Download functions-linear-answers

1.
(a)
Domain x < 3 (accept –4  x < 3) Range y  2 (accept –2  y  2) (A2)(A2) (C4)
Note: Award (A1) for x  3 and (A1) for y < 2. If the domain
and range are reversed award [0 marks] in this part of the
question. Allow for other notation such as [–, 3] or [, 3] for
domain and [–, 2] for range.
(b)
Domain {–3, –2, –1, 0, 1, 2, 3} Range {1, 2, 3, 4}
(A2)(A2) (C4)
Note: Award (A2) ft, (A2) ft if domain and range are reversed.
Award (A1) if 1 number is omitted from the domain and (A1) if
1 number is omitted from the range.
Award (A0) if more than 1 number is omitted from the domain
and (A0) if more than 1 number is omitted from the range.
Award (A0) for –3  x  3 and 1  y  4.
[8]
2.
(a)
2J + 3C = 5.95
(A2) (C2)
(b)
2×2.15 + 3C = 5.95
3C = 1.65
C = 0.55
55 (pence) or £0.55
(M1)
(M1)
(A1)
(C2)
[4]
3.
(a)
c – 0.10k + 1.40
(b)
(i)
(ii)
c = 0.10(7) + 1.40 (allow follow through from part (a))
= 0.70 + 1.40
= $2.10
2.40 = 0.10k + 1.40 (allow follow through from part (a))
1.00 = 0.10k
10 = k
10 km
(A1)
(A1)
(M1)
(A1)
[4]
4.
(a)
III
(A1)
(b)
I
(A1)
(c)
II
(A1)
(d)
IV
(A1)
[4]
1
5.
(a)
(b)
A; y = 0, 3x = 24  x = 8
A(8, 0)
(A1)
B; x = 0, 4y = 24  y = 6
B(0, 6)
(A1)
2
(A1)
2
M; xm =
80
06
= 4, ym =
=3
2
2
M(4, 3)
(c)
(d)
(A1)
3  2 5

