Post on 08-Mar-2018
1
FEB
EXAM 1
2016 SEC 4 ADDITIONAL MATHEMATICS CW & HW
1 Find the values of k for which the line 16 xy is a tangent to the curve
22 7 xky . Find also the coordinates of the point at which this tangent touches
the curve. [5]
2
2
Given that q
2sin and both cos and tan are negative, find an expression, in
terms of q , for (i) tan , [2]
(ii) 2sin . [3]
3
3
(i) The graph of cxy 4 passes through the point (2, 3). Find the possible
values of the constant c .
[3]
(ii) Solve the inequality 1174 x . [2]
4
4 The height, h m, of a stone t seconds after it has been thrown vertically upwards
from ground level is given by 2btath , where a and b are constants.
t 2 4 6 8 10
h 14 -4 -45 -136 -250
The table shows experimental values of the variables t and h, but an error has
been made in recording one of the values of h.
(i) Express the given equation in the form suitable for drawing a straight line
graph and, using graph paper, draw the graph for the values given. [4]
Use the graph to (ii) correct the reading of h for which an error has been made, [2] (iii) estimate the value of a and b. [3] A second stone is thrown into the air from ground level. The height, h m, is directly
proportional to t and h = 60 m when t = 10 s. (iv) Draw a line on your graph to illustrate the motion of the second stone. [1] (v) Hence find the time when the two stones meet. [1]
7
6 A cubic polynomial )(xf is such that the roots of 0)( xf are 3
1, 5 and -2 and it
gives a remainder of 24 when it is divided by )1( x . (i) Express )(xf as a cubic polynomial in x with integer coefficients. [4]
(ii) Hence, solve 0)2( xf . [2]
8
7 (i) Prove the identity xecx
x
x
x cos2
cos1
sin
sin
cos1
. [3]
(ii) Hence, find all the angles between 0 and 180 which satisfy the equation
70tan2cos1
2sin
2sin
2cos1
x
x
x
x.
[3]
9
8 A curve has the equation 23
4
x
xy .
(i) Find an expression for dx
dy and explain why the curve has no turning points. [3]
(ii) Find the gradient of the curve when y = 0. [2] (iii) Given that y is increasing at the rate of 0.28 units per second at the instant
when x = 2, find the rate of change of x at this instant. [3]
10
9 In the expansion of
9
2
2
px
xwhere p is a positive constant, the term independent
of x is 5376. (i) Show that p = 4. [4] (ii) With this value of p, find the coefficient of 9x in the expansion of
9
2 99 12
px x
x.
[4]
11
10 A line 022 yx cuts a curve yx
2
2
11 at two points A and B .
(i) Find the coordinates of A and B . [4]
(ii) Show that the perpendicular bisector of AB cuts the y -axis at
4
3,0 . [4]
12
EXAM 1 ANSWERS
1
.1,3
1 ispoint of sCoordinate
1,3
1
013
0169
meet, line and curvewhen
4
0)1)(7(436
04 curve, o tangent tis line
0167
716
2
2
2
2
22
22
yx
x
xx
k
k
acb
xxk
xkxy
M1 M1
A1 M1 A1 [5]
2
(i)
tantan
4
22q
4
22
q
M1
A1
(ii) sin2coscos2sin2sin
2sin cossin2
q
q
q
422
2
2
2 44
q
q
M1 M1 A1 [5]
13
3 (i) ycx 4 or ycx 4
324 c or 324 c
5c or 11
M1 A1, A1
(ii) 1174 x
117411 x
5.41 x
M1 A1 [5]
4 (i)
btat
h
btath
2
t 2 4 6 8 10
h 14 -4 -45 -136 -250
t
h
7 -1 -7.5 -17 -25
15
10
5
-5
-10
-15
-20
-10 10 20 30 40
B: (0.01, 13.68)
E: (2.47, 4.12)
Slope CD = -3.88
A: (5.89, -9.14)
E
B
B1 B1 for table of values B2 for straight line graph (i)
(ii)
54
9
h
t
h
B2
(iii) From graph, a = y-intercept = 14 b = gradient = 4 B1 M1, A1
(iv) Refer to line drawn above. B1
(v) t = 2.5 s B1
(i)
(iv)
14
5 Given 3loglog3
log 333 xx
3 3 33
33 3
3
3 3 3 3
3 3 3 3
3 3
3
1
33
log 3 log log log 3 [ 1]
log2log 3 log [ 1]
log 3
log 3 2log 3 log log
1log 3 log 3 log log [ 1]
2
1 log 2log
1log [ 1]
3
3 3 [ 1]
x x M
xx A
x x
x x M
x x
x A
x A
M1 A1
M1
M1
A1 [5]
6
(i) Let 253
1)(
xxxaxf
1033
1 2
xxxa
3
109
3
10 23 xxxa
Since 24)1( f ,
243
109
3
101
a
248 a 3a
3
109
3
103)( 23 xxxxf
103103 23 xxx
M1 M1
M1
A1
(ii) 0)2( xf
022523
123
xxx
0433
53
xxx
,3x 3
5, 4
M1
A1 [6]
15
7
(i)
xx
xxx
x
x
x
x
cos1)(sin
sincoscos21
cos1
sin
sin
cos1 22
xx
x
cos1)(sin
1cos21
xx
x
cos1)(sin
cos12
ecxcos2
M1 M1
A1
(ii)
70tan2cos1
2sin
2sin
2cos1
x
x
x
x
70tan2cos2 xec
70tan
22sin x
3.133,7.462x
7.66,4.23x
M1 M1
A1 [6]
8
(i) 23
4
x
xy
2
2
2
23
14
23
12323
23
34123
x
x
xx
x
xx
dx
dy
Since 0dx
dy, ∴ the curve has no turning point.
