Additional Mathematics 4047/02
O-LEVEL A-MATHS 2017 – PAPER 2
Question 1
[ Ans:
k =2 ]
y = e− x x 2
dy
= e − x ( 2 x ) + ( −e − x ) x 2 = e − x ( 2 x − x 2 )
dx
d2y
= e − x ( 2 − 2 x ) + ( −e − x )( 2 x − x 2 ) = e − x ( 2 − 4 x + x 2 )
dx 2
 d2y

dy
k = ex  2 + 2 + y 
dx
 dx

= e x e − x ( 2 − 4 x + x 2 ) + 2e − x ( 2 x − x 2 ) + e − x x 2 
= ( 2 − 4 x + x2 ) + 2 ( 2 x − x2 ) + x2
= 2 − 4x + x2 + 4x − 2x2 + x2
=2
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O-Level A-Maths 2017 – Paper 2
Page 1 of 11
Additional Mathematics 4047/02
Question 2
[ Ans: (i) show (ii)
(i)
5
]
6
d
( tan x − x )
dx
= sec2 x − 1
= tan 2 x (shown)

(ii)
a = 4, b = −
 ( sec
3
2
x + 5 tan 2 x ) dx
6

=  tan x + 5 ( tan x − x )  3
6

3

6
=  6 tan x − 5 x 



   
  
= 6 tan − 5    − 6 tan − 5   
3
6
 3  
 6 

5
6 5
=6 3−
3
−
3
+
6
6 3 5
−
3
6
5
=4 3−
6
=6 3−
 a = 4 and b = −
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5
6
O-Level A-Maths 2017 – Paper 2
Page 2 of 11
Additional Mathematics 4047/02
Question 3
[ Ans: (i) explain (ii) p = 2 ]
(i)
( r + 1) th term in the binomial expansion of  px3 + 1 

9− r  1 
9
=   ( px3 )  
 x
r
9
=   p 9− r x 27 −3r x − r
r
9
x
r
9
=   p 9− r x 27 −4 r
r
 power of a general term in the binomial expansion is of the form ( 27 − 4r ) , where
r is an integer.
27 is an odd number and 4r is an even number for all integer values of r ,
( 27 − 4r ) cannot take the form of any even numbers.  there is no even powers of x
Since
9
1

in the expansion of  px 3 +  .
x

(ii) Let
27 − 4r = 11  r = 4
9
 Coefficient of x11 =   p 9− 4 = 126 p 5
4
 
Let
27 − 4r = 7  r = 5
9
 Coefficient of x 7 =   p 9−5 = 126 p 4
5
126 p5 = 2 (126 p 4 )
p5 − 2 p 4 = 0
p4 ( p − 2) = 0
p = 0 (NA) or p = 2
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O-Level A-Maths 2017 – Paper 2
Page 3 of 11
Additional Mathematics 4047/02
Question 4
[ Ans: (i) show (ii) less ]
(i)
y=
6
+ x = 6x
x
1
−
2
+x
 1 −3 
dy
3
= 6 − x 2  +1 = − 3 +1
dx
 2

