Suggested Solutions to 2010 NJC SH2 H2 Math Prelim Paper 1
1
The sum of the digits in a three-digit-number is 15. Reversing the digits in that number
decreases its value by 594. Also, the sum of the tenth digit and four times the unit digit is
five more than the hundredth digit. Find the number.
1Let the unit digit be z.
Let the tenth digit be y.
Let the hundredth digit be x.
x  y  z  15
 (1)
(100 x  10 y  z )  (100 z  10 y  x)  594
 99 x  99 z  594  (2)
y  4z  x  5
  x  y  4z  5
 (3)
Using GC to solve the equations simultaneously,
x  8, y  5, z  2 .
Thus the number is 852.
Page 1 of 29
[4]
Suggested Solutions to 2010 NJC SH2 H2 Math Prelim Paper 1
2
Find the value of p such that

p
0
3 x dx 
2
.
ln 3
[2]
The graph of y  3x for 0  x  1 , is shown in the diagram below. Rectangles, each of width
1
, where n is an integer are drawn under the curve.
n
y
y  3x
…
0
1
n
2
n
n3
n
3
n
n  2 n 1
n
n
x
1
 
1
2 3n
2
(i)
Show that the total area of all n rectangles, A is given by
(ii)
State the limit of A as n   .

p
0
3 x dx 

1

n 3n  1
.
[2]
[1]
2
ln 3
p
 3x 
2
 ln 3   ln 3

0
1
2
3 p  30  

ln 3
ln 3
p
3 1  2
3p  3
 p 1
2
Suggested Solutions to 2010 NJC SH2 H2 Math Prelim Paper 1
2(i)
Total area of all the n rectangles, A

1  1n  1  n2  1  n3 
3   3   3  
n  n  n 
2
3
1  1n
n
n
 3  3  3 
n
1 n 
  3n 
n 

3 

n
n
  1 n 
1  3n
  1 

3n   
  1

n  n
3 1 




 1
2  3n 
 

 1 
n  3 n  1


2(ii)
As n   ,
limit of A = area under the curve y  3x for 0  x  1
1
  3 x dx
0

2
ln 3
3
Suggested Solutions to 2010 NJC SH2 H2 Math Prelim Paper 1
3
The points A and B have position vectors a and b respectively, relative to the origin O such
that a = b . The point P with position vector p lies on AB such that b • p = a • p.
(i)
Show that AB is perpendicular to OP.
(ii)
Determine the position vector of the point D in terms of a and b, where D is the
(iii)
3(i)
reflection of O about the line AB.
[2]
Give the geometrical meaning of a  b .
[1]
AB  OP   b  a  • p
=b•p–a•p
= a • p – a • p (since b • p = a • p)
=0
Hence, AB is perpendicular to OP.
OR
b•p=a•p
b•p–a•p=0
b
 a • p = 0
AB  OP  0
Hence, AB is perpendicular to OP.
3(ii)
[2]
Since a  b , then P must be the midpoint of AB.
Using ratio theorem, OP 
1
a  b 
2
Thus, OD  2OP
1

 2  a  b  
2

 ab
4
Suggested Solutions to 2010 NJC SH2 H2 Math Prelim Paper 1
3(iii)
a  b represents the
(1) area of rhombus OADB or OBDA.
(or)
(2) magnitude of a vector which is perpendicular to a and b.
4
π
. P is a fixed point on the curve
2
and Q is a point on the curve with coordinates
The diagram shows the graph of y  sin x for 0  x 
with coordinates
 x1,sin x1 
 x   ,sin  x    , where 
1
1
is measured in radians.
y
Q
y  sin x


P 
x
O
sin x1 (cos   1)  cos x1 sin 
(i)
Prove that the gradient of PQ is
(ii)
Given that  is sufficiently small for  3 and higher powers of  to be neglected,
express the gradient of PQ as a linear expression in terms of  .
[2]

.
[2]
Verify that the gradient of PQ approximates to the gradient of the tangent at P when
[2]
 tends to zero.
5
Suggested Solutions to 2010 NJC SH2 H2 Math Prelim Paper 1
4(i)
Gradient of PQ

