2012 h2 other jc prelim (selected)
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CJC H2 Mathematics
Keeping Warm During the Holidays
Here are some problems for you to try during the December break. If you do 2 a day from 8 Dec onward,
you will be done in time for Christmas, and Santa will know you’ve been nice, and your present will be a
smooth start to JC2 Maths…
Binomial Expansion
(SAJC/2012 Promo/3)
Expand ( )3
29 2x+ in ascending powers of x , up to and including the term in 3
x . [2]
State the range of values of x for which the expansion is valid. [1]
By using the substitution 2x = , find an approximation for 13 , leaving your answer as a fraction in its
lowest terms. [2]
Ans: 2 3
27 13 54 1458
x x x + + −
;
9 9 9
2 2 2x or x< − < < ;
1265
351
System of Linear Equations
(IJC/2012 Promo/3)
The equation of a curve is given by 3 2y ax bx cx d= + + + , where a, b, c and d are constants. It is known
that the curve has a minimum point at ( )3, 6− . When the curve is translated 2 units in the direction of y-
axis, the curve passes through the points ( )0,5 and 20
1,3
− . Determine the equation of the curve. [5]
Ans: 333
1 23 +−−= xxxy
Equations/Inequalities
(PJC/2012 Promo/2)
Without the use of graphic calculator, solve the inequality 066
1522
2
≥++
−−
xx
xx [3]
Hence solve the inequality
2
2
2 150.
6 6
x x
x x
− −≥
+ + [3]
Ans: or or ;
Recurrence Relations
(IJC/2012 Promo/5)
A sequence is such that 1
3
2u = and
( )( )
1
1
3 2 1
1
n
n n
nu u
n n
−
−
−= +
+ for 2n ≥ .
(i) Write down the values of 2u , 3u and 4u . [1]
(ii) Make a conjecture for nu , where 1n ≥ . [1]
(iii) Prove your conjecture in part (ii) using mathematical induction. [4]
(iv) Hence determine if the sequence is convergent. [1]
3 3x < − − 3 3 3x− ≤ < − + 5x ≥ 5 or 5x x≥ ≤ −
Ans: (i) 2 3u = , 3
27
4u = and
4
81
5u = (ii)
3
1
n
nun
=+
(iv) not convergent
(ACJC/2012 Promo/13)
A sequence of real numbers 1 2 3, , ,u u u K satisfies the recurrence relation
1
( 3)
2 2n
n
n
n uu
u n+
+=
+ +.
As n → ∞ , nu L→ .
(i) Find the exact value(s) of L. [2]
(ii) For the case where 1 3u = , prove by induction that, for 1,n ≥
2
2 1n
nu
n
+=
−. [4]
Write down the limit of this sequence. [1]
Ans: (i) 1
0 or 2
L L= = , (ii) 1
2
APGP
(IJC/2012 Promo/12)
(a) A factory manufactures light bulbs. Using a newly bought machine, the number of light bulbs
manufactured on the first day is 4130. Due to wear and tear, the number of light bulbs manufactured
on each subsequent day is 11 less than that on the previous day. The machine will be deemed
uneconomical when the number of light bulbs manufactured is less than 100. The machine will then
be condemned on that day after use.
(i) How many light bulbs are manufactured on the 85th day? [2]
(ii) Find the total number of light bulbs manufactured by the machine when it is condemned. [4]
(b) A bank offers a cash loan of $10,000. To make the loan attractive, the bank offers the following
repayment plan.
Repay a fixed amount of $x to the bank on the 15th of every month. At the end of each month, the
bank will add an interest at a fixed rate of 5% on the remaining amount owed. When the amount owed
is less than $x, only the balance will have to be paid on the 15th of the following month.
John takes up the loan on 1st October 2012.
(i) How much will he owe the bank on 31st October 2012 after the interest has been added? Leave
your answer in terms of x. [1]
(ii) Show that the total amount of money John owes the bank at the end of n months is given by
( ) ( )$ 10000 1.05 21 1.05 1n nx − − . [3]
(iii) If John repays $500 every month to the bank, find the total number of months for the loan to be
repaid fully. [3]
Ans: (a) (i) 3206 (ii) 777,032 (b) (i) ( )( )$ 10000 1.05x− (iii) 63
Sigma/Method of Difference
(SAJC/2012 Promo/5)
(i) Show that ( )
( )( )2 2 11 2 3
1 2 1 2
r
r r r r r r
++ − =
+ + + + [1]
(ii) By using the method of differences, find ( )( )1
2 1
1 2
n
r
r
r r r=
+
+ +∑ . [3]
(iii) Hence, find the value of ( )( )2
2 1
1 2r
r
r r r
∞
=
+
+ +∑ . [3]
Ans: (ii) 5 1 3
4 2( 1) 2( 2)n n− −
+ + (iii)
4
3
(PJC/2012 Promo/4)
(i) Express ( )( )
2
2 4r r+ + in partial fractions. [2]
(ii) Find 1
1
( 2)( 4)
n
r r r= + +∑ , leaving your answer in the form of ( )fa n− , where a is a constant. [3]
(iii) Find
1
2
1
( 2)( 4)
n
r r r
−
= + +∑ . [2]
Ans: (i)1 1
2 4r r−
+ + (ii)
7 1 1 1
24 2 3 4n n
− + + +
(iii) 9 1 1 1
40 2 2 3n n
− + + +
Transformations
(IJC/2012 Promo/6)
(a) The graph of f ( )y x= has a non-stationary point of inflexion at the origin and the equations of the
asymptotes are 2y = ± and 2x = ± .
