mjc jc 2 h2 maths 2011 mid year exam solutions paper 1
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1
MJC 2011 H2 MATHS JC2 MID YEAR EXAMINATION PAPER 1
Qn Solution Q1 Inequalities (i) Method 1
Since 2 1 1x + ≥ , Therefore ( )2ln 1 0x + ≥
Method 2 From graph ( )2ln 1 0x + ≥
(ii)
From (i) since ( )2ln 1 0x + ≥ therefore need to solve ( )2 3
30 and 0
2x
xx x−
> ≠−
( )
2 3
30
2x
x x−
>−
( )( )2
30, 0, 2
2x
xx x
−⇒ > ≠
−
Therefore 3x > or 2x < , 0x ≠ Alternative solution: 0 or 0 2 or 3x x x< < < >
0 2 − + + 3 +
y
x O
2
Qn Solution Q2 Transformation of Graphs (a)
3 61 e
2xy − −=
( )
3 6
3 2
e
e
x
x
y − −
− +
=
=
1Q−
3e xy −=
1P−
3e xy =
3e xy∴ =
(b)
8
O 2 x
y
( )2 4,−8y =
2x =
0y =
1R−
3
Qn Solution Q3 Maclaurin Series
1tany x−=
( )
( ) ( )
( ) ( ) ( ) ( )
( )
2
2
22
2
3 2 22
3 2 2
3 22
3 2
d 1d 1
d1 1dd d1 2 0 (proven)d dd d d d1 2 2 2 0d d d dd d d1 4 2 0d d d
yx x
yxxy yx x
x xy y y yx x x
x x x xy y yx x
x x x
=+
+ =
+ + =
+ + + + =
+ + + =
When x = 0,
2
2
3
3
0d 1dd 0dd 2d
yyxy
xy
x
=
=
=
= −
1 3
3
2tan ...3!
1 ...3
y x x x
x x
− −= = + +
= − +
Required area 12 1
012 3
01
2 4 2
0
tan d
1 d3
12 3 4
1 1 08 19223 or 0.120 (3s.f.)
192
x x
x x x
x x
−=
≈ −
= −
= − −
=
∫
∫
Working is not required, i.e. can use G.C.
directly.
4
Qn Solution Q4 Curve Sketching (a)
(b)
11b
−
y
x O
11=x
axy =
y x= y
x O
11=x
1 10
1110
−
( )7.84,4.68
( )14,2,17.3
11by ax
x= +
−
5
Qn Solution Q5 Techniques of Integration (By parts) (i)
( )( )
1 2
2
2 2
2 2
2
2
d sin 1d
1 1 2 121 1
1 11
2 1 (shown)
x x xx
x x xx x
x xx
x
− + −
= + ⋅ − + −− −
− + −=
−
= −
(ii)
( )
( )
22
2 2
2
2
1 1111 1
shown1
xx
x xx
x
− − +− − + =
− −
=−
( )
( )
( )
22
2 2
2
2 2
2
2
Alternative method:
1 1
1 11 11 1
11 shown1
xxx x
x
x x
xx
− −= −
− −
− −= − −
− −
= − − +−
(iii)
( )
1
2 1 2
2
2 1 2
2
2 1 1 2 1
2 1 2
4 sin d
12 sin 2 d1
12 sin 2 1 d1
2 sin sin 1 2sin
2 1 sin 1
x x x
x x x xx
x x x xx
x x x x x x c
x x x x c
−
−
−
− − −
−
= − ⋅−
= + − −
−
= + + − − +
= − + − +
∫
∫
∫
6
Qn Solution Q6 Application of Differentiation (exc. Tangent/Normal) (a) 2 tan 3 5x y xy+ + =
Using implicit differentiation, 2 d d2 sec 3 3 0
d dy yx y x yx x
+ + + =
( ) ( )2d 3 sec 2 3dy x y x yx
+ = − +
( )( )2
2 3dd 3 sec
x yyx x y
+= −
+
(b)
5 1using similar triangles, 15 3
13
rh
r h
= =
⇒ =
2
2
3
13
1 13 3127
V r h
h h
h
π
π
π
=
=
=
( )
( )( )
2
2
2
1
d 1d 9d d dd d d
9 4
9 42
9 cms
V hhh h Vt V t
h
π
π
π
π−
=
= ×
= × −
= × −
= −
7
Qn Solution 7 Loci & Integration application (Vol)
( ) ( )2 2
2 i 2
2 1 2
z
x y
− − =
− + − =
4 i 3iz z− + = −
( )
4 0 1 3Midpoint of line-segment ,2 2
2,1
+ − + =
=
1 3Gradient of line-segment 4 01
− −=
−= −
Equation of perpendicular bisector is
( )( )
1 1 2
1 shown
y x
y x
− = −
= −
( ) ( )( ) ( )
( )
2 2
2 2
2
2 1 2
2 2 1
2 2 1
x y
x y
x y
− + − =
− = − −
= ± − −
( ) ( )( )22 2 2
0
3
Volume generated 1 2 2 1 d
23.9 units (3 s.f)
y y yπ= + − − − −
=
∫
Re
Im
O 1 2 4
−1
3
1
2
2
3
8
Qn Solution Q8 AP/GP
( ) ( ) ( )( )
2 11
12
Interest owed to Bank A 0.01 0.01 1.1 0.01 1.1 0.01 1.1
1.1 10.01
1.1 1
0.21384 21.4% (3 s.f.)
= + + + + +
−=
− = ≈
( ) ( ) ( )
[ ]
Interest owed to Bank B 0.01 0.01 0.01 0.01 0.02 0.01 0.1112 0.01 0.01 0.112
0.12 0.66
k k k
k
k
= + + + + + + +
= + +
= +
Alternative method:
( ) ( ) ( )
( ) ( )
Interest owed to Bank B 0.01 0