M1
A1
B1
(ii) when y = 0,
4
23
40
x
x
x
∴ gradient of the curve 14
1
14
142
M1 A1
16
8
(iii)
when x = 2, 32
7
8
142
dx
dy
dt
dx
dx
dy
dt
dy
∴
sunits
dt
dx
/28.1
32
728.0
M1
M1
A1 [8]
9
(i)
M1
A1
M1
AG1
(ii)
B1
M1 M1 A1 [8]
9th 9 21 term
2
rr
r
kr C x
x
9 18 2
9 18 2
9 18 3
18 3 0
9 6 6
7 6
6 6
6
6
6
2
2
2 [M1]
3 18
6 [ 1]
2 5376 [M1]
84 2 5376
5376
84 2
4096
4 [1]
rr r
r
r r r r
r
r r r
r
r
C x p x
C p x x
C p x
x x
r
r A
T C p
p
p
p
p
9
92 9 2 1 94
9 1 2 9 12
x x x x xx
9 6 32 1 9 9 2 1
4 3
9
9 9 9
9 9
9
9
For 2 , term in 2
672 [B1]
term in 672 9 5376 [M1]
48384 672
47712
coefficient of 47712 [A1]
x x x T C x x
x
x x x
x x
x
x
1
FEB
EXAM 2
2016 SEC 4 ADDITIONAL MATHEMATICS CW & HW
1 Given that 2 1 42 8 2 2 ,x x x find the exact value(s) of x. [5]
2
2 The equation of a circle C is 04422 xyx .
(i) Find the centre and radius of the circle. [3] (ii) The circle C is reflected in the y -axis to obtain the circle D . Write down
the equation of the circle D . [1]
(iii) The two circles intersect at points P and Q , as shown in the diagram
above. Find the length of line segment PQ . [4] (iv) Two line segments are then drawn from P to meet each of the circle at X
and Y respectively. Find the coordinates of X and Y , given that PX and PY are diameters of the circles C and D respectively. [6]
x
y
Circle C Circle D
P
Q X Y
4
3 (i) State the amplitude of the curve
xy
2
1sin31 . [1]
(ii) Sketch the graph of
xy
2
1sin31 for the interval 20 x . [2]
(iii) By drawing a suitable line on the same diagram in part (ii), determine how
many solutions there are of the equation
xx
2
1sin6210 for the
interval 20 x . [3]
5
4 (i) Solve the equation xxx 3sin2sin8sin for 1800 x . [4]
(ii) Find all angle(s) between 0 and 2
for which 5cos6sin4 22 . [3]
6
5 A particle moves in a straight line so that its velocity, v m/s, is given by tt 216 2 , where t is the time in seconds after passing O. Find an expression in terms of t for
(i) its acceleration, [1] (ii) its displacement from O. [1] Calculate (iii) the value of t at which the particle passes through O again, [2] (iv) the minimum velocity of the particle, [3] (v) the total distance traveled by the particle in the interval t = 0 and t = 10. [5]
8
6 The diagram shows a rectangle PQRS . PQT is an
isosceles triangle with PT = 14 cm and PTQ = 2 radians. TV is parallel to PS , and
TV = 7 cm.
(i) Show that the perimeter, W cm, of the rectangle PQRS is given by
14cos28sin56 .
[3]
(ii) Express W in the form ba sin , where 0a and 900 . [2]
(iii) Find the value of for which W = 73 cm. [2]
2
7 cm
P Q
S R
T
V
14 cm
9
7 The equation 052 2 xx has roots and and the equation 042 pxx
has roots
k and
k. Find the value of k and of p.
[8]
10
8
The diagram shows two circles with centres O and P intersecting at two points A
and B. P is a point on the circumference of the circle with centre O. A straight line through B meets the circles at D and C. The line CP meets AB at E and CP produced meets AD at M. Prove that
(i) APAC = AECP, [4] (ii) CM is perpendicular to AD. [5]
A
B
O P
M E
C
D
12
9
(i) Given that 1221
2
b
a, where a and b are integers, find, without using
a calculator, the value of a and of b.
[4]
(ii) Given that ln x2y = a and ln y
x = b, express ln
2x
y in terms of a and b. [5]
14
10 The diagram shows part of the graph of the curves
2sin
xy and
12
cos3
xy for x0 .