x2
At
M,
dy
=0
dx
3
− 3 +1 = 0
x2
3
x2 = 3
2
 32 
2
3
x  =3  x =9
 
 the x -coordinate of M satisfies the equation x3 = 9 . (shown)
(ii) At A (1, 7 ) ,
dy
3
= − 3 + 1 = −2
dx
12
Equation of tangent:
y − 7 = −2 ( x − 1)  y = −2 x + 9 (1)
At B ( 4, 7 ) ,
dy
3
5
= − 3 +1 =
dx
8
42
Equation of tangent:
y −7 =
At
P , (1) =
5
5
9
( x − 4)  y = x +
8
8
2
(2)
(2)
5
9
x+
8
2
21
9
12
x =  x = = 1.7143
8
2
7
−2 x + 9 =
x -coordinate of
M = 3 9 = 2.0801
 the x -coordinate of P is less than the x -coordinate of M .
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O-Level A-Maths 2017 – Paper 2
Page 4 of 11
Additional Mathematics 4047/02
Queston 5
[ Ans: (i)
(i) log5 ( x − 1) − log 5 ( x + 1) = 1 + log 5
x = 6 (ii) y = 0.039 or y = 26 ]
1
7
1
 x −1 
log 5 
 = log 5 5 + log 5
7
 x +1 
5
 x −1 
log 5 
 = log 5
7
 x +1 
x −1 5
=
x +1 7
7 x − 7 = 5x + 5
2 x = 12  x = 6
(ii) log y 100 = lg y
lg100
= lg y
lg y
( lg y ) = lg102
2
( lg y ) = 2 lg10 = 2
2
lg y = − 2 or
2
y = 10−
2
2
or
10
y = 0.039 or y = 26 (to
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2
s.f.)
O-Level A-Maths 2017 – Paper 2
Page 5 of 11
Additional Mathematics 4047/02
Question 6
[ Ans: (i)
(i)
 1 7
5 (ii) P  − ,  (iii) L is a vertical line ]
 3 3
y = 9 x 2 + ( 2m + 1) x + 1 + c (1)
y = mx + c (2)
At P , (1)
=
(2)
9 x 2 + ( 2m + 1) x + 1 + c = mx + c
9 x 2 + ( m + 1) x + 1 = 0 (3)
Let Discriminant
=0
( m + 1) − 4 ( 9 )(1) = 0
2
( m + 1) = 36
2
m + 1 = −6 or m + 1 = 6
m = −7 (NA) or m = 5
(ii) For
m = 5 and at ( −2,19 ) ,
2
(1)
19 = 9 ( −2 ) +  2 ( 5 ) + 1 ( −2 ) + 1 + c
c=4
(3)
9 x 2 + ( 5 + 1) x + 1 = 0
9 x 2 + ( 5 + 1) x + 1 = 0
9 x2 + 6 x + 1 = 0
( 3x + 1)
2
=0
1
3
7
 1
y = 5 −  + 4 =
3
 3
 1 7
P− , 
 3 3
x=−
(iii) It can be deduced that
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L
is a vertical line.
O-Level A-Maths 2017 – Paper 2
Page 6 of 11
Additional Mathematics 4047/02
Question 7
[ Ans: (a)(i)
(a) (i) When
−0.000121 (ii) 38.0% (b) 4.1 ]
t = 0,
P = 100e0 = 100
When
t = 5730 ,
1
P = 100
2
100e
− k ( 5730 )
e−5730 k
= 50
1
=
2
1
2
1
1
k =−
ln = −0.000121
5730 2
−5730k = ln
(ii) P = 100e
When
1
1

− −
ln t
 5730 2 
1
 1
ln t

2
= 100e 5730
t = 8000 ,
P = 100e
1
 1
ln ( 8000 )

 5730 2 
= 38.0
 38.0 % of carbon-14 would indicate a fossil age of 8000 years.
(b) When
S = 2.4 ,
lg
I
= 2.4
c
I
= 102.4  I = 102.4 c
c
When I = 50 (102.4 c ) ,
S = lg
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50 (102.4 c )
c
= lg 50 (102.4 ) = 4.1 (to 1 d.p.)
O-Level A-Maths 2017 – Paper 2
Page 7 of 11
Additional Mathematics 4047/02
Question 8
[ Ans: (a)
 1



0.4 units per second (b)(i) explain (ii)  − ,8  ]
2
(a) y = ln ( 3x − 1)
dy
1
3
=
( 3) =
dx 3x − 1
3x − 1
When
x =7,
dy
3
3
=
=
dx 3 ( 7 ) − 1 20
dy dy dx
= 
dt dx dt
3 dx
dx
0.06 =  
= 0.4
20 dt
dt
(b) (i)
y = 8 − ( 2 x + 1)
3
dy
2
2
= −3 ( 2 x + 1) ( 2 ) = −6 ( 2 x + 1)
dx
Let
dy
= 0,
dx
−6 ( 2 x + 1) = 0
2
2x +1 = 0  x = −
1
2
 the curve has only one stationary point.
1
2
When x = − ,
1
 
