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NANYANG JUNIOR COLLEGE
JC2 PRELIMINARY EXAMINATION
Higher 2
MATHEMATICS 9740/01
Paper 1 11 Sep 2012
3 hours
Additional Materials: Answer Papers
List of Formulae (MF15)
READ THESE INSTRUCTIONS FIRST
Write your name and class on every script you hand in.
Write in dark blue or black pen on both sides of the paper.
You may use a soft pencil for any diagrams or graphs.
Do not use staples, paper clips, highlighters, glue or correction fluid.
Answer all the questions.
Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case of angles in
degrees, unless a different level of accuracy is specified in the question.
You are expected to use a graphic calculator.
Unsupported answers from a graphic calculator are allowed unless a question specifically states otherwise.
Where unsupported answers from a graphic calculator are not allowed in a question, you are required to
present the mathematical steps using mathematical notations and not calculator commands.
You are reminded of the need for clear presentation in your answers.
At the end of the examination, fasten all your work securely together.
The number of marks is given in brackets [ ] at the end of each question or part question.
This document consists of 6 printed pages.
2
2012 NYJC 9740/01 [Turn Over
1 It is given that 3 2f ( )x ax bx cx d , where a, b, c and d are constants.
The curve C with equation f ( )y x passes through (0, –1) and has a maximum point at (–1, 1). The
area bounded by C, the x-axis and the lines x = 2 and 3x is 31
4 units
2. Given that f(x) > 0 for
2 3x , find the values of a, b, c and d. [5]
2 The vectors a and b are given by
(sin ) (cos )a i j k and (sin ) (cos )b = i j k , where 0 .
Find an expression for a b in terms of , where 12( ) . [5]
Deduce that the angle between a and b is given by 2sin sin 1 cos . [2]
3 The equation of a curve is given by 3 2 34 3 2x x y y . Find
d
d
y
x in terms of x and y, simplifying
your answer. [2]
The curve meets the line y x at point P. Find
(i) the coordinates of P and [2]
(ii) the equation of the tangent at P. [2]
The tangent to the curve at P cuts the x-axis at R and the normal to the curve at P cuts the y-axis at Q.
If O denotes the origin, use the results above to give a geometrical description of the quadrilateral
OQPR. [1]
3
2012 NYJC 9740/01 [Turn Over
4 (a) Solve the inequality 1 2sin 0x forπ
02
x . Hence, evaluate the exact value of
π
2
0
1 2sin dx x . [4]
(b) Use the substitution x = sin to show that 2
cos 1 d d
2cos 2 1 1 4x
x
.
Given that 0
cos π d
42cos2 1
, find the value of , given that 02
. [4]
5 A curve f ( )y x undergoes in succession, the following transformations:
A: A reflection in the y-axis
B: A translation of 2 units in the direction of the x-axis
C: A translation of a units in the direction of the y-axis, where 1a
(i) The equation of the resulting curve is g( )y x , where1 3
g( )5
ax
x
. Determine the equation
of the curve f ( )y x , in terms of a and x . [4]
(ii) Sketch, on the same diagram, the graphs of g( )y x and 1g ( )y x , indicating clearly the
equations of the asymptotes and the axial intercepts. [4]
6 (a) In a triangle with vertices A, B and C, angle BAC is a right-angle and angle ABC = 3
x .
(i) Show that 1 3 tan
3 tan
AB x
AC x
. [1]
(ii) Hence, show that when x is small enough for x2 and higher powers of x to be neglected,
thenAB
a bxAC
, where a and b are exact constants to be determined. [3]
4
2012 NYJC 9740/01 [Turn Over
(b) A curve is defined by the equation
2 2d1 1
d
yx xy x
x
and (0, 1) is a point on the curve.
(i) Find the Maclaurin’s expansion of y up to and including the term in x2. [3]
(ii) Hence, find the series expansion of e y , up to and including the term in x2. [3]
7 (a) An arithmetic progression has first term a and common difference d, where a and d are non-
zero. The first, third and seventh terms of the arithmetic progression are three consecutive
positive terms of a geometric progression with common ratio r.
(i) Show that 2r . [3]
(ii) The first term of the geometric progression is one-tenth that of the first term of the
arithmetic progression. Find the smallest value of n such that the sum of the first n terms
of the geometric progression exceeds the sum of the first 2n terms of the arithmetic
progression. [3]
(b) Each time that a ball falls vertically on to a horizontal floor, it rebounds to three-fifth of the
height from which it fell. It is initially dropped from a point h m above the floor.