(i) Show that the point of intersection of the 2 curves is
5.0,
3
. [4]
(ii) Given that the curve 12
cos3
xy cuts the x -axis at the point P. Show
that the x -coordinate of P is approximately 1.91. [2]
(iii) The line 3
x is drawn to divide the area enclosed by the 2 curves and the
x -axis into 2 regions, A and B . A student claimed that the regions A and B
are of the same size by just looking at it. Determine if this claim is true, explaining your argument clearly. [5]
2sin
xy
12
cos3
xy
x
y
P 0
16
EXAM 2 ANSWERS
1 2 1 42 8 2 2x x x
Let 2 .xy
2 4
2
2 8 2
2 17 8 0
2 1 8 0
y y y
y y
y y
1
2y or 8y
12 2x or 32 2x
1x or 3x
M1
M1 for sub. 2xy
M1 for factorisation
A1, A1 [5]
2
[M1]
[A1]
[A1]
04022222 yxyx
Centre of circle = (2, 0)
Radius = 42022
= 228
[B1]
[M1]
[A1]
OR
[B1]
19
3 (i) Amplitude = 3 B1
(ii)
x
y
-2
-1
0
1
2
3
4
5
B1 for ½ cycle of sine curve B1 for correct position
(iii)
xx
2
1sin6210
x
x
2
1sin31
25 Draw
25
xy
No. of solutions = 2
M1 for getting eqn of line M1 for correct line drawn A1 [6]
(ii)
(iii)
20
4 (i) xxx 3sin2sin8sin
xxxxx 3sin282
1sin28
2
1cos2
xxx 3sin3sin5cos2
015cos23sin xx
03sin x or 5.05cos x 540,360,1803x 780,660,420,300,605x
180,120,60,0x 156,132,84,60,12x
180,156,132,120,84,60,12,0x
M1
M1 M1
A1
(ii) 5cos6sin4 22
4
5.0sin
1sin2
5sin66sin4
5sin16sin4
2
22
22
M1 M1 A1 [7]
5 (i)
2112
216 2
ta
ttv
B1 (ii)
2
212
0,0,0when
2
212
216
23
23
2
tts
cst
ct
ts
ttv
B1
(iii)
st
tt
tt
s
4
15
4
21
02
212
02
212
again O passed particle when 0
2
23
M1 A1
21
5 (iv)
smv
st
t
/8
318
4
721
4
76
4
7
12
21
02112
0a velocity,min
2
M1 M1 for finding t A1
(v)
m
ms
s
ms
t
t
75.1035)28
742(950 distance Total
950)100(2
21)10(2
10 t tos2
7 tfrom travelleddistance
8
742
2
7
2
21
2
72
2
7 t to0 t from travelleddistance
2
7,0
07-2t3t
021t-6
0,when v
3
23
2
M1
M1 M1
M1 A1 [12]
22
6 (i)
14cos28sin56
cos1472sin282
cos147
sin28sin142
W
W
PS
PQ
M1
M1
M1
AG
(ii)
57.2656
28tan
61.622856
1
22
a
146.26sin6.62 P
A2
(iii) 731457.26sin61.62
61.62
5957.26sin
45.7057.26
9.43 , 83.0º
M1
A1 [7]
7
5.2
5.0
441
1600
21
191
21
4025.5)5.2(4
25.55.225.0
2
4
2
2
222
22
p
pkkk
k
kkk
B1 B1 M1 M1 A1 A1
M1 A1 [8]
(shown)
23
8 (i) 1. Join PB
2. PA=PB ( radii of circle with centre P)
3. PBEPAE (base angles, isos )
4. PCAPBE ( s in the same seg)
5. PCAPAE
6. CPAAPE (common angle)
7. APE is similar to CPA (AA)
8. AC
AE
CP
AP
9. CPAEACAP (proven)
M1
M1
M1
A1
(ii) 1. ADBAPB 2 ( at centre = 2 at circumference)
2. 180PBEPAEAPB ( s sum of )
3. 18022 PAEADB
4. 90PAEADB
5. BCEPAE ( s in the same seg)
6. 90BCEADB
7. 180CMDBCEADB
8. 90CMD CM is perpendicular to AD .
