into un = a + (n − 1)d
(M1)
e.g. u101 = 2 + 100 × 3
u101 = 302
A1
N3
[3 marks]
Examiners report
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 un = 3n − 1 .
47b. Consider the arithmetic sequence 2, 5, 8, 11,… .
[3 marks]
Find the value of n so that un = 152 .
Markscheme
(M1)
correct approach
e.g. 152 = 2 + (n − 1) × 3
correct simplification
(A1)
e.g. 150 = (n − 1) × 3 , 50 = n − 1 , 152 = −1 + 3n
n = 51
A1
N2
[3 marks]
Examiners report
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 un = 3n − 1 .
Let f(x) = 3(x + 1)2 − 12 .
48. Show that f(x) = 3x2 + 6x − 9 .
[2 marks]
Markscheme
f(x) = 3(x2 + 2x + 1) − 12
= 3x2 + 6x + 3 − 12
=
3x2
+ 6x − 9
AG
A1
A1
N0
[2 marks]
Examiners report
This problem was generally well done. The “show that” question in part (a) was done correctly by most candidates, with a few
attempting to show it by working backwards, which earned no marks.
Let S be the total area of the two segments shaded in the diagram below.
49. Show that S = 2(π − 2 sin θ) .
[3 marks]
Markscheme
area semicircle = 12 × π(2)2 (= 2π)
A1
area ΔAPB = 2 sin θ + 2 sin θ (= 4 sin θ)
A1
S = area of semicircle − area ΔAPB (= 2π − 4 sin θ)
S = 2(π − 2 sin θ)
AG
M1
N0
[3 marks]
Examiners report
Many candidates seemed to see why S = 2(π − 2 sin θ) but the arguments presented for showing why this result was true were not
very convincing in many cases. Explicit evidence of why the area of the semicircle was 2π was often missing as was an explanation
for 2(2 sin θ) and for subtraction.
Consider the infinite geometric sequence 3000, − 1800, 1080, − 648,… .
50a. Find the common ratio.
[2 marks]
Markscheme
evidence of dividing two terms
e.g.
− 1800
3000
r = −0.6
,
(M1)
− 1800
1080
A1
N2
[2 marks]
Examiners report
This question was generally well done by most candidates.
50b. Find the 10th term.
[2 marks]
Markscheme
evidence of substituting into the formula for the 10th term
(M1)
9
e.g. u10 = 3000(−0.6)
u10 = 30.2 (accept the exact value −30.233088)
[2 marks]
A1
N2
Examiners report
This question was generally well done by most candidates, although quite a few showed difficulty answering part (b) exactly or to
three significant figures.
50c. Find the exact sum of the infinite sequence.
[2 marks]
Markscheme
evidence of substituting into the formula for the infinite sum
(M1)
e.g. S = 3000
1.6
S = 1875
A1
N2
[2 marks]
Examiners report
This question was generally well done by most candidates, although quite a few showed difficulty answering part (b) exactly or to
three significant figures. Some candidates reversed the division of terms to obtain a ratio of − 53 . Of these, most did not recognize this
ratio as an inappropriate value when finding the sum in part (c).
51.
8
Find the term x3 in the expansion of ( 23 x − 3) .
[5 marks]
Markscheme
evidence of using binomial expansion
(M1)
8
8
e.g. selecting correct term, a8 b0 + ( ) a7 b + ( ) a6 b2 + …
1
2
evidence of calculating the factors, in any order
e.g. 56 ,
22
33
A1A1A1
8
, −35 , ( ) ( 23 x) (−3)5
5
3
−4032x3 (accept = −4030x3 to 3 s.f.)
A1
N2
[5 marks]
Examiners report
Candidates produced mixed results in this question. Many showed a binomial expansion in some form, although simply writing rows
of Pascal’s triangle is insufficient evidence. A common error was to answer with the coefficient of the term, and many neglected the
use of brackets when showing working. Although sloppy notation was not penalized if candidates achieved a correct result, for some
the missing brackets led to a wrong answer.
A city is concerned about pollution, and decides to look at the number of people using taxis. At the end of the year 2000, there were 280 taxis
in the city. After n years the number of taxis, T, in the city is given by
T = 280 × 1.12n .
52. (i)
Find the number of taxis in the city at the end of 2005.
