2013 Higher Paper 1
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Paper 1: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 Paper 2: 1 2 3 4 5 6 7 8 9 Question 1 Question 2 Gradient of tangent at (5, 12) Using gives Question 3 Question 4
The graph of this function has an amplitude of 4, a period of 180 , and is shifted down by 1 unit. So, the solution is A.
g(f(x)) = g(x
2+ 1) = 3(x
2+ 1) − 4
= 3x
2+ 3 − 4
= 3x
2− 1
y = x
2− 4x + 7
dy
dx
= 2x − 4
= 2 ∙ 5 − 4 = 6
y − b = m(x − a)
y − 12 = 6(x − 5)
y − 12 = 6x − 30
y = 6x − 28
a = 2, b = 4, c = 5
b
2− 4ac = 4
2− 4 ∙ 2 ∙ 5
= 16 − 40
= − 24
°
Rearranging gives
Gradient of this line is
Since L is parallel it also has gradient
Using with gives
Rearranging gives
Question 6 Let
The remainder is 4 when is divided by
Question 7
5x + 3y − 6 = 0
y = − 5
3
x + 2
− 5
3
− 5
3
y − b = m(x − a)
(−2, − 1)
y − (−1) = − 5
3
(x − (−2))
y = − 5
3
x − 8
3
f(x) = x
3+ 3x
2− 5x − 6
f(2) = 2
3+ 3 ∙ 2
2− 5 ∙ 2 − 6 = 4
f(x)
(x − 2)
∫ x(3x + 2) dx
= ∫(3x
2+ 2x) dx
= x
3+ x
2+ c
Question 8
Since the sequence does have a limit as n The answer is C) Only statement 2) is correct
Question 9 Question 10 Question 11
Compared to the graph of this graph is inverted and shifted ‘k’ units to the right By inspection B is the correct solution
u
n+1= 0.1u
n+ 8
u
1= 0.1u
0+ 8
11 = 0.1u
0+ 8
3 = 0.1u
030 = u
0−1 < 0.1 < 1
→ ∞
sin2x = 2sinxcosx
= 2 ∙ 1
5
∙ 2
5
= 4
5
cos(270 − a) = cos270cosa + sin270sina
= 0 ∙ cosa + ( − 1) ∙ sina
= − sina
y = − f(x − k)
Question 13 Let
So cannot be in the domain of
Question 14
where is the angle between and itself
Question 15
so is our reference angle
Using CAST tan is negative in quadrants 2 & 4 So, the solutions are and
However, these solutions are for giving
f_ + g_ =
5
4
5
f_ + g_ = 5
2+ 4
2+ 5
2= 64
= 8
x
2− 7x + 12 = 0
(x − 4)(x − 3) = 0
x = 3, x = 4
x = 3, x = 4
f(x)
a
_ . ( a
_ + b)
_ = a
_ . a
_ + a
_ . b
_
= a
_ a
_ cosθ + 5
θ
_
a
= 3 ∙ 3 ∙ cos0 + 5
= 9 + 5
= 14
tan π
4
= 1 π
4
π − π
4
= 3π
4
π + π
4
= 5π
4
x
2
x = 3π
2
, x = 5π
2
But is out of our range so
Question 16
Question 17
Rewriting this gives
This shows that the roots are at So, by inspection,
So
Substituting the point (1, 3) gives
Question 18
5π
2
x = 3π
2
∫ (1 − 6x)
− 1 2dx
= (1 − 6x)
1 2 1 2∙ ( − 6)
+ c
= − 1
3 (
1 − 6x)
1 2+ c
y = kx(x + a)
2y = k(x − 0)(x + a)(x + a)
x = 0, x = − a
a = 2
y = kx(x + 2)
23 = k(1 + 2)
23 = 9k
k = 1
3
y = sin(x
2− 3)
dy
dx
= cos(x
2− 3) ∙ 2x
= 2xcos(x
2− 3)
Multiplying through by -1 gives
Consider
These are the roots of the quadratic. By considering the graph we see It is negative (i.e. for
The solutions are all such that
Question 20