4(ii)
sin( x1   )  sin x1
( x1   )  x1

sin x1 cos   cos x1 sin   sin x1

sin x1 (cos   1)  cos x1 sin 


(shown)
When  is small,
Gradient of PQ

sin x1 (cos   1)  cos x1 sin 

 1  
sin x1 1   2   1    cos x1 
 2  


1
 cos x1   sin x1
2
4(iii)
As  tends to zero,
gradient of PQ =
1


lim  cos x1   sin x1 
 0
2


 cos x1
gradient of tangent at point P

d
 sin x 
dx
x  x1
 cos x1
6
Suggested Solutions to 2010 NJC SH2 H2 Math Prelim Paper 1
5
Given that z  3  4i  5 , illustrate the locus of the point P representing the complex
number z in an Argand diagram.
[1]
Hence, find the least exact value of z  1 .
[2]
The locus of point Q, representing the complex number w, satisfies the relation
w  2i  w  ai , where a is a real number, a  2 . Find the range of values of a such that
the loci of P and Q meet more than once.
[3]
5
Im (z)
9
  3, 4 
O
Re (z)
–1
B1: circle centred at (– 3, 4) (Note the circle passes through origin.)
Least exact value of z  1
 (3  1) 2  (4  0) 2  5
 32  5
or
4 2 5
7
Suggested Solutions to 2010 NJC SH2 H2 Math Prelim Paper 1
w  2i  w  ai represents the perpendicular bisector y 
a2
2
Im (z)
9
y
a2
2
  3, 4 
O
–1
For the line to meet the circle more than once,
a2
9
2
2  a  2  18
1 
4  a  16 (ans)
8
Re (z)
Suggested Solutions to 2010 NJC SH2 H2 Math Prelim Paper 1
6
By using the substitution y  vx , find the general solution of the differential equation
x
(i)
dy
 3x  y  2 .
dx
[4]
State the equation of the locus where the stationary points of the solution curves lie.
[1]
(ii)
Sketch, on a single diagram, the graph of the locus found in part (i) and 2 members
of the family of solution curves where the arbitrary constant in the solution is
non-zero.
6
y  vx 
[3]
dy
dv
v x
dx
dx
dy
 3x  y  2
dx
dv
 x(v  x )  3x  vx  2
dx
dv 3x  2


dx
x2
x
3 2
 dx
x x2
2
 v  3 ln | x |   C
x
 y  3x ln | x | 2  Cx
  dv  
6 (i)
dy
 0  y  3 x  2
dx
9
Suggested Solutions to 2010 NJC SH2 H2 Math Prelim Paper 1
6 (ii)
(ii)
C 0
C0
2
y  3 x  2
Note: (0,2) satisfies the differential equation. (0,2) is a singular solution.
7(a) By considering x  2  A(2  2 x)  B where A and B are real constants, or otherwise, find

x2
x  2x  8
2
dx.
[5]
1
5
(b) Show that  e x cos 2 x dx  e x (cos 2 x  2 sin 2 x)  c , where c is an arbitrary constant.
Hence, find
7(a)
e
x
cos 2 x dx.
[5]
x  2  A(2  2 x)  B
By comparing coefficient of
x :1  2 A  A  
1
2
constant:  2  2 A  B  B  1
10
Suggested Solutions to 2010 NJC SH2 H2 Math Prelim Paper 1
x2

dx
x  2x  8
1
 (2 x  2)  1
 2
dx
 x2  2 x  8
1
2 x  2
1
 
dx  
dx
2
2
2
x  2x  8
3  ( x  1) 2
2

1
 x 1 
(2  x 2  2 x  8)  sin 1 
C
2
 3 
 x 1 
   x 2  2 x  8  sin 1 
C
 3 
In absence of C, deduct from presentation marks.
7 (b)
e
x
cos 2 x dx  e x cos 2 x   e x  2 sin 2 x  dx
 e x cos 2 x  2  e x sin 2 x  2  e x cos 2 x dx 


 e x cos 2 x  2e x sin 2 x  4  e x cos 2 x dx
 5 e x cos 2 x dx  e x cos 2 x  2e x sin 2 x
e
e
x
cos 2 x dx 
x
cos 2 x dx
1 x
e (cos 2 x  2 sin 2 x)  c (shown)
5
1 x
e (1  cos 2 x) dx
2
1
1
  e x dx   e x cos 2 x dx
2
2
1
1
 e x  e x (cos 2 x  2 sin 2 x)  c
2
10

(ans)
11
Suggested Solutions to 2010 NJC SH2 H2 Math Prelim Paper 1
8 (a)
The sum, S n , of the first n terms of an arithmetic progression is given by Sn  n 2  2n .
Write down the expression for Sn  Sn1 . Hence, find the value of the common difference.
[3]
8(a)
Given Sn  n 2  2n .