M1 M1 M1 M1
A1 [9]
9
(i) Given 12
21
2
b
a
2)1()12(2
21)2(22
)21)(12(2
bba
bba
ba
Hence by comparing coefficients, we obtain
12 ba -------- (1)
11 b -------- (2)
Solving, we obtain b = - 2 and a = - 3
M1 M1 A1, A1
24
OR
2 2 12 1 [ 1]
1 2 2 1
2 2 22 1
2 1
1 2 3 0 [ 1]
ab M
a ab
a b a A
Hence by comparing coefficients, we obtain
01ba -------- (1)
03 a -------- (2) [A1]
Solving, we obtain b = - 2 and a = - 3 [A1]
M1 M1 A1, A1
9 (ii) ln x2y = a
ln x2 + ln y = a
2 ln x + ln y = a ------ (1)
ln x
y = b
ln x – ln y = b ------- (2)
Solving simultaneously eqns (1) and (2),
ln x = 1
3(a+b),
ln y = 1
23
a b
ln 2x
y = ln y – ln x 2 = ln y – 2 ln x
= 1
23
a b – 2 [1
3(a+b) ]
= ba 43
1
M1
M1
A1 M1
A1 [9]
26
10
The claim is not true as area of region A area of region B. [A1] [11]
2
[M1]
[A1]
[A1]
04022222 yxyx
Centre of circle = (2, 0)
Radius = 42022
= 228
[B1]
[M1]
[A1]
OR
[B1]
FEB
EXAM 3
2016 SEC 4 ADDITIONAL MATHEMATICS CW & HW
1 Find the range of values of k if 32 xy intersects 42 kyx at two distinct points. [5]
2
2 The roots of the equation 0235 2 xx are and .
(a) State the value of and . [2]
(b) Find the quadratic equation in x whose roots are 2 and 2 . [5]
3
3 A wire is bent to form a rectangle of area 253 cm2. If the breadth of the rectangle is
12 cm, find the length of the wire, giving your answer in surds. [4]
7
7 Write down and simplify, in ascending powers of x, the first three terms of the expansion of
(a)
5
21
x (b) 523 x
Hence, or otherwise, obtain the first three terms of the expansion of
5
2
23
x
x and use it
to estimate the value of 594.2 correct to 3 decimal places. [8]
8
8 A circle with centre C passes through points A 7,1 and B 8,0 .
(i) Explain why the perpendicular bisector of AB will pass through C. [1]
(ii) Given further that the line 22 xy passes through the centre of the circle,
show that the coordinates of C is 4,3 . [4]
(iii) Hence find the equation of the circle. [2]
9
y
x
4
2
4
3
1
0
1
cbxay sin
2
9 The following diagram shows the graph of cbxay sin for x0 .
(a) State the possible values of a and the value of b and of c. [2]
(b) Copy the graph on your answer sheet. By sketching an additional graph on the same
axes and domain, find the number of solutions of 01cos2sin cxbxa . [4]
10
10 Answer the whole of this question on a sheet of graph paper.
The table below shows the experimental values of two variables x and y.
x 1 2 3 4 5
y 1.19 0.87 0.69 0.57 0.48
It is known that x and y are related by an equation of the form nx
my
where m and n are
unknown constants. Draw a graph of y against xy and use the graph to estimate
(i) the value of m and of n,
(ii) the value of x when 75.0y . [8]
12
EXAM 3 ANSWERS
1
)2(4
)1(32
32
2
kyx
xy
xy
Sub (1) into (2):
0342
432
4)32(
2
2
2
kkxx
kkxx
xkx
For 2 distinct roots,
1or4
0)1)(4(
043
012164
0)34)(1(4)2(
04
2
2
2
2
kk
kk
kk
kk
kk
acb
M1
M1
M1
M1
A1
5
2(a)
2
5
2
3
2
3
5,3,2
0235 2
a
c
a
b
cba
xx
B1
B1
2
2(b)
0125308
08
125
4
15
8
125
2
5
4
15
2
3
2
5
)(
2
2
3
3
3322
22
xx
xx
M1
A1
M1
A1
A1
5
13
3
cm227
12
2510323
12
12
12
253rectangle ofLength
cm 2216
228
212227
Perimeter wireofLength
M1
A1
M1
A1
4
4
22
2
22
22
2
1
3
13
12
)1(3
4116
3
12
3
7(2),into2Sub
242:)2()1(
)2(3
793342224,2When
)1(3
59334,0When
334116,1When
3)1(3)1(34116
11)1(3
4116
xxx
xx
BA
AA
BABAx
BABAx
CCx
CxBxAxx
x
C
x
BA
x
xx
M1
M1
M1
M1
A1
5
5
)4(67
3
333,(2)From
)3(16
16
4log
4loglog
2log2
log,(1)From
)2(9813
)1(2loglog
67
634
2
2
22
2
22
22
32
3
24
yx
yx
y
x
y
x
yx
yx
yx
y
yx
yx
M1
M1
M1
M1
M1
8
14
2
1and4
42
167),4(into
2
1Sub
2
1)(
8
7
0)12)(78(
07616
6716
,(4)into(3)Sub
2
2
yx
xy
yorNAy
yy
yy
yy
M1
A1
A1
6
Let 1437163)( 23 xxxxf
7or3
1or2
7or3
1or2
014)(37)(16)(3
01437163
01437163
7or3
1or2
)1)(13)(2(0
)7223)(2()(
22
442
2737
,oftcoefficiencomparingBy
)73)(2(1437163
).(offactorais)2(
014)2(37)2(16)2(3)2(
23
23
23
2
223
23
xxx
xxx
xxx
xxx
xxx
xxx
xxx
xxxxf
b
b
b
x
bxxxxxx
xfx
f
M1
M1
M1
M1
A1
M1
A1
7
15
7
375.219
...)1.0(2
675)1.0(
2
40524394.2
),1( into1.0ngsubstitutiBy
)1(...2
675
2
405243
23
...2
12152025
2
12151080810243
...1080810243...2
5
2
51
232
1
232
12
3
...1080810243
...2)3(2)3(323)(
...2
5
2
51
...22
12
1)(
25
2
5
2
222
22
5
5
55
2
2
23
2
54
1
555
2
2
2
5
1
5
5
x
xxxx
xxxxx
xxxx
xx
xx
xx
xx
xCxCxb
xx
xC
xC
xa
M1
A1
M1
A1
M1
A1
M1
A1
8
8(i)
By the property of circle, the perpendicular bisector of chord passes through
centre of circle. Therefore, the perpendicular bisector of chord AB passes
through C.