40 4
L2: gradient =
y=
5
x – 2 (or equivalent)
4
(i)
M(4, 3), C(0, –2)
(A1)
(A1)
(4  0) 2  (3  (2)) 2
MC =
(M1)
= 41
= 6.40
(ii)
(A1)
A(8, 0), C(0, –2)
AC =
8 2  (2) 2
(M1)
= 68
= 8.25
(e)
2
(A1)
4
(i)
M
 41
C
 68
cos M =
=
5
5 2  ( 41 ) 2  ( 68 ) 2
2  5 41
25  41  68
10 41
CM̂A = 91.8° (3 s.f.)
A
(M1)
(M1)
(A1)
2
Area of CMA =
(ii)
1
41 × 5 sin 91.8°
2
= 15.99991171...
= 16.0 (3 s.f.)
(M1)
(A1)
5
[15]
6.
(a)
Gradient of l 2 
=
(M1)
2
5
(A1)
5
2
Gradient of l1 
(b)
0  (2)
50
(A1) (C3)
5
x+7
2
2y = –5x + 14
5x + 2y – 14 = 0
y=
(A1)(A1)
(A1)(A1)(A1) (C5)
[8]
7.
(a)
(b)
(i)
Domain:
(A2)
(ii)
Range:
(i)
Domain: {x –360°  x  360°}
Accept –360  x  360
(A2)
(ii)
Range: {y–1.5  y  1.5}
Accept –1.5  y  1.5
(A2) (C4)
{y y  2} accept y  2
(A2) (C4)
[8]
8.
Equation
Diagram number
y=c
2
(A2)
y = –x + c
3
(A2)
y = 3x + c
4
(A2)
1
x+c
3
1
(A2) (C8)
y=
[8]
3
9.
1
.
(A1)(A1) (C2)
4
Note: Award (A0)(A1)ft if the order of the gradients is reversed
or both signs are wrong or both are reciprocals of the correct
answer.
(a)
L1 has gradient 2 and L2 has gradient 
(b)
L2 is drawn incorrectly.
(c)
The product of the gradients is 2 × 
(d)
The drawing should show a straight line passing through
x and y intercepts at (4, 0) and (0, 1) respectively.
(A1)(A1) (C2)
Note: Award (A1) for each intercept. If these are wrong but
1
gradient is  then (A1). If correct line is very poorly drawn
4
then (A1).
(A2) (C2)
1
1
(M1)(A1) (C2)
   –1.
4
2
Note: Award (M1) for looking at product of gradients,
(A1) for comparing something to –1.
[8]
10.
(a)
0.40
(A2) (C2)
(b)
0.55 + 0.50 = 1.05
(A1)(A1)(A1) (C3)
Note: Award (A1) for 0.55, (A1) for 0.50, (A1) for correct total
of amounts given.
(c)
0.80 + 1.40 = 2.20
(A1)(A1)(A1) (C3)
Note: Award (A1) for 0.80, (A1) for 1.40, (A1) for correct total
of amounts given.
[8]
11.
(a)
(b)
5(50  32)
9
= 10°C.
C=
(M1)
(A2) (C3)
Put C = –273
(A1)
5( F  32)
so – 273 =
(M1)
9
Hence 9 × –273 = 5(F – 32)
(M1)
F = –491.4 + 32 = –459.4 (accept –459).
(M1)(A1) (C5)
Note: (M1) is for adding 32, even if the other number is
incorrect.
[8]
4
12.
(a)
(b)
(i)
{–3, –2, –1, 0, 1, 2, 3}
Notes: Award (A1) for set brackets.
Award (A1) for all and only correct numbers.
(ii)
{0, 1, 4, 9}
Notes: Award (A1) for all and only correct numbers.
If domain and range reversed, can follow through in (ii).
(A1)
(iii)
f(x) = x2
Note: Allow any other rule that works.
(A2) (C5)
[1, ] or {x 
x  1}
(A1)(A1)
(A1) (C1)
[6]
13.
(a)
–2
–1
0
1
2
3
–3
–5
3
13
For six single lines going to correct y (y-value can be repeated)
Correct diagram (y-values not repeated)
(M1)
(A1) (C2)
(b)
x  {–2, –1, 0, 1, 2, 3}
Note: Award (A1) if one value omitted.
(A2) (C2)
(c)
y  {–3, –5, 3, 13}
(A2) (C2)
[6]
5
14.
(a)
10
y
9
8
f(x)
7
g(x)
6
B
f(x)
5
4
3
2
A
1
0
1
2
3
4
5
6N 7
8
9
10
11
12
13 14
x
(A5)
5
Note: Award (A1) for a correctly labelled graph, (A1) for
correct scales, (A1) for line f(x) = 6 – x drawn correctly, (A1)
1
for line f(x) = x – 6 drawn correctly, (A1) for g(x) = x drawn
2
correctly.
(b)
(c)
(d)
(i)
Points named on the graph (A and B can be inversed)
(ii)
A(4, 2), B(12, 6)
 12  4 6  2 
,
Midpoint = 

2 
 2
= (8, 4)
Note: Allow (A2) for reading from the graph but both
coordinates must be correct.
Gradient =
40
=2
86
y = mx + c
0=2×6+c
c = –12
Equation is y = 2x – 12 (or correct alternatives).
Ft from candidate’s previous work.
(A1)
(A1)(A1)
3
(M1)
(A1)
2
(A1)
(M1)
(A1)
(A1)
4
[14]
6
15.
(a)
(A1)
(A1)
(A1)
Notes: Accept any symbol for ticks.
Do not penalize if the other boxes are left blank.
(b)
(C3)
(i)
(A1)(A1)
Notes: Award (A1) for structure and layout of
mapping diagram,
(A1) for correct values.
(ii)
Range = {2, 9, 14}
Note: Brackets not required.
(C2)
(A1)(ft)
(C1)
[6]
7