Sn  Sn1  n2  2n   n  1  2  n  1
2

 n 2  2n   n  1  2n  2
2
 2n  3
Since the progression is AP,
common difference, d  Tn  Tn1
 2n  3   2  n  1  3
 2.
OR
d  T2  T1
 1   1
2
12
Suggested Solutions to 2010 NJC SH2 H2 Math Prelim Paper 1
(b)
A metal screw of length L (measured in millimetre) is driven into a concrete wall by an
electrical screwdriver, such that its distance driven into the wall is proportional to the angle
turned by the screwdriver.
Wall
L
Due to some reasons, every subsequent turn by this electrical screwdriver can only achieve
an 80% of the angle turned previously.
(i)
Given that the initial angle turned by the screwdriver is  radians, write down the
expressions for the first 3 distances driven into the concrete wall, leaving your
answers in terms of  and k, where k is the constant of proportionality.
(ii)
Find the total distance driven into the concrete walls after n turns, leaving your
answer in the form ak 1  b n  , where a and b are constants to be determined.
(iii)
[2]
[3]
Given that k  2 and assuming that the total distance driven could never exceed the
length of the screw, find the minimum length of the metal screw, giving your answer
in terms of  .
[2]
13
Suggested Solutions to 2010 NJC SH2 H2 Math Prelim Paper 1
8(b)
Let d be the distance driven for every turn.
(i)
n  1 , d  k
 8 
n  2 , d    k
 10 
or 0.8k
2
8
n  3 , d    k
 10 
(ii)
or 0.64k
Total distance driven into the wall after n turns
2
 8
 8
 8
 k    k    k  ...   
 10 
 10 
 10 
n 1
k
n 1
  8   8 2
8 
 k 1        ...    
  10   10 
 10  

  8 n
 1  
10
 k   

8
 1   10 
 







  4 n 
 5k 1    
 5 


a 5, b 
(iii)
4
5
Distance driven in the long run
  4 n 
 lim 5k 1    
 5 
n 


 5k
Given k  2 , minimum length of the metal screw is 10 unit.
14
Suggested Solutions to 2010 NJC SH2 H2 Math Prelim Paper 1
9
0
 
Relative to the origin O, the point A has position vector given by OA   1  and A lies on
0
 
1
 
the plane 1 with equation defined by r •  3   3 . Another plane  2 has equation y  x .
 2
 
The planes 1 and  2 intersect at line l.
(i)
Find the vector equation of the line l .
[1]
(ii)
Show that the cosine of the acute angle between the planes 1 and  2 is
(iii)
Find the position vector of the foot of perpendicular, OF , from point A to the line l.
7
.
7
Hence, find the exact length of projection of AF onto the plane  2 .
(iv)
[2]
[5]
Another plane  3 has equation px  qy  1 , where p and q are real constants. Find
the condition in which p and q must satisfy such that the planes 1 ,  2 and  3
intersect at exactly one point.
9 (i)
[2]
1
 
r •  3   3  x  3y  2z  3
 2
 
 2 : x  y  0z  0
3
 1
4
 2
 


3
1


Using GC: l: r 
  ,  R
4
 2
 


 0 
 1 
 


OR
15
Suggested Solutions to 2010 NJC SH2 H2 Math Prelim Paper 1
3
4
1
 
3
 
r       1 ,  
4
 2 
 
 
0
 
 
9 (ii)
For the plane  2 :
y  x  x  y  0z  0
 1
 
Let n 2 be normal vector to plane  2 , then n 2   1 
0
 
 1   1
  
 3  1 
 2  0 
2
1
7
 cos       


(shown)
7
14 2
28
7
9 (iii)
Since point F lies on line l,
3
 1 3  
4
 2  4  2 
 

 

3
1 3  


Let OF 
for some  .
  