B1
1
8(ii)
473(1),into3Sub
3
93
227(2),into(1)Sub
)2(22
)1(7
72
1
2
15,
2
15and
2
1When
1 AB ofbisector lar perpendicu ofGradient
1)1(0
78 ofGradient
2
15,
2
1
2
87,
2
01 ofpoint Mid
yx
x
x
xx
xy
xy
ccyx
AB
AB
The coordinates of C are 4,3 .
M1
M1
M1
A1
4
16
8(iii)
Radius = units5)48()03( 22
Equation of circle:
08625)4()3(
5)4()3(
2222
222
yyxxoryx
yx
M1
A1
2
9(a)
Possible values of 3or3 a
Value of 2b and 2c
B0.5,B0.5
B0.5,B0.5
2
9(b)
1cos2sin
01cos2sin
xcbxa
cxbxa
Award 2 marks for correct graph drawn (with labels).
Number of solutions = 3
M1
B2
B1
4
y
x
4
2
4
3
1
0
1
2
cbxay sin
1cos2 xy
3
17
10
(i)
(ii)
n
mxy
ny
mxyny
mnyxy
nx
my
1
1 2 3
0.5
1
1.5
2
xy
y
Award 1 point for each of the following:
Correct labelling of the graph, x-axis and y-axis
Appropriate use of scale and plotting of all points
Drawing of best fit line
gradient n
1 73.1578.0
1 n
n 05.0
y-intercept 85.1 20.385.1 mn
m 05.0
From the graph, 95.1xy when 75.0y .
05.060.2
95.175.0
x
x
xy 1.19 1.74 2.07 2.28 2.40
y 1.19 0.87 0.69 0.57 0.48
M1
M1
M3
A1
A1
A1
8
FEB
EXAM 4
2016 SEC 4 ADDITIONAL MATHEMATICS CW & HW
1. Find the coordinates of the points of intersection A and B of the line 2 2 0x y and
the curve 1 2 1
2x y . Hence show that the distance AB is 3 5 units. [5]
2. Given that the expression 12532 23 xxx is exactly divisible by
42 2 pxx .
(a) Find the value of p. [3]
(b) Hence solve the equation )3)(12(12532 23 xxxxx . [5]
3. Given the quadratic equation 2 3 3 8 0x kx k , where k is a real number,
(a) if the equation has equal real roots, find the values of k, [4]
(b) if the equation has complex roots, find the range of values of k. [2]
4. (a) Solve the equation 2 1 12 7(2 ) 36x x , leaving your answers in two decimal
places. [4]
(b) Without using a calculator, find the value of p for which
4 5(log 25)(log ) 3p . [3]
(c) Simplify 1
27 73 2
. [3]
5. (a) Express 3 2
2
2 3 9 2
4
x x x
x
in partial fractions. [5]
(b) Prove the identity 21 cos
cos cot1 cos
xec x x
x
[4]
6. (a) Sketch the following graphs cos 1y x and sin 2y x on the same axes, for
0 360x . [3]
(b) Hence, find the number of solutions for the following equation [1]
sin2 cos 1x x .
7. (a) Given that 4 x 6, find the value of x for which
3 tan x + cot x = 5 cosec x. [5]
(b) Express 9sin3 7cos3 in the form sin 3R where R is a positive
number and α is acute. Hence, or otherwise, solve the equation
9sin3 7cos3 3 for 0° ≤ ≤ 180°. [6]
EXAM 4 ANSWERS
1. 2x + y + 2 = 0 -----
1 2 1
2x y -----
: y = – 2x – 2 -----
: 2y + 4x = xy -----
Sub into ,
2(– 2x – 2) + 4x = x(– 2x – 2) [M1]
– 4x – 4 + 4x = – 2x2 – 2x
2x2 + 2x – 4 = 0 [M1]
(2x + 4)(x – 1) = 0
x = – 2 or 1
y = 2 or – 4
(x, y) = (– 2, 2), (1, – 4) [A2]
2 2( 2 1) (2 4) 45 3 5AB [B1]
2(a) )3)(42(12532 223 xpxxxxx
Coeff of x2 : 3 = 2(3) +p(1) [M1]
p = – 3 [A1]
2(b) )3)(12(12532 23 xxxxx
2x3 + 3x2 – 5x + 12 – (2x + 1)(x + 3) = 0
(2x2 – 3x + 4)(x + 3) – (2x + 1)(x + 3) = 0
(x + 3)[2x2 – 3x + 4 – 2x – 1] = 0 [M1]
(x + 3)(2x2 – 5x + 3) = 0
(x + 3)(x – 1)(2x – 3) = 0 [M1]
x = – 3, 1, 1.5 [A3]
3(a) 2 3 3 8 0x kx k
x2 – 3kx + (3k+ 8) = 0
if equation has equal real roots, b2 – 4ac = 0 [M1]
(–3k)2 – 4(1)(3k + 8) = 0
9k2 – 12k – 32 = 0 [M1]
(3k – 8)(3k + 4) = 0
4 8 or
3 3k k [A2]
3(b) if complex roots, b2 – 4ac < 0 [M1]
4 8
3 3x [A1]
4(a) 2 1 12 7(2 ) 36x x
22x. 21 = 7(2x. 2) + 36 [M1]
Let y = 2x.
