Higher Prelim Paper 1 2015
Marking Scheme
Give 1 mark for each
1(a)
ans:
proof
Illustration(s) for awarding each mark
(3 marks)
2 2 3  29 30
●1
●
2
●3
(b)
2
process – synthetic division for example
●1
completes synthetic division
●
conclusion
●3
2 7  15
0
since remainder = 0, (x – 2) is a factor
●1
●2
(x – 2)(2x2 + 7x – 15)
(x – 2)(2x – 3)(x + 5) [must include (x – 2)]
ans: (x – 2)(2x – 3)(x + 5)
●1 finds quotient
●2 factorises fully
ans:
x2 
1
x2
(3 marks)
4
4
 30
substitutes
●1
●2
removes brackets
●2
●3
1
(x  )2  2
x
1
x2  2  2  2
x
1
x2  2
x
(3 marks)
7
3
●1
finds gradient of given line
●1
m
●2
finds perpendicular gradient
●2
m perp 
●3
substitutes into equation and rearranges
●3
ans: 1/8
14
Pegasys Extension Test 2010-11 Q10
●1
ans: 7y – 3x + 13 = 0
30
(2 marks)
●3 states answer
3
2 2 3  29
2
3
7
3
y  1  ( x  2)
7
(3 marks)
1 2
x
2
●1
prepares to differentiate
●1
f ( x) 
●2
differentiates
●2
f ' ( x)  1x 3  
●3
subs and evaluates
●3
1
x3
1
1
f ' (2)  

3
8
(2)
Give 1 mark for each
5
ans:
C 2 or second circle
Illustration(s) for awarding each mark
(5 marks)
SQA 2006 Higher P2 Question 4
●1
●2
●3
state centre of C1
equates x coordinates to find k
finds radius of C1
●1
●2
●3
C1  (3, 4)
k 6
R1  5
●4
uses radius formula for R2
●4
R2  (3) 2  (4) 2 12 or equivalent
●5
find R2 and compare with R1
●5
37  5 or C 2
y
6
ans:
●
●2
●3
1
graph drawn
(3 marks)
B’
●
correct shape
correct image for A annotated
correct image for B annotated
8
1
●2
●3
A’
(3,4)
o
7
ans:
●1
●2
●3
8
ans:
●1
●2
●3
●4
9
64
●1
●2
●3
differentiates and equates to 0
subs and solves for a
x4 – 2x²
[34 – 2(3)²] – [(–1)4 – 2(–1)²]
64
(4 marks)
●1 b 2  4ac  0 for real roots
●2 (3 p) 2  (4(4 p  1).1)  0 ; 9 p 2  16 p  4  0
●3 diagram drawn
●4 (9 p  2)( p  2)  0  p   92 or p  2
knows condition for real roots
calculates discriminant
strategy for solving
factorising to answers
ans: a = –4
●1
●2
(3 marks)
integrates
subs values
evaluates
p   92 , p  2
x
(2 marks)
●1
●2
2x + a = 0
2(2) + a = 0; a = –4
Give 1 mark for each
10(a) ans: y = 2x -1
(b)
SQA 2006 Higher P1 Question1
finds coordinates of D
●1
●2
finds gradient of BD
●2
●3
states equation of BD
●3
ans: y = -3x + 9
finds gradient of BC
●1
●2
●3
finds perpendicular gradient
finds equation of BC
●2
●3
ans:
●1
●2
●3
●3
12(a) ans:
●1
●2
●3
●4
ans:
●1
●2
●3
●4
(4 marks)
12  20 p  q
10  12 p  q
p  0.25
q7
SQA 2005 Higher P2 Question 2
15
●1 cos p  178 sin p  17
stated or implied by ●4
●2 cosq  108 sin q  106 same order as ●3
●3 sin p cos q  cos p sin q explicitly stated
15
48
168
84
 108  178 106 = 120
●4 17
170  170  170  85
interpret diagram
interpret diagram
expand sin (A + B)
substitute and complete
3 1
●1 2x – 1 = - 3x + 9 or equivalent
●2 x = 2
●3 y = 3
(4 marks)
set up one equation
set up second equation
solve for one variable
solve for second variable
proof
25 1

72 3
m   3
y 12   3( x 1) or equivalent
m BC 
(3 marks)
starts solving system of equations
Finds value of x
Finds value of y
p = 0·25
D = (3,5)
55
m BD 
2
3 2
y  5  2( x  3) or equivalent
(3 marks)
●1
●1
●2
●3
(b)
(3 marks)
●1
10(c) ans: (2,3)
11
Illustration(s) for awarding each mark
(3 marks)
2 2
●1
any expression equivalent to sin750
●1
●2
correct exact values
●2
sin (45  30)  or equivalent
1

2
●3
correct answer
●3
3 1
2 2
3
2

1
2

1
2
Give 1 mark for each
13(a) ans:
●1
●2
●3
(b)
ans:
●1
14
ans:
a = 2, b = 1 , c = - 5
Illustration(s) for awarding each mark
(3 marks)
●1
●2
●3
find a
find b
find c
(-1,-5)
a2
b 1
c5
(1 mark)
●1
interpret equation of parabola
proof
SQA 2006 Higher P1 Question 8
(5 marks)
(-1,-5)
SQA 2005 Higher P2 Question 8
●1
●2
●3
equate for intersection
use double angle formula
factorise
●1
●2
●3
k sin 2 x  sin x
k  2sin x cos x
sin x ( 2k cos x 1)
●4
process two solutions
●4
sin x  0
●5
complete proof
●4
sin x  0
1
cos x 
for A and C
2k
1
2k
x  0 ,  , 2 ie at O,B and D
cos x 
Total: 60 marks