The equation of the line is since gradient and y-intercept
Rewriting as an exponential gives
Question 21
Now complete the square on the inside of the bracket
1 − 2x − 3x
2> 0
3x
2+ 2x − 1 < 0
(3x − 1)(x + 1) < 0
(3x − 1)(x + 1) = 0
x = − 1, x = 1
3
y < 0)
−1 < x < 1
3
x
−1 < x < 1
3
log
3y = 2x
= 2
= 0
log
3y = 2x
y = 3
2x2x
2+ 12x + 1
= 2(x
2+ 6x) + 1
= 2[(x + 3)
2− 9] + 1
= 2(x + 3)
2− 18 + 1
= 2(x + 3)
2− 17
Question 22
a) Centre , Radius
b) Let C be the centre of
Therefore since perpendicular gradients
Using with (3, 2) gives
c) Radius of
Since the centre of we have
Expanding brackets and collecting terms gives
d) Substituting into gives
Expanding and simplifying gives
Using the discriminant with gives
Since the discriminant equals 0 there is only one solution Therefore, there is only one point at which the line Intersects with . In other words, it is a tangent
x
2+ y
2+ 2x + 4y − 27 = 0
= ( − 1, − 2)
= 1
2+ 2
2+ 27 = 32
C
1m
cp= 1
m
tan= − 1
y − b = m(x − a)
y − 2 = − 1(x − 2)
y = − x + 5
c
2=
32
2
= 8
c
2= (10, − 1)
(x − 10)
2+ (y + 1)
2= ( 8)
2(x − 10)
2+ (y + 1)
2= 8
x
2+ y
2− 20x + 2y + 93 = 0
y = − x + 5
c
2x
2+ (5 − x)
2− 20(5 − x) + 2y + 93 = 0
x
2− 16x + 64 = 0
a = 1, b = − 16, c = 64
b
2− 4ac = (−16)
2− 4 ∙ 1 ∙ 64 = 0
y = − x + 5
c
2a) Let
By inspection we have and
Giving by exact values
Since we want to be such that both and are positive, by CAST, must be in quadrant 1. So, is correct
Thus
b)
from a)
has a maximum value of 10
So has a maximum value of 14
3sinx − cosx = ksin(x − a) = ksinxcosa − kcosxsina
= kcosasinx − ksinacosx
3 = kcosa
1 = ksina
k = ( 3)
2+ 1
2= 2
ksina
kcosa
= 1
3
= tana
a = 30°
a
sina
cosa
a
30°
3sinx − cosx = 2sin(x − 30)°
4 + 5cosx − 5 3sinx = 4 − 5( 3sinx − cosx)
= 4 − 5 ∙ 2sin(x − 30)
= 4 − 10sin(x − 30)
= − 10sin(x − 30) + 4
−10sin(x − 30)
Question 24
a) i)
This shows that & are parallel but since T is a common point, It follows that are collinear.
ii) The ratio is 2 : 3
b) Since C lies on the x-axis it has a y co-ordinate of 0 and a z co-ordinate of 0. Let
Then
since they are perpendicular So,
→
AT = 10
10
4
→
TB = 15
15
6
→
AT = 2 55
2
→
TB = 3 55
2
1
2
AT = 1
→
3
TB
→
3→
AT = 2 →
TB
→
AT →
TB
A, T B
C = (c, 0, 0)
→
TC = C − T = c − 3
0
0
→
TB ∙ →
TC = 0
→
TB ∙ →
TC = 15(c − 3) + (15 ∙ 0) + (6 ∙ 0) = 0
15c − 45 = 0
c = 3
c = (3, 0, 0)
Question 1
Substituting and gives
Substituting and gives
Now we have simultaneous equations to solve for (1)
(2) (2) – (1) gives
Substituting into (2) gives
So,
Question 2 a)
since are perpendicular
Using with we have
u
1= 4, u
2= 7, u
3= 16
u
n+1= mu
n+ c
u
1u
2u
2= mu
1+ c
7 = 4m + c
u
2u
3u
3= mu
2+ c
16 = 7m + c
m c
7 = 4m + c
16 = 7m + c
9 = 3m
