IFYMB004 Mathematics
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THE NCUK INTERNATIONAL FOUNDATION YEAR
IFYMB004 Mathematics (Business)
Examination
MARK SCHEME
Question 1
3600 (1 + )3 (= 4049.51 … ) or equivalent
Subtracts 3600 from their 4049.51…
Anything rounding to 450 (pounds)
Question 2
(a) 0.57
A B
0.25 |
(B1) for any 2 correct entries; (B2) for all entries correct and diagram enclosed in a rectangle. Condone missing E.
(c) 0.68
Question 3
(a) Finds gradient of line l2 (= − )
Correct follow through equation in any form e.g. y − 8 = their − (X + 2) y = − X + 7 [must be in this form]
(b) Finds coordinates of P and Q [ (−6, 0) and (14, 0) respectively]
× [their 14 − their (−6)] × 8
= 80 (units2)
Question 4
(a) 6C2 × k4 × ( )2 (M1) 6C3 × k 3 × ( )3 (M1) Accept equivalent
binomial coefficients, allow presence of x but do not accept reversed powers.
Multiplies second expression by 4 or divides first by 4, and solves [there must now be no x present]
k = ±4 [both answers needed]
Please note: following the syllabus change, if the binomial coefficients are written the wrong way round, this is no longer penalised e.g. 2C6 will not lose the mark.
(b) Please note: this is a ‘prove that’ question so all working must be seen.
[S2n] = [2a + (2n − 1)d] seen
[4Sn] = 4 × [2a + (n − 1)d] , seen and then sets equal to S2n
Expands and simplifies (at least one correct line of intermediate working must be seen)
Reaches d = 2a having scored all 3 M marks and no errors seen.
Question 5
(a)
xy |
30 |
234 |
648 |
435 |
136 |
∑ xy = 1483 |
(b) sxx = 114
sxy = 44.6
y − 12 = ⋯ (x − 21) or seen
y − 12 = (x − 21) or y = 0.391x + 3.78 (anything rounding to 0.391
and anything rounding to 3.78)
(c) Any valid answer e.g. 32 is in the range of x values
Any valid answer, e.g. 32 is close to the (upper) end of the range; or equation has been formed on only 5 [or (very) few] readings.
Question 6
(a) Circle; centre (0, 0) [Condone centre O]; radius 10 (units)
Any two correct statements (B1); all three correct statements (B2).
(b) Expresses x in terms of y or vice versa (x = or y = )
Substitutes into second equation and forms a quadratic equation
(y2 + 16y + 64 = 0 or x 2 − 12x + 36 = 0)
Solves. This is dependent on the previous M mark.
Solves for x or y and substitutes into either equation to find a value of the other unknown.
x = 6, y = −8 [Accept in coordinate form] (A1)
(c) The line is a tangent to the curve/circle at [the point] (6, −8). [The word ‘tangent’ must be seen or clearly implied e.g. ’the line touches the curve…’ The coordinates must also be seen. Allow follow through on their solution to part b) provided the line is still a tangent to the curve.]
Question 7
(a) Uses two versions of the cosine formula on side BD and sets equal to each
other [62 + 82 − 2 × 6 × 8 × cos e = 82 + 102 − 2 × 8 × 10 × cos ] Rearranges and simplifies
cos = (2 + 3 cos e) [Accept only in this form]
(b) Uses sine formula [ = ]
sin q = or equivalent but must be in this form.
c) Correct order of solving (arctan, subtracts 40 and then divides by 2) seen or implied anywhere
Finds one solution for 2e which does not have to be in the range (probably anything rounding to −26°)
Realises search is from 0 to 720 degrees
e = anything rounding to 77, 167, 257 and 347 (degrees) Any two correct (A1) all correct (A2) One mark is lost for any extra solutions in the range. Ignore solutions outside the range.
Question 8
(a) Attempts to differentiate [sight of ke2x (k ≠ 1) or − 8 is sufficient for this [M1*] mark.] [2e2x − 8]
Sets equal to 0 (can be implied) [M1]
Finds a value for x [M1]
Substitutes their x into original equation and finds a value for y [M1]
(ln 2, 4 − ln 256) or equivalent (e.g. 8 ln 2 for ln 256) but must be in exact [A1]
form.
(b) Changes the base of one of the logs [ logx3 = or log 3x = ]
Forms a quadratic equation
[ ( log3x)2 + log 3x − 6 = 0 or 6( logx3)2 − logx3 − 1 = 0]
Solves. This is dependent on the previous M mark.
[ log3x = −3, 2 or logx3 = − , ]
Removes logs correctly
x = , 9 Allow anything rounding to 0.037(0)
Question 9
(a) 6xℎ + 2x2 = 4.86
ℎ = or equivalent
Please note: this is a ‘show that’ question so all working must be seen. V = 2x2 ℎ
Substitutes their ℎ into expression for volume [V = 2x2 ( ) ] V = 1.62x − x 3 (both M marks scored and no errors seen).
Part b) is on the next page.
Question 9 – (continued)
(b) Attempts to differentiate (sight of 1.62 or X 2 is sufficient for this mark)
[1.62 − 2X2]
Sets equal to 0 (can be implied)
Solves. This is dependent on the previous two M marks.