(i) Find the distance travelled by the ball just before it strikes the floor for the third time in
terms of h. [1]
(ii) Show that the total distance travelled by the ball cannot exceed 4h m. [3]
8 (a) Solve the equation iz5 = 32, giving your roots in the form re
i, where r > 0 and < .
Sketch on an Argand diagram the points P1, P2, P3, P4 and P5 representing these roots, where P1
represents the root with the smallest argument and P1 P2 P3 P4 P5 is a polygon described in an
anticlockwise sense. Find the area of P1 P2 P3 P4 P5. [6]
(b) On an Argand diagram sketch clearly the locus of P where P represents the complex number z
such that z satisfies both |z – 2 – 2i| 1 and arg(z – 1) = arg(1 + 3i). Find the range of values
of arg(z – 3 – 2i), given that |z – 2 – 2i| 1 and arg(z – 1) = arg(1 + 3i). [4]
5
2012 NYJC 9740/01 [Turn Over
9 (a) A sequence of negative real numbers 1 2 3, , ,...x x x satisfies the relation
1 1 2n nx x , for n ≥ 1.
Given that the sequence converges to l, find the exact value of l. [2]
(b) (i) By expressing
2
2r r in the form
2
A B
r r
, where A and B are real constants to be
determined, show that 1
1 3 2 3
2 4 2 1 2
n
r
n
r r n n
. [3]
(ii) Prove the result in (i) using mathematical induction. [4]
(iii) Deduce the value of 2
1
2r r r
. [2]
10 The points P and Q have position vectors 3 4i j k and 5 7 6i j k respectively.
(i) The plane passes through Q and is perpendicular to PQ. The equation of is
,r a b c where , and vectors b and c are perpendicular to each other. Write
down a suitable vector a and explain why 3 2i j can be taken as b. Find a suitable vector c.
[5]
The point R has position vector 6 43 8i j k . The line l passing through the points P and R intersect
at point S with position vector s.
(ii) Explain why the lines with equations r a b and r s c , where , , will intersect.
[2]
(iii) Find the position vector of point S and determine whether point S lies on PR produced. [5]
6
2012 NYJC 9740/01 [Turn Over
11 In a city of 5 million people, the number of customers of a new company increases at a rate
proportional to the number of people who are not its customers. The company determines that in
order for its business to be profitable, it must have at least 1 million customers at any point in time.
The company has 1 million customers at the end of the third year of operation, starting from a
customer base of 0. Assume that the population remains unchanged at any point in time.
(i) State a differential equation involving y and t, where y is the number of customers (in millions)
and t is the number of years from the time the company starts its operation. Hence show that
5 1 kty e , where k is a constant to be determined. [5]
(ii) Sketch the graph of y against t. [2]
At the end of the 6th year of operation, the company has a customer base of 1.8 million. However,
due to new competitors, the growth rate of its number of customers is now modelled by the equation
2
2
d0.1
d
y
t .
It is known that under this model, its number of customers would drop to 1.65 million at the end of
the following year of operation.
(iii) Determine in which year the company’s business will become unprofitable. [5]
End of Paper
1
2012 NYJC H2 Math Preliminary exam Paper 1 solutions
1 Let the equation of C be 3 2y ax bx cx d
Subst (0, –1) into equation, 1d
Subst ( –1, 1) into equation, 1
2 (1)
a b c d
a b c
At ( –1, 1), d
0 3 2 0 (2)d
ya b c
x
3
3 2
2
311 d
4ax bx cx x
3
4 3 2
2
31
4 3 2 4
a b cx x x x
81 9 8 319 3 4 2 2
4 2 3 4
a c bb a c
65 19 5 35(3)
4 3 2 4a b c
Solving eqn (1), (2) and (3), 1, 0, 3a b c
3 3 1y x x
2 1 1
sin sin
cos cos
a b
sin cos cos sin
cos cos
sin sin
sin
2sin sin2 2
2cos sin2 2
sin 2
2sin sin2
2cos sin2
2
a b = 2 2 2 2sin 2 4sin (sin cos ) , where 2
2 2sin 2 4sin
2 2 2
2
4sin cos 4sin
2sin 1 cos
Using sina b a b
22sin 1 cos = 2 2 2 21 sin cos 1 sin cos sin
2 2 sin .