2
x
−
1
2
1
 
2
dy
dx
−
0
−
Shape
\
_
\
 the stationary point at x = −
+
1
is a point of inflexion.
2
1
2
(ii) When x = − ,
3
  1 
y = 8 −  2  −  + 1 = 8
  2 
 1 
 the coordinates of the stationary point is  − ,8  .
 2 
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O-Level A-Maths 2017 – Paper 2
Page 8 of 11
Additional Mathematics 4047/02
Question 9
 2 1
 5 5
[ Ans: (i) show (ii) D  − ,  (iii)
(i) Gradient of
AB =
p −1
p −1
=
0 − ( −2 )
2
Gradient of
BC =
3− p
= 3− p
1− 0
Since
6.8 units2 ]
ABO = CBO ,
p −1
= − (3 − p )
2
p − 1 = −6 + 2 p
p = 5 (shown)
(ii) Gradient of
AB = 5 − 1 = 2
2
 gradient of AD = −
1
2
Equation of line passing through
AD :
1
y − 1 = −  x − ( −2 )
2
1
y = − x (1)
2
Gradient of
CD = Gradient of AB = 2
Equation of line passing through
y − 3 = 2 ( x − 1)
y = 2 x + 1 (2)
CD :
1
− x = 2x +1
2
5
2
− x =1 x = −
2
5
1 2 1
2
y = − −  =
Sub. x = − into (1)
2 5 5
5
(1)
=
(2)
 2 1
D− , 
 5 5
(iii) Area of trapezium
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ABCD =
1
2
2
1 0 −2
1 17  51  
5
=  −  −   = 6.8
1
2  5  5 
3 5 1
5
−2 −
1
O-Level A-Maths 2017 – Paper 2
Page 9 of 11
Additional Mathematics 4047/02
Question 10
[ Ans: (i)(a)
2 , 360
(b) 1 ,
180 (ii) 218.2,321.8 (iii) sketch (iv) 0  x  218.2 or
321.8  x  360 ]
(i) (a) Amplitude = 2
Period = 360
(b) Amplitude
Period =
(ii)
=1
360
= 180
2
y1 = y2
2sin x + 1 = − cos 2 x
2sin x + 1 = − (1 − 2sin 2 x )
2sin 2 x − 2sin x − 2 = 0
sin 2 x − sin x − 1 = 0
sin x =
sin x =
− ( −1) 
( −1) − 4 (1)( −1)
2 (1)
2
=
1 5
2
1− 5
( −1  sin x  1 )
2
  1 − 5 
  = 38.173

  2  
Basic  = sin −1  − 
x = 180 + 38.173,360 − 38.173
= 218.2,321.8
(iii)
(iv)
y1 − y2  0  y1  y2
From the graphs in (iii),
0  x  218.2 or 321.8 
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x  360
O-Level A-Maths 2017 – Paper 2
Page 10 of 11
Additional Mathematics 4047/02
Question 11
[ Ans: (i) 4 cos  , 4 sin  (ii) show (iii)
(i) Let a be the shortest distance from
Let
b be the shortest distance from
P
P
544 cos ( − 31.0 ) (iv) 80.9 ]
OX .
to OY .
to
𝑏
𝑎
𝑎
a
 a = 4cos 
4
b
sin  =  b = 4sin 
4
cos  =
(ii) By similar triangles,
3− a b
=
3
5
15 − 5a = 3b
15 − 5 ( 4 cos  ) = 3 ( 4sin  )
20 cos  + 12sin  = 15 (shown)
(iii) Let 20 cos  + 12sin  = R cos ( −  ) = R cos  cos  + R sin  sin 
 R cos  = 20 (1)
R sin  = 12 (2)
2
2
2
R = 202 + 122  R = 544
(1) + ( 2)
( 2)
(1)
tan  =
12
12
  = tan −1
= 30.964
20
20
 20 cos  + 12sin  = 544 cos ( − 31.0 )
(iv) 20 cos  + 12sin  = 15
544 cos ( − 30.964 ) = 15
15
544
15
Basic angle = cos −1
= 49.975
544
 − 30.964 = 49.975
cos ( − 30.964 ) =
 = 80.9
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O-Level A-Maths 2017 – Paper 2
Page 11 of 11