On separate diagrams, sketch the graphs of
(i) 2 f ( )y x= , [2]
(ii) f ( )y x= . [2]
y
y = 2
x = 2
x
y = f(x)
x = –2
y = –2
O
(b) The diagram below shows the graph of h( )y x= .
Given that h ( ) 0x′ < for 0, ≠ℜ∈ xx , sketch the graph of h( )y x= . [2]
Ans: a)(i) (ii)
(b)
(PJC/2012 Promo/6)
The diagram shows the graph of . The graph crosses the x-axis at , and has two
turning points at and . The asymptotes of the graph are
( )fy x= 2x = − 2.5x =
(0, 3)− (3, 4)− 1, 2 and 2.x x y= − = = −
y
x
O
y = a
b
On separate diagrams, sketch the graphs of
(i) ( )2 f 1y x= + , [3]
(ii) [3]
Ans: (i) (ii)
Functions
(IJC/2012 Promo/9)
The function f is defined by
21
f : 1 , 1x xx
− >
a .
(i) By using differentiation, show that f(x) increases as x increases. [3]
(ii) Find ( )1f x− and its domain. [4]
The function g is defined by
1g : , 1 0
1x x
x− < <
+a
(iii) Find fg and state the domain and range of fg. [3]
Ans: (ii) 1 1
f ( )1
xx
− =−
, 1fD (0,1)− =
(iii) 2fg : , 1 0x x x− < <a , ( )fgD 1,0= − , fgR (0,1)=
(ACJC/2012 Promo/6)
Functions f and g are defined as follows:
( )
2f : 2 1,
g : ln , 1 0
x x x x a
x x x
− + + <
− − < <
a
a
(i) Write down the largest value of a, such that the inverse function f-1 exists, and define f-1
in a similar form. [4]
(ii) Using the value of a found in (i), show that the composite function 1f g−
exists. Define 1f g−
in
similar form and find its range exactly. [3]
Ans: (i) 1, )2,(,21:f 1 −∞∈−−−xxx a
(ii) )0,1(,)ln(21:gf 1 −∈−−−−xxx a , )21,( −−∞
( )1
fy
x=
Vectors (lines)
(IJC/2012 Promo/10)
Relative to the origin O, the points A, B and C have position vectors 6 2+ −i j k , 3 2 5− + −i j k and
3 2 2− +i j k respectively. The line L passes through the points A and C.
(i) Find a vector equation of L. [2]
(ii) Find the exact length of projection of ABuuur
onto L. [3]
(iii) Hence or otherwise, find the shortest distance from B to L, leaving your answer in exact form.[2]
D is a point such that ABCD is a parallelogram. Use a vector product to find the exact area of
the parallelogram. [3]
Ans: (i)
1 1
6 4
2 2
λ
= + − −
r (ii) 6
21(iii)
115
7, 10 33
(MJC/2012 Promo/6)
Relative to the origin O, points A and B have position vectors 4 3i j+ and 5 7i j+ respectively. The line 1l
passes through point B and is parallel to the vector 2i j kα+ + . The line 2l has Cartesian equation
1 2 , 4.
2 1
x zy
− −= =
(i) Given that 1l and 2l are perpendicular, show that the value of a is 2. [2]
(ii) Find the length of projection of BAuuur
onto line 1l . Hence find the vector BNuuur
, where N is the
foot of perpendicular from A to line 1l . [4]
(iii) Point 'A is the reflection of point A in the line 1l and point C is such that 'AA CB forms a
parallelogram. Find the exact area of the parallelogram 'AA CB . [3]
Ans: (ii) 3 units;
−
−
−
=
2
2
1
BN (iii) 212 2 units
Vectors (planes)
(NYJC/2012 Promo/12)
1Π and 2Π are parallel planes with equation 2
1
2
1
=
⋅r and 8
1
2
1
=
⋅r respectively. The line l passes
through point A(1, 0, 1) and is parallel to the vector +j k .