2sin sin 1 cos
3
3 2 3
2 2 2
2 2 2
(i) 4 3 2
Differentiating wr.t. :
d d12 6 3 3
d d
d3 3 12 6
d
x x y y
x
y yx xy x y
x x
yx y x xy
x
2
2 2
d 4 2
d
2 2
y x xy
x y x
x x y
y x y x
3 2 3
3
Curve meets when:
4 3 2
2 2
1 and 1
Thus, coordinates of is 1,1
y x
x x x x
x
x y
P
d
(ii) At 1,1 , is undefined. d
y
x
Equation of tangent at : 1
is a square.
P x
OQPR
3
4 (a)
11 2sin 0 sin
2x x
π 0
6x .
π π π
2 6 2
π0 0
6
1 2sin d 1 2sin d 1 2sin dx x x x x x
= π π
6 2
π0
6
1 2sin d 1 2sin dx x x x
= π π
6 2π 0
6
2cos 2cosx x x x
= π2 3 1
6 .
(b)
2
cos cos d d
2cos2 1 2 1 2sin 1
= 2
1cos d
1 4sin
= 2
1 d
1 4x
x using x = sin
sin
20 0
cos 1 d d
2cos2 1 1 4x
x
= sin
0 2
1 d
12
4
x
x
= sin
1
0
1sin 2
2x
= 11sin 2sin
2
0
cos π d
42cos 2 1
11sin 2sin
2 =
π
4
1 πsin 2sin
2
2sin 1
1 π
sin =2 6
.
4
5 1 3
5
ay
x
1 1 3C
5
ay a
x
:
1 1 3B
2 5
ay a
x
:( )
1 1 3A
2 5
1 3 1 3 3
3 3
1
3
ay a
x
a a ax aa
x x
ax
x
:( )
(13a)/5
(13a)/5
y = x
1g ( )y x y = 5
x = 5
y = 0
x = 0
1 3g( )
5
ay x
x
y
x
5
6 (a)(i) tan
3
ACx
AB
tan tan3
1 tan tan3
x
x
1 3 tan
3 tan
AB x
AC x
(ii)1 3
3
AB x
AC x
when x is small
1
1 3 3x x
1
11 3 1
3 3
xx
11 3 1
3 3
xx
11 3 ...
3 3
xx
1 4...
33x
Hence, 1 4
,33
a b
(b)(i) 2 2d1 1
d
yx xy x
x
2
2
2 2
d d d1 2
d d d 1
y y y xx x x y
x x x x
2
2
2 2
d d1 3
d d 1
y y xx x y
x x x
When x = 0, y = 1, d
1d
y
x ,
2
2
d1
d
y
x
2
1 ...2
xy x
(ii)
2 2
12 2e e e ex x
x xy
6
22 21
e 12 2 2
x xx x
2
21e 1
2 2
xx x
e 1 x
7 (a)(i)
2 6
2
a d a dr
a a d
2
2 6a d a a d
24 2d ad
1
2d a ( 0d )
12
22
a ar
a
(ii)
0.1 2 1 2
2 2 12 1 2 2
na n aa n
2 1 20 5 2 1n n n
2 1 20 5 2 1 0n n n
Smallest 11n
(b)(i)
1st hit
2
nd hit 3
rd hit ……………
Distance travelled by the ball when it strikes the floor for the third time
h
3
5h
23
5h
7
23 3
2 25 5
h h h
2.92h
(ii) Total distance travelled S
2 3
2 0.6 2 0.6 2 0.6h h h h
2 32 0.6 0.6 0.6
0.62
1 0.6
h h
h h
4h
(shown)
8 (a)z
5 = 32i = 32
22
i n
e
, where n = 0, 1, 2.
z = 10
2
52i
n
e
, where n = 0, 1, 2.
z = 9 3 7
10 10 10 1022 , 2 , 2 , 2 , 2 .