2y2 – 14y – 36 = 0
(2y + 4)(y – 9) = 0 [M1]
y = – 2 or 9
2x = – 2 or 2x = 9 [M1]
(rejected) x = lg 9 / lg 2
x = 3.17 (2 dec pl) [A1]
4(b) 4 5(log 25)(log ) 3p
log 25 log
3log 4 log5
p [M1]
2log5 log
32log 2 log5
p
log p = 3 log 2 [M1]
p = 23 = 8 [A1]
4(c) 1
27 73 2
= 1 2 3
3 3 72 3 2 3
[M1]
= 2 3
3 3 74 3
[M1]
= 9 2 3 [A1]
5(a)
3 2
2
2 3 9 2
4
x x x
x
= 2
102 3
4
xx
x
= 10
2 3( 2)( 2)
xx
x x
[M1]
Let 10
( 2)( 2) 2 2
x A B
x x x x
[M1]
x – 10 = A(x – 2) + B(x + 2)
when x = 2, – 8 = 4B
B = – 2 [A1]
when x = – 2, – 12 = – 4A
A = 3 [A1]
3 2
2
2 3 9 2
4
x x x
x
= 3 2
2 32 2
xx x
[A1]
5(b) 21 cos
cos cot1 cos
xec x x
x
RHS =
21 cos
sin sin
x
x x
[M1]
=
21 cos
sin
x
x
=
2
2
1 cos
sin
x
x
[M1]
=
2(1 cos )
1 cos 1 cos
x
x x
[M1]
= 1 cos
1 cos
x
x
= RHS [A1]
6(a)
6(b) From the graph, 2 solutions [B1]
7(a) 3 tan x + cot x = 5 cosec x
sin cos 5
3cos sin sin
x x
x x x [M1]
3sin2 x + cos2 x = 5cos x
3(1 – cos2 x) + cos2 x = 5cos x
2cos2 x + 5cos x – 3 = 0 [M1]
(2cos x – 1)(cos x + 3) = 0
cos x = 0.5 or cos x = – 3 [M2]
basic = 1.05 (rejected)
Soln in 1st and 4th quadrant,
x = 1.05, 5.23 [A1]
(rejected)
Correct Graph – M2
Correct Range and Axis – M1
7(b) 9sin3 7cos3 = sin 3R
2 29 7 11.402R [M1]
7
tan9
037.87 [M1]
9sin3 7cos3 = 011.402sin 3 37.87 [A1]
011.402sin 3 37.87 = 3
0sin(3 37.87 ) 0.263 [M1]
3 - 37.870 = 15.250, 164.750,
375.250, 524.750
735.250, 884.750
3 = 53.120, 202.620, 413.120, 562.620,
773.120, 922.620
= 17.70, 67.50, 137.70 (1 dec place) [A2]
1
FEB
EXAM 5
2016 SEC 4 ADDITIONAL MATHEMATICS CW & HW
1. (a) (i) Expand
9
14
x
up to the first 3 terms. [2]
(ii) Hence, given that 9
2 28 2 3 1 8 .......4
xx x hx kx
,
find the values of h and k. [3]
(b) Evaluate the coefficient of x7 in the binomial expansion of
14
2
2
1
xx . [3]
2
2. (a) If α and β are the roots of the equation 22 9 0x hx and β = 2α, calculate
the values of h. [3]
(b) Given that α and β are the roots of the equation 23 5 1 0x x , form another
equation whose roots are 2
1
and
2
1
. [4]
3
3. The line 7 34y x cuts the circle 2 2 2 4 20 0x y x y at two points, A and B.
Find
(a) the coordinates of A and B, [5]
(b) the equation of the perpendicular bisector of AB and show that it passes
through the centre of the circle. [6]
4
4. The diagram shows a prism such that each cross-section is a quadrant of a circle of
radius x cm, with angle at the centre equal to 90. The cross-sections are OAB and
PDC where A, B, C, D lie on the curved surface of the prism and the vertical line OP
is the intersection of the vertical plane faces OADP and OBCP. The cross-sections are
horizontal and y cm apart.