(ii) Find the year in which the number of taxis is double the number of taxis there were at the end of 2000.
[6 marks]
Markscheme
(i) n = 5
(A1)
T = 280 × 1.125
T = 493
A1
N2
(A1)
(ii) evidence of doubling
e.g. 560
A1
setting up equation
e.g. 280 × 1.12n = 560, 1.12n = 2
n = 6.116 …
(A1)
in the year 2007
A1
N3
[6 marks]
Examiners report
A number of candidates found this question very accessible. In part (a), many correctly solved for n, but often incorrectly answered
with the year 2006, thus misinterpreting that 6.12 years after the end of 2000 is in the year 2007.
In a geometric series, u1 =
1
81
and u4 = 13 .
53a. Find the value of r .
[3 marks]
Markscheme
evidence of substituting into formula for n th term of GP
e.g. u4 =
1 3
r
81
setting up correct equation
r=3
(M1)
A1
1 3
r
81
=
1
3
A1
N2
[3 marks]
Examiners report
Part (a) was well done.
53b. Find the smallest value of n for which Sn > 40 .
[4 marks]
Markscheme
METHOD 1
setting up an inequality (accept an equation)
e.g.
1
(3n−1)
81
2
> 40 ,
1
(1− 3n)
81
−2
M1
> 40 , 3n > 6481
M1
evidence of solving
e.g. graph, taking logs
n > 7.9888 …
(A1)
∴n=8
N2
A1
METHOD 2
if n = 7 , sum = 13.49 … ; if n = 8 , sum = 40.49 …
n = 8 (is the smallest value)
A2
A2
N2
[4 marks]
Examiners report
In part (b) a good number of candidates did not realize that they could use logs to solve the problem, nor did they make good use of
their GDCs. Some students did use a trial and error approach to check various values however, in many cases, they only checked one
of the "crossover" values. Most candidates had difficulty with notation, opting to set up an equation rather than an inequality.
54a.
7
[1 mark]
Expand ∑ 2r as the sum of four terms.
r=4
Markscheme
7
∑ 2r = 24 + 25 + 26 + 27 (accept 16 + 32 + 64 + 128 )
r=4
A1
N1
[1 mark]
Examiners report
This question proved difficult for many candidates. A number of students seemed unfamiliar with sigma notation. Many were
successful with part (a), although some listed terms or found an overall sum with no working.
54b.
(i)
30
Find the value of ∑ 2r .
r=4
∞
(ii) Explain why ∑ 2r cannot be evaluated.
r=4
[6 marks]
Markscheme
(i) METHOD 1
(M1)
recognizing a GP
u1 = 24 , r = 2 , n = 27
(A1)
correct substitution into formula for sum
e.g. S27 =
(A1)
24(227−1)
2−1
S27 = 2147483632
A1
N4
METHOD 2
30
30
3
r=4
r=1
r=1
recognizing ∑ = ∑ − ∑
(M1)
recognizing GP with u1 = 2 , r = 2 , n = 30
(A1)
correct substitution into formula for sum
S30 =
2(230−1)
2−1
(A1)
= 214783646
30
∑ 2r = 2147483646 − (2 + 4 + 8)
r=4
= 2147483632
A1
N4
(ii) valid reason (e.g. infinite GP, diverging series), and r ≥ 1 (accept r > 1 )
R1R1
N2
[6 marks]
Examiners report
The results for part (b) were much more varied. Many candidates did not realize that n was 27 and used 30 instead. Very few
candidates gave a complete explanation why the infinite series could not be evaluated; candidates often claimed that the value could
not be found because there were an infinite number of terms.
In an arithmetic series, the first term is −7 and the sum of the first 20 terms is 620.
55a. Find the common difference.
[3 marks]
Markscheme
attempt to substitute into sum formula for AP
e.g. S20 =
20
(2(−7) + 19d)
2
,
20
(−7 + u20 )
2
setting up correct equation using sum formula
e.g.
20
(2(−7) + 19d
2
d=4
A1
M1
A1
= 620
N2
[3 marks]
Examiners report
[N/A]
55b. Find the value of the 78th term.
[2 marks]
Markscheme
correct substitution −7 + 77(4)
u78 = 301
A1
A1
N2
[2 marks]
Examiners report
[N/A]
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