i i i ii
e e e e e
Area of P1 P2 P3 P4 P5 = 21 25( (2) sin )
2 5
9.51 units2
(b) |z – 2 – 2i| 1 |z – (2 + 2i)| 1
P3
P4
P5
P1 P2
/10
2/5
0
2/5
0
2/5
0 2/5
0
2/5
0
2
2
2
2 2
Re(z)
Im(z)
8
= tan-1
(1/3)
= 2
= ( (/2))/2 0.46365
3arg( 3 2 )
4
or arg( 3 2 ) ( )
3arg( 3 2 )
4
or arg( 3 2 ) 2.68
z i
z i
z i
z i
9 (a) When n , nx l and 1nx l ,
1 2l l
2 1 2l l
2 2 1 0l l
2 4 4 1
2l
2 2 2
2
1 2
Since 0nx , 1 2l
(b)(i)
2
2 2
A B
r r r r
Im(z)
Re(z)
1 2 3
1
2
3
(2,2)
1
(2,3)
locus
of P
(3,2)
tan1
3 or equiv
9
By cover-up rule or otherwise, A = 1, B = –1
Hence 1 1
1 1 1 1
2 2 2
n n
r rr r r r
1 1
1 3
1 1
2 4
1 1
3 51
21 1
2
1 1
1 1
1 1
2
n n
n n
n n
1 1 1 11
2 2 1 2
1 3 2 1
2 2 1 2
3 2 3
4 2 1 2
n n
n n
n n
n
n n
(ii)Let Pn denote the proposition 1
1 3 2 3
2 4 2 1 2
n
r
n
r r n n
for n
When n = 1, LHS =
1
1
1 1 1
2 1 3 3r r r
RHS =
3 2 3 3 5 4 1
4 2 1 1 1 2 4 12 12 3
P1 is true
Assume that Pk is true for some k , i.e. 1
1 3 2 3
2 4 2 1 2
k
r
k
r r k k
To prove Pk+1 is also true i.e.
1
1
1 3 2 5
2 4 2 2 3
k
r
k
r r k k
LHS =
1
1 1
1 1 1
2 2 1 1 2
k k
r rr r r r k k
3 2 3 1
4 2 1 2 1 3
k
k k k k
10
2 3 3 2 23
4 2 1 2 3
k k k
k k k
23 2 9 9 2 4
4 2 1 2 2 3
k k k
k k k
23 2 7 5
4 2 1 2 3
1 2 53
4 2 1 2 3
3 2 5
4 2 2 2 3
k k
k k k
k k
k k k
k
k k
Hence Pk+1 is true
Since P1 is true and Pk is true Pk+1 is true, by mathematical induction, Pn is true for all
n
(iii) 2 1
1 1 1
2 2 1 3r rr r r r
3 1
4 3
5
12
10 (i) Since plane contains point Q, therefore 5 7 6a i j k
2
3
7
PQ OQ OP
3 2
2 3 6 6 0
0 7
Since 3 2i j is perpendicular to the normal of the plane, 3 2i j can be taken as b.
3 2 14
2 3 21
0 7 13
c
(ii) Points Q and S lie on plane and since vectors b and c are parallel to , the 2 lines
are coplanar. Therefore they will intersect.
11
iii) line PR :
3 3
4 39
1 9
r
plane :
2 5 2
3 7 3 10 21 42 73
7 6 7
r
since line PR intersect plane,
3 3 2
4 39 3 73
1 9 7
6 6 12 117 7 63 73
1186 62
3
3 1 4
4 13 17
1 3 2
OS
4 3 11
17 4 133
2 1 3
PS PR
Therefore point S does not lie on PR produced.
11 (i)
d5
d
yk y
t
1
d d5
y k ty
ln 5 y kt c
5
kt cy e
5 kty Ae where e cA
25 ty Ae
When 0t , 0y , hence 5A
5 1 kty e
When 3t , 1y ,
31 5 1 ke
(*)
12
3 4
5
ke
1 4
ln 0.07443 5
k (3s.f.)
0.07445 1 ty e
(*) Alternative working:
When 0t , 0y , hence ln5c
When 3t , 1y ,
1 4
ln 0.07443 5
k (3s.f.)
Hence 0.0744 ln55
ty e
0.0744 ln5
5t
y e
since 5 0y
0.0744 ln5 0.07445 5 5t ty e e
(ii)
(iii) 2
2
d0.1
d
y
t
d
0.1d
yt B
t
20.05y t Bt C
When 0t , 1.8y , hence 1.8C
When 1t , 1.65y , hence 0.1B
20.05 0.1 1.8y t t
Using GC, when 3t , 1.05y
when 4t , 0.6y
6 4 10
The company’s business will become unprofitable in the 10th
year.
(#) Alternative working:
t
y
5y
0
(#)
13
When 6t , 1.8y , hence 1.8 1.8 6B C
6 3.6B C ------(1)
When 7t , 1.65y , hence 1.65 2.45 7B C
7 4.1B C ------(2)
Hence 0.5B , 0.6C
20.05 0.5 0.6y t t
Using GC, when 9t , 1.05y
when 10t , 0.6y
The company’s business will become unprofitable in the 10th
year.