(a) Given that the volume of the prism is 20 cm3, express y in terms of x.
Hence show that the total surface area, A cm2, of the prism is given by
2 40 4
2
xA
x
. [4]
(b) Find the value of x for which A has a stationary value, [2]
(c) Find the stationary value of A, [1]
(d) Determine if the stationary value of A is a maximum or a minimum. [2]
(e) Given also that the total surface area, A, is increasing at a constant rate of
3 cm2 s-1, find the rate at which x is changing when x 4. [3]
y cm
O A
B
C
D P
x cm
6
5. (a) Solve the following equations for 0 x 360:
(i) 3cos 2sin2 0x x , [3]
(ii) 2sec 4 tan 6x x . [4]
(b) Prove the identity
sin 7 sin 5
tancos 7 cos5
x xx
x x
. [3]
7
6. Differentiate the following with respect to x :
(a) (3 2x3)10 [2]
(b) 122 xex [3]
(c)
4
32ln
x
x, leaving your answer to the simplest form. [4]
8
7. Variables x and y are related by the equation x
bxay , where a and b are
constants. The table below shows measured values of x and y.
x 1 2 3 4 5 6
y 5.7 5.6 5.9 6.2 6.6 6.9
(a) On a graph paper, plot xy against x, using a scale 2 cm to represent 1 unit
on the x axis and 1 cm to represent 1 unit on the xy axis. Draw a straight
line graph to represent the equation x
bxay . [3]
(b) Use your graph to estimate the value of a and of b. [3]
(c) On the same diagram, draw the line representing the equation x
xy
3 and
hence find the value for which a
bx
3. [2]
10
8. A particle moves along a straight line so that its displacement, s metres, from a fixed
point P is given by 3 214 15
3s t t t , where t is the time in seconds after passing P.
Find the
(a) initial velocity and acceleration of the particle, [4]
(b) minimum velocity, [2]
(c) range of values of t for which the velocity is negative. [2]
11
9. The diagram shows the curve 1
42
x
xy crosses the x-axis at P.
(a) Find the coordinates of P. [1]
(b) Find the equation of normal to the curve at P. [4]
P
R
S
y
x
Q
12
10. The solution to this question by accurate scale drawing will not be accepted.
The diagram shows a rectangle ABCD. The coordinates of A and D are A(4, 1) and
D(16, 5) and the equation of AC is y = x – 3.
Find
(a) the equation of CD, [2]
(b) the coordinates of C and B, [3]
(c) the length of AB, [1]
(d) the area of the rectangle ABCD. [2]
y
x A(4, 1)
D(16, 5)
C
B
O
13
EXAM 5 ANSWERS
1(a) 9
2 28 2 3 1 8 .......4
xx x hx kx
(i)
9 29 91 1
4 4 4
x x x
+ ……. [M2]
(ii) 2
2 9 98 2 3 1 ...
4 4
x xx x
=
22 29
8 18 18 2 3 ...2
xx x x x
=
2218 16 ...
2
xx [M2]
21
16 and 2
h k [A1]
1(b)
14
2
2
1
xx
General term or 1th
r term of the expansion
= 14
14 2 1
2
rr
rC xx
= 14 28 31
2
r
r
rC x
[M1]
For the term in 7x ,
7 28 3rx x
7 = 28 – 3r
r = 7 [M1]
Coeff of
7
7 14
7
1 1326
2 16x C
[A1]
2(a) 22 9 0x hx
2
h
9
2 [M1]
Given β = 2α
2α2 = 9
2 α =
3
2 [M1]
2 6 9h [A1]
14
2(b) 23 5 1 0x x
5
3
1
3 [M1]
Sum of new roots
=
2 2
22 2
1 1
=
2
2
2
[M1]
= 31
Product of new roots
=
2
19
[M1]
New equation is 2 31 9 0x x [A1]
3(a) 7 34y x ---
2 2 2 4 20 0x y x y ---
Sub y = 34 – 7x into eqn : [M1]
22 34 7 2 4 34 7 20 0x x x x
2 21156 476 49 2 136 28 20 0x x x x x
250 450 1000 0x x
2 9 20 0x x [M1]
5 4 0x x
x = 4 or 5