X = 0.9 or equivalent
Attempts to differentiate a second time (sight of X is sufficient for this mark)
d 2 V
This is negative [when X = 0.9] so there is a maximum. Allow follow through
on their provided it gives a maximum.
or for the final three marks:
takes a numerical value between 0 and 0.9 and shows > 0 (M1*) takes a numerical value above 0.9 and shows < 0 (M1*)
thus there is a maximum when X = 0.9. (A1ft) Allow follow through on their
dV
provided it gives a maximum.
d
Question 10
(a) Finds area under line y = X + 4 by either using a trapezium
[ (4 + 7) × 6 = 33] or by integrating X + 4 [ X 2 + 4X] between limits 0 and 6 [(9 + 24) − 0 = 33]
Attempts to integrate X 2 − 16X + 67 (sight of X 3, X 2 or X is sufficient for this mark) [ X 3 − 8X2 + 67X] (This mark is not lost if the limits are incorrect).
Substitutes correct limits into their integrated expression and subtracts the right way round. [(243 − 648 + 603) − (72 − 288 + 402) = 198 − 186 = 12]
Adds their areas
45
(b) Expands the integrand [ − + 9t2 ]
= − t −3 − 6 ln t + 3t3 + c or equivalent.
(A1) Any two parts correct; (A2) all correct and + c
Question 11
(a)
FIRST GAME
SECOND GAME
W
2p
W
1 − 2p
W ᶥ
W
1 − p
W ᶥ
1 − p
W ᶥ
W denotes ‘wins game’
Set of branches under first game correct
Set of branches under second game correct
(b) P(he wins second game) = p × 2p + (1 − p) × p = p2 + p [this can be
seen at any stage in the answer]
P(wins first game given he wins second) = =
Solves. This is dependent on the previous M mark.
p = or equivalent.
[If 0 is also given without being discarded, this mark is lost. Placing in
brackets is sufficient to show non-inclusion]
Question 12
(a)
Mid-value, x |
Frequency f |
x × f |
Cum. Freq. |
1 |
5 |
5 |
5 |
3 |
12 |
36 |
17 |
5 |
24 |
120 |
41 |
7 |
31 |
217 |
72 |
9 |
18 |
162 |
90 |
11 |
12 |
132 |
102 |
13 |
6 |
78 |
108 |
15 |
4 |
60 |
112 |
Forms x × f column
Divides their ∑ (x × f) (810) by 112
Anything rounding to 7.23
(b) Finds cumulative frequencies
Plots correct curve (a sketch is on page 13). 1 mark lost for each omitted/incorrect plot; 1 mark lost for each point missed by the curve by at least 1 mm (but allow ft for any incorrect plots); 1 mark lost if either axis is not labelled correctly.
(Please note: a maximum total of 3 marks can be lost i.e. there are no negative scores. If the candidate plots the mid-values instead of the upper values in each interval, this will score A0.)
If graph paper is not used, award 1 mark out of the A3 if a reasonable curve is drawn. If a cumulative frequency polygon is drawn, award up to 2 marks out of the A3.
(c) Please note: in part c, marks are given for values taken from the candidate’s
curve, and there must be some evidence that this has been done. Reads off their median (around 7)
Reads off their LQ and UQ (around 5 and 9.4 respectively)
Subtracts their LQ from their UQ (around 4.4)
Question 13
(a) Sight of either P(≤ 11) ≈ 0.431 or P(≤ 12) ≈ 0.579
x = 12
Special case: if x = 12 is given with no working, award 1 mark out of 2.
(b) z = (= 1.24)
Finds percentage below 351 g (≈ 89.25%)
Realises symmetry or repeats process for 289
[Allow any other valid method]
Anything rounding to 78.5% (must be a percentage)
(c) E(X) = 2 × 0.07 + 4 × 0. 15 + 5 × 0.25 + 8 × 0. 16 + 10 × 0. 17 + 20 × 0.2
= 8.97
E ( ) = × 0.07 + × 0. 15 + × 0.25 + × 0. 16 + × 0. 17 + × 0.2
= 0.1695 (can be implied)
= 0.170 to 3 significant figures. Do not accept 0.17 . Allow follow through provided a more accurate answer is seen earlier.
Question 14
(a) Uses Chain Rule [sight of is sufficient for this mark.] [M1]
Uses Product Rule [sight of 4x × their or 4x ln 4 × ln(x2 + 1) is [M1]
sufficient for this mark].
4x × + 4x ln 4 × ln(x2 + 1) [A1]
{There is no need to simplify the answer, but if a candidate attempts to do so and makes a mistake, this mark is not lost unless a serious index or logarithmic error is made [e.g. ln 4(x2 + 1) is seen] in which case the mark is lost.}
(b) Uses Quotient Rule with correct bottom line and one part of the top line [M1*]
correct
dy 3(2x − 1) − 2(3x + 1)
dx (2x − 1)2
[A1]
There is no need to simplify for this mark
=
which is never 0 so there are no stationary values (or similar words) . [The
answer must be simplified to earn this final mark. Allow follow through [A1ft] provided their is never zero.]
(c) cos x + sin y + 3e3y − 2e2x = 0
Correct implicit differentiation (sight of sin y or 3e3y is sufficient for this [M1]
mark)
Assembles dy available [M1]
only if there are at least two terms)
dy 2e2x − cos x
dx sin y + 3e3y
[A1]
Question 15
(a) Degree of numerator must be less than degree of denominator (or similar
words) and states that A = 4.
Writes as + or equivalent
giving 4x + 5 = B(x + 2) + C(x − 1)
3 |
+ |
1 |
(A1) for each correct partial fraction |
x − 1 |
x + 2 |
(b) Uses integration by parts in the right direction
2023-06-02