(i) Verify that A lies on the plane 1Π . [1]
(ii) Find the distance between 1Π and 2Π . [2]
(iii) Show that the acute angle between l and 1Π is 60o. Hence find an inequality that the acute
angle between any line on 1Π and l satisfies. [4]
(iv) Obtain the coordinates of the point of intersection, B, between l and 2Π . [3]
P is a point between 1Π and 2Π such that the ratio of the distance from P to 1Π and the distance
from P to 2Π is 3:1. Find the equation of the plane that P lies on. [2]
If 3Π is a plane that contains line l , state with reason, the number of points of intersection of 1Π ,
2Π and 3Π . [1]
Ans: (ii) 6 (iii) 60 90α≤ ≤o o (iv) B(1, 2, 3),
113
· 22
1
=
r
(SRJC/2012 Promo/12)
With respect to the origin O, the points A and B have position vectors i – 2k and 3i + 4j + 2k respectively.
The line l and the plane 1Π have equations as shown below:
l: r = (i – 2k) + λ(α i + 4j + β k) where λ ∈ ℝ;
1 :Π
r = (i + j + k) + s (2i – 2j + 2k) + t (3i – k) where s, t ∈ ℝ.
The numbers α and β are fixed constants.
(i) Given that the line l is perpendicular to the plane 1Π , show that α = 1 and β = 3. [2]
(ii) Find the point of intersection between l and 1Π and hence, determine the shortest distance
from the point A to the plane 1Π . [4]
(iii) The plane 2Π is perpendicular to 1Π and contains the point A and B. Find the equation of 2Π
in scalar product form. [3]
(iv) A third plane 3Π contains the line l . State, with valid reason(s), the number of points of
intersection of the three planes. [2]
Ans: (ii) (1.5, 2, – 0.5), 262
1 (iii) 6
2
1
2
=
−
⋅r (iv) 1
Techniques of differentiation
(ACJC/2012 Promo/2)
Differentiate the following with respect to x.
(i) )(sincos 1x
− where
2
3
2
ππ<< x [2]
(ii) x
x
−−
+
e1
1eln [3]
Ans: (i) 1; (ii)
−−
+ 1
1
12
1xx
x
ee
e
(RI/2012 Promo/12a)
The equation of a curve is given by ecose2 =+ xxyy
. Find x
y
d
d in terms of x and y. [3]
Ans: 2d e sin
d 2 e cos
y
y
y x y
x xy x
−=
+
Applications of Differentiation
(IJC/2012 Promo/4)
A closed circular cylinder of radius x is inscribed in a right circular cone of radius 2h and vertical height 3h,
where h is a constant. One circular end of the cylinder lies on the base of the cone and the circumference of
the other circular end is in contact with the inner surface of the cone.
(i) Show that 2 13 ( )
2V x h xπ= − , where V is the volume of the cylinder. [2]
(ii) If x is made to vary, find the maximum volume of the cylinder in terms of h. [4]
Ans: (ii) 3
9
16hπ
(IJC/2012 Promo/11)
A curve has parametric equations θ3sin2=x θcos3=y , for πθ ≤≤0 .
(a) Sketch the graph of the curve, showing clearly the points of intersection with the axes if any. [2]
(b) Find the equations of the tangent and normal to the curve at the point P where 6
πθ = . The
tangent and normal at P meet the y-axis at A and B respectively. Hence find the exact area of
triangle ABP. [7]
(c) Given that y
xw = and θ is decreasing at a constant rate of 0.1 rad s-1, find the value of
d
d
w
t
when 6
πθ = . [5]
Ans: (a)
(b) 3
35
3
2+−= xy ,
8
311
2
3+= xy ,
192
37 (c)
18
1−
2h
3h
x
Maclaurin’s Series
(IJC/2012 Promo/7)
Given that ( )1tan ln 1y x− = − , show that
(i) 2d(1 ) 1 0
d
yx y
x− + + = , [2]
(ii) ( ) ( )23 2
3 2
d d d1 2 1 2 0
d d d
y y yx y
x x x
− + − + =
. [3]
Find the Maclaurin series for y, up to and including the term in 3x . [3]
Write down the equation of the tangent to the curve ( )tan ln 1y x= − at the point where x = 0. [1]
Ans: 32
3
2
2
1xxxy −−−= , xy −=
(NJC/2012 Promo/10b)
Given that 11 tan ,y x
−= + show that
( )
22
22 2
d d.
d d 1
y y xy
x x x
+ = − +
By further differentiation, obtain the Maclaurin’s series for y, up to and including the term in 3x . [5]
Deduce the series expansion of 11 tan
,1
x
x
−+
− as far as the term in 2x . [2]
Ans: 2 31 1 5
12 8 48
y x x x≈ + − − ; 1
21 tan 11
1 2
xx x
x
−+= + +
−