4
 2 4 2 
 

 

 0 
 1    
 

 

3 
 3  
4 2
 42 
0






3     1 

Then AF 

 1   
4 2    4 2

  0  




  




16
Suggested Solutions to 2010 NJC SH2 H2 Math Prelim Paper 1
 1
 2 


1

Now, AF  l  AF • 
0
 2


 1 


 3    1
 4  2   2 

 

 1    •  1   0
 4 2  2

 

    1 

 


1
6
3 1  2
 4  12   3 

  
3 1  2

OF 


 4 12   3 

  
 1   1 
 6  6
 3 1   2 
 4  12   3 


 
1 1   1

AF   
 
 4 12   3 


 
1

  1 
6

  6 
2
2
2
7
 2  1 1
AF           
12
 3  3  6
1
A
l

2
F
17
Suggested Solutions to 2010 NJC SH2 H2 Math Prelim Paper 1
Hence, exact length of projection from AF to the plane  2
 AF cos 

7  7


12  7 

1
1
or

12 2 3
3
6
OR
exact length of projection from AF to the plane  2
 AF 
9 (iv)
n2
n2
3
 1
4
 2
 


3
1


OP 
   , for some  
4
 2
 


 0 
 1 
 


 3 has equation px  qy  1 .
 p
 
3 : r •  q   1
0
 
18
Suggested Solutions to 2010 NJC SH2 H2 Math Prelim Paper 1
For the three planes to intersect exactly a point, l is not parallel to  3 , then:
 1
 2 

  p
 1  • q   0
 2  

  0 
1





1
1
p q 0
2
2
 p   q (ans)
10(a) Given that 2  3i is a solution to the equation
z 2  (a  i) z*  16  bi  0 ,
where z* is conjugate of complex number z, a and b are real constants, solve for the values
of a and b.
10(a)
[3]
z 2  (a  i) z*  16  bi  0
 2  3i 
2
 (a  i)  2  3i   16  bi  0
 5  12i  2a  3ai  2i  3  16  bi  0
 8  2a  10  3a  b  i  0
By comparing real and imaginary coefficient,
Real: 8  2a  0  a  4 (ans)
Im :10  3a  b  0  b  22 (ans)
19
Suggested Solutions to 2010 NJC SH2 H2 Math Prelim Paper 1
(b)
(i)
Solve the equation z 5  1  0 , expressing the solutions in the form rei , where r  0
and π    π . Show the roots of the equation on an Argand diagram.
10(b)
i

  i
 2cos   e 2 .
2
(ii)
For π    π , show that 1  e
(iii)
Deduce the roots of  w  1  1  0 in the exponential form.
5
z5  1  0
z5  1  0
z 5  1
z 5  eiπ
z5  e 
i 2 k 1 π
ze
i 2 k 1 π
5
iπ
5
, k  0, 1,  2

iπ
5
i3π
5
z  1, e , e ,e ,e

i3π
5
Im( z )
1


–1
1

O

[4]
1

–1
20
Re( z )
[2]
[3]
Suggested Solutions to 2010 NJC SH2 H2 Math Prelim Paper 1
(i)
i   i 
i 
1  e i  e 2  e 2  e 2 


i 
 
 
 
 
 e 2  cos     i sin    + cos    i sin   
 2
 2
2
 2 


  i
 2 cos   e 2 (shown)
2
(ii)
Replace complex number z with w  1 ,
ze
i  2 k 1 π
5
 w 1  e
 w  1 e
i  2 k 1 π
5
i  2 k 1 π
5
From (i),
 w  1 e
i  2 k 1 π
5
  2k  1 π  i
 2 cos 
e
10


 2k 1 π
10
for k  0, 1,  2 .
21
Suggested Solutions to 2010 NJC SH2 H2 Math Prelim Paper 1
11
The curve C has equation
f(x) 
ax 2  bx  1
, where a, b and c are real constants.
xc
Given that the line y  2 x  1 is an asymptote of C, find the value of a and show that
b  2c 1.
[3]
(i) For c  1 , using algebraic method, prove that the curve C cannot lie between 2
values, which are to be determined.
(ii) Sketch the graph of f(x) 
[3]
2x2  x  1
, showing clearly its asymptotes, the coordinates of
x 1
the axial intercepts, and turning point(s) (if any).
[3]
Hence, state the range of x for which f(x) is concaving downwards.
[1]
(iii) Given that the line y  kx  k  3 ,where k is a real constant, passes through the
intersection of the asymptotes of C, deduce the range of k where
2 x2  x  1  (kx  k  3)( x  1)
has 2 real solutions.
[1]
22
Suggested Solutions to 2010 NJC SH2 H2 Math Prelim Paper 1
11
f(x) 
ax 2  bx  1
1  bc  ac 2
 (ax  b  ac) 
xc
xc
ax  b  ac  2 x  1
 a  2 (ans) and b  ac  1  b  2c  1 (shown)
(i)
Given c  1, f(x)  2 x  1 
2
x 1
2
x 1
( x  1) y  (2 x  1)( x  1)  2
Let y  2 x  1 
2 x 2  (1  y ) x  (1  y )  0
For all real values of x,
D<0
 (1  y ) 2  4(2)(1  y)  0
 ( y  7)( y  1)  0
7 < y  1 (ans)
23
Suggested Solutions to 2010 NJC SH2 H2 Math Prelim Paper 1
(iii)
y
y  2x 1
 0,1
x
 2, 7 
x  1
B1: Shape of the curve
f is concaving downwards for x  1.
(iv)
2 x 2  x  1  (kx  k  3)( x  1)
2x2  x  1