y = 6 or – 1 [A2]
A(4, 6) and B(5, – 1) [A1]
3(b) Gradient of line AB = 1 6
75 4
Gradient of line = 1
7 [M1]
Midpoint of AB = 4 5 6 1 9 5
, ,2 2 2 2
[M1]
Eqn of bisector of AB is
5 1 9
2 7 2y x
15
7 13y x [M1]
2 2 2 4 20 0x y x y ---
Centre of Circle is 2g = – 2, – g = 1
2f = – 4, – f = 2
[M2]
(1, 2)
Sub (1, 2) into eqn :
LHS = RHS [A1]
4(a) volume of prism = 20π cm3
base area of prism
= 21
4x [M1]
volume of prism = 21
4x y = 20π
2
80y
x [A1]
Curved surface area ABCD = 1
24
xy
= 2
1 802
4x
x
= 40
x
cm2
Surface Area of OADP and OBCP
= 2xy = 160
xcm2
Surface Area of quadrant OAB and PCD
=
2
2
xcm2
Total Surface Area of Prism, A
=
2
2
x+
40
x
+ 2xy
=
2 40 160
2
x
x x
A =
2 40( 4)
2
x
x
cm2 (shown)
16
4(b) For stationary value, 0dA
dx
2
40 4dAx
dx x
= 0 [M1]
2
40 4x
x
3
40 4x
x = 4.50 cm [A1]
4(c) Stationary value of A = 95.3 cm2 [A1]
4(d)
2
40 4dAx
dx x
2
2 3
80 4d A
dx x
[M1]
At A = 95.3 cm2, 2
2 3
80 4d A
dx x
>0
minimum point. [A1]
4(e) Given 2 13 cm
dAs
dt
dA dA dt
dx dt dx [M1]
when x = 4,
2
40 4dAx
dx x
= – 5.288
5.288
1.7623
dt
dx
[M1]
0.567dx
dt cm/s [A1]
17
5(a)(i) 3cos 2sin2 0x x
3cos 4sin cos 0x x x
cos (3 4sin ) 0x x [M1]
cos 0x or 3 4sin 0x
3
sin4
x
90 , 270x 228.6 , 311.4x
[A2]
5(a)(ii) 2sec 4 tan 6x x
21 tan 4 tan 6x x [M1]
2tan 4 tan 5 0x x
tan 5 tan 1 0x x [M1]
tan x = 5 or tan x = – 1
basic = 78.69 basic = 45
x = 135 ,78.7 ,315 ,258.7 [A2]
5(b) sin 7 sin 5
tancos 7 cos5
x xx
x x
LHS = sin 7 sin 5
cos 7 cos5
x x
x x
=
7 5 7 52cos sin
2 27 5 7 5
2cos cos2 2
x x x x
x x x x
[M2]
= sin
tancos
xx RHS
x [A1]
6(a) 10
33 2d
xdx
= 9
3 210 3 2 6x x [M1]
= 9
2 360 3 2x x [A1]
6(b) 2 2 1xdx e
dx
= 2 2 1 2 1 2x xd d
x e e xdx dx
[M1]
= 2 2 1 2 12 2x xx e e x [M1]
= 2 12 ( 1) xx x e [A1]
18
6(c) 2 3
In4
d x
dx x
= [ln 2 3 ln 4 ]d
x xdx
[M1]
= 2 1
2 3 4x x
[M2]
=
11
2 3 4x x
[A1]
7(a)
x
bxay
y x ax b
Y = mX + C
y x 5.70 7.92 10.22 12.4 14.76 16.90
x 1 2 3 4 5 6
y 5.7 5.6 5.9 6.2 6.6 6.9
[M1]
Correct Graph – B1
Correct Points – B1
7(b) a = gradient = 16.9 5.7
2.246 1
[B2]
b = y-intercept = 3.45 [B1]
19
7(c) Draw the line Y = 3X [B1]
a
bx
3
3x – ax = b
3x = ax + b
Points of intersection of the line Y = 3X and Y = aX + b
(4.6, 13.8) [A1]
8. 3 21
4 153
s t t t
(a) initial velocity =
0t
ds
dt
= 2 8 15t t [M1]
= 15 m/s [A1]
initial acceleration =
2
2
0t
d s
dt
= 2 8t [M1]
= – 8 m/s2 [A1]
20
8(b) Minimum velocity
2
20
d s
dt
2t – 8 = 0
t = 4 s [M1]
min velocity = 2 8 15t t = – 1 m/s [A1]
8(c) 2 8 15t t < 0
5 3 0t t [M1]
3 5t [A1]
9. 1
42
x
xy
(a) when y = 0, x = – 2 P(– 2, 0) [B1]
(b) Gradient at point P
= 2 4
1
d x
dx x
=
2
1 2 4 2 4 1
1
d dx x x x
dx dx
x
=
2
6
1x
[M1]
= 2
3 [M1]
Normal gradient = 3
2 [M1]
Eqn of normal at point P:
3 3
2 32 2
y x x [A1]
10(a) Gradient of AD = 5 1 1
16 4 3
Gradient of line to AD = – 3 [M1]
Eqn of CD,
5 3( 16)y x
3 53y x [A1]
21
10(b) 3 53y x ---
y = x – 3 ---
x – 3 = – 3x + 53
4x = 56
x = 14
y = 11
C(14, 11) [A1]
Midpoint of AC and BD is the same.
16 5 4 14 1 11
, ,2 2 2 2
x y
[M1]
x = 2, y = 7
B(2, 7) [A1]
10(c) Length of AB
= 2 2
4 2 1 7 2 10 6.32 units
[B1]
10(d) Area of ABCD
= 4 2 14 16 41
1 7 11 5 12 [M1]
= 80
= 80 units2 [A1]