S4MA Set C Paper 2 Answer
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Sec 4 Maths
Exam papers with worked solutions
SET C
PAPER 2
Answer
Compiled by
THE MATHS CAFE
1|Page
S4MA Set C Paper 2 Answer
www.pmc.sg
Q
Steps / Answers
(a)
(i)
1
Type of
Marks
p = 2 p 2  15
 12
 1

 p  3  p 2  5   0






1
(b)
(i)
(b)
(ii)
2 3   2
3x2
x 1
2
(b)
(i)
(b)
(ii)
(a)
3
M1
for correct substitution
M1
for correct factorisation
M1
A1
4
3 x 1  2 43 x  2
(x – 1)lg 3 = (2 – 3x)lg 2
0.782
M1
M1
M1
A1
M1
M1
A1
log 3 y  log 3 x  2
y
2
x
y = 9x
M
e
2 tan x + 4 = sec2x
2 tan x + 4 = 1 + tan2x
(tan x + 1)(tan x – 3) = 0
71.6, 135, 251.6, 315
C
s
M1
M1
A1
for identifying common base
for using law of indices
for using log a x r  r log a x
4
3
for correct substitution
3
M1
M1
M1
A1
4
basic shape
correct amplitude
correct no. of cycles
G1
G1
G1
3
x
2
correct table of values
8 solutions
M1
2x = 6 – 3y
(6 – 3y + 1)2 + 6(y – 2)2 = 49
(5y – 2)(y – 4) = 0
M1
M1
M1
 2 2
 2 , and  3,4 
 5 5
M1
h
T
y=2–
e
fa
5
h
ta
p+q=3
3p + q =5
p = 1, q = 2
log 3
(a)
Remarks
M1
1
lg p  lg 5
2
25
3.22
(a)
(ii)
Total
Marks
M1
A1
for using log a x  log a y  log a
1
cos
2
for using sec x = 1 + tan2x
for correct factorisation
for using sec =
for correct line
3
for correct new equation
for correct substituition
for correct factorisation
2|Page
x
y
S4MA Set C Paper 2 Answer
www.pmc.sg
6.49
(b)
A1
M1
for correct expansion
 14  6 5 
 14  6 5 
1






14  6 5  14  6 5  14  6 5  14  6 5 
M1
for rationalization
Denominator = 16
1.75
M1
A1
for simplification
Ax – 2A + B
A=1
B=2
M1
A1
A1
1
96 5 5

1
96 5 5
1
(a)

4
 1
2 


dx
2
 x  2  x  2 
ln (x – 2)
2 x  22 1
1 1
(a)
5
(b)
M1
A1
r2 = 144 – h2
1
 144  h 2 h
3
1
48h - h 3
3
M1
48 - h 2
M1
M1
M1
A1
 2x  1 
8 e     c
  2 
x
2
6
(a)
(ii)
(b)
(i)
4
for recognizing application of PF
in integration
M1
M1
M1
[ln 2 – 1] – [ln 1 – 2]
1.69
48 - h 2 = 0
6.93
max value
(a)
(i)
5
M1
A1
for 1st integral
for 2nd integral
for using limits
8
for using PT
for correct substitution
3
for differentiation
for using stationary values = 0
4
M1, M1
for correct integration
for constant, c.
2e  c
A1
1

6  sin 2 x 
2

M1
for correct integration


3sin 2  sin 
2

3
M1
for using limits
e2x (1) + x(2e2x)
e2x (1 + 2x)
M1
A1
A1
3
3
2
for using product rule
allow B2 if e2x (1) + x(2e2x) seen
3|Page
S4MA Set C Paper 2 Answer
www.pmc.sg
(b)(i
i)
e2x (1 + 2x) = 0
0.5
(b)
(iii)
 e 1  2 x dx  xe
 e dx   2 xe dx
2x
2x
 2 xe
 4 xe
(a)
(b)
(a)
(b)
M1
A1
M1
2x
2x
M1
M1
1
dx  xe 2 x  e 2 x  c
2
dx  2 xe 2 x  e 2 x  c
A1
2x2 + (4 – k)x + (2 – 3k) = 0
(4 – k)2 – 4(2)(2 – 3k) > 0
k( k + 16) > 0
k < 16 and k > 0
M1
M1
M1
A1
 +  = 2 and  = 0.5
product of roots = 9
sum of roots = 2
x2 – 2x – 9 = 0
M1
M1
M1
A1
2x
 = 1.1071
M
e
(c)
20.9 cm
(d)
sin (  + 1.1071) =
 = 1.30
(e)
(a)
9
(b)
3
2 5
h
T
ABCD = 4 cos  (6 + 2 sin )
XAB = 4 cos  sin 
4 cos  (6 + 2 sin ) = 5(4 cos  sin
) + 6
4 cos  = sin 2 + 1
3
x  0
4
4y + 16 = 3x
4
y – 6 =   x  5
3
4x + 3y = 38
y – ( 4) =
solving sim eqns
(8,2)
x0
y   4 
 8,
2
2
2
M1
A1
M1
A1
A1
e
fa
4
C
s
h
ta
AB = cos , AD = 6 + 2 sin 
P = 4 sin  + 8 cos  = 12
dy
 0.
dx
for showing integration is reverse
of differentiation
for splitting up integrals
for indication of 2xe2x be subject
2x
P = 12 + 4 5 sin ( + 1.1071)
8
2
for using
4
4
2
2
1
M1
A1
2
M1
M1
A1
3
allow B1 for y =
3
x4
4
A1
M1
A1
3
M1
M1
M1
4|Page
S4MA Set C Paper 2 Answer
www.pmc.sg
(c)
(a)
(b)
10
(16, 8)
A1
1 5 0 16 5
2 6 4 8 6
M1
50
A1
2
lg y = 2x lg h + lg k
grad = 2 lg h
y-intercept = lg k
A1
A1
2
new table of values
k = 13.2
h = 1.13
appropriate scale
straight line that intersects vertical
axis
(c)
(i)
x = 24
(c)
(ii)
lg y = 0.73
y = 5.37
4
for correct substitution of formula
M1
M1
M1
M1
M1
5
A1
1
M1
A1
2
5|Page