m = 3
m = 3
16 = 7 ∙ 3 + c
c = − 5
m = 3, c = − 5
m
PQ= 6 − 2
5 − 7
= − 2
m
QR= 1
2
PQ QR
y − b = m(x − a)
Q = (5, 6)
b)
Equating QR with PT gives
Multiply through by 6 to simplify
Substituting into gives
c) So, So,
y − 6 = 1
2
(x − 5)
y = 1
2
x + 7
2
x + 3y = 13
y = − 1
3
x + 13
3
1
2
x + 7
2
= − 1
3
x + 13
3
3x + 21 = − 2x + 26
x = 1
x = 1
y = 1
2
x + 7
2
y = 1
2
+ 7
2
= 4
T = (1, 4)
→
QT = (
−4
2 )
R = (1 − 4, 4 − 2) = ( − 3, 2)
→
QP = (
2
−4 )
S = (−3 + 2, 2 − 4) = ( − 1, − 2)
a) Let
Using synthetic division gives
Note that does not factorise
b) Let
Stationary Points occur where giving
from a)
The solution to this equation is since has no solutions (
Thus has only one stationary point
Question 4
Equation the line and curves equations gives
f(x) = x
3+ 3x
2+ x − 5
1 1 3 1 -5 1 4 5 1 4 5 0f(x) = (x − 1)(x
2+ 4x + 5)
x
2+ 4x + 5
g(x) = x
4+ 4x
3+ 2x
2− 20x + 3
g′(x) = 4x
3+ 12x
2+ 4x − 20
g′(x) = 0
4x
3+ 12x
2+ 4x − 20 = 0
x
3+ 3x
2+ x − 5 = 0
(x − 1)(x
2+ 4x + 5) = 0
x = 1
x
2+ 4x + 5
b
2− 4ac = − 4 < 0)
g(x)
2x + 3 = x
3+ 3x
2+ 2x + 3
x
3+ 3x
2= 0
x
2(x + 3) = 0
When So Area Question 5
x = 0, x = − 3
x = − 3, y = 2 ∙ (−3) + 3 = − 3
B = ( − 3, − 3)
= ∫
−30(x
3+ 3x
2+ 2x + 3) − (2x + 3) dx
= ∫
−30(x
3+ 3x
2)dx
=
[
x
44
+ x
3]
0 −3= (0) − (
81
4
− 27) =
27
4
units
2log
5(3 − 2x) + log
5(2 + x) = 1
log
5[(3 − 2x)(2 + x)] = 1
5
1= (3 − 2x)(2 + x)
5 = 6 − x − 2x
22x
2+ x − 1 = 0
(2x − 1)(x + 1) = 0
x = − 1, x = 1
2
(
There are no other solutions in
Question 7
a) by inspection of the diagram Area
Substituting into gives
b) i)
∫
a 05sin3x dx = 10
3
[−
5
3
cos3x]
a 0= 10
3
− 5
3
cos3a) − ( − 5
3
cos0) = 10
3
− 5
3
cos3a + 5
3
= 10
3
− 5
3
cos3a = 5
3
cos3a = − 1
3a = π, a = π
3
0 ≤ a < π
L = 3x + 4y
= 24 = 2xy
y = 24
2x
= 12
x
y = 12
x
L
L = 3x + 4 ∙ 12
x
= 3x + 48
x
L(x) = 3x + 48
x
= 3x + 48x
−1L′(x) = 3 − 48x
−2
Minimum values occur where
Since is a length it cannot be negative, so
gives a minimum value of
ii) Minimum length occurs where Minimum cost
= 3 − 48
x
2L′(x) = 0
3 − 48
x
2= 0
3x
2− 48 = 0
x
2− 16 = 0
(x + 4)(x − 4) = 0
x
x = 4
x = 4
L(x)
x = 4
L(4) = 3 ∙ 4 + 48
4
= 24m
= 24 ∙ £8.25 = £198
Separating into two equations gives Question 9 a)
Since the concentration has halved
to 2 significant figures
sin2x = 2 cos
2x
2sinxcosx = 2 cos
2x
2sinxcosx−2 cos
2x = 0
2cosx(sinx − cosx) = 0
2cosx = 0
sinx − cosx = 0
cosx = 0
sinx = cosx
x = π
2
, 3π
2
cosx
sinx
= 1
tanx = 1
x = π
4
, 5π
4
p
t= p
0e
−ktp
t=
p
2
0p
02
= p
0e
−25k0.5 = e
−25klog
e0.5 = log
ee
−25klog
e0.5 = − 25k
k =
log
e0.5
−25
k = 0.028
b)
Let and Substitute to give Rounding gives So, is 11% of
Therefore, the concentration has decreased by 89%