 kx  k  3
x 1
The line kx  (k  3) passes through the point ( 1, 3) , which is the intersection of the
asymptotes. Since the oblique asymptote passes through the point ( 1, 3) and using the
graph in (iii), the gradient of the line kx  k  3 has to be more than 2 for the above equation
to have 2 real solutions.
Hence, k  2 .
24
Suggested Solutions to 2010 NJC SH2 H2 Math Prelim Paper 1
12
A curve C has parametric equations
x  2 2 and y   3  2 , where  is a real parameter.
Sketch the curve C for 0    3 .
[1]
The tangent to C at point P  2 2 ,  3  2  cuts the y-axis at point Q. Show that the equation
of the tangent at P may be written as
4 y  3 x  2  4   3  .
[2]
(a) (i) C cuts the y-axis at the point R. Find the area of triangle PQR, A, in terms of  . [2]
(ii) If x increases at a rate of 4 units per second when   2 , find the rate of change of
A at that instant.
(b)
[3]
Calculate the exact area of the region bounded by the curve C, the tangent to C at
point P when   2 and the y-axis.
12
18, 29
 0, 2 
When λ = 0,
x  2  0   0 ; y   03   2  2
When λ = 3,
x  2  3  18 ; y   3  2  29
2
2
3
25
[5]
Suggested Solutions to 2010 NJC SH2 H2 Math Prelim Paper 1
dx
dy
 4 ,
 3 2
d
d
dy 3 2 3


dx 4
4
Equation of tangent:
3
x  2 2 

4
3
4 y  4    2   3 x  6 3
y   3  2 
4 y  3 x  6 3  4 3  8
4 y  3 x  8  2 3
4 y  3 x  2  4   3  (shown)
(a)
(i)
P  2 2 ,  3  2 
R  0, 2
 4  3 
Q  0,

2 

When x = 0, 4 y  3  0   2  4   3   2  4   3   y 
4  3
2

4  3 
Coordinates of Q is  0 ,
.
2 

Area of triangle PQR,
A=
3
1
4  3 
5
2
2  44 
2

2



=
units2






2
2 
2

 2
26
Suggested Solutions to 2010 NJC SH2 H2 Math Prelim Paper 1
(ii)
Given:
A
5
2
dx
 4 when   2 .
dt

dA 5 4

d
2
By chain rule,
(Find

dA dA d


.
dt d dt
d
dx dx d
d


 4  4  2 
) For   2 ,
dt
dt d dt
dt
d 1

dt 2
dA 5  2 

 0.5  20 units2 /sec .
dt
2
4
(b)
R  0, 2
P 8 , 10
Q  0, 2
For   2 ，equation of tangent at P is 2 y  3 x  4.
Coordinates of P is 8,10  .
Coordinates of Q is  0, 2  .
27
Suggested Solutions to 2010 NJC SH2 H2 Math Prelim Paper 1
Method 1
Area
8
8
0
0
  y dx  
1
 3x  4  dx
2
8

1  3x 2
     2   4  d  
 4x
0
2 2
0
2
3
2
  4 4 +8 d 
0
1
96  32  0
2
2
 4 5


 4 2   32
 5
0
 208 

 0   32
 5

48
3

or 9 or 9.6 (ans)
5
5
OR Method 2
10
1
2 y  4  dy   x dy

2 3
2
Area of region  
10
Area bounded by the tangent at P and y-axis

10
2
1
 2 y  4  dy
3
10
1
  y 2  4 y 
2
3
1
 140   4  
3
 48
Area bounded by the curve and y-axis
10
  x dy
2
2
6 
  2  3  d    5 
0
 5 0
2
4
  6 d
2
2
2
0
28
Suggested Solutions to 2010 NJC SH2 H2 Math Prelim Paper 1
6
 32  0 
5
192

5

Area of region  48 
192 48
3

or 9 or 9.6 (ans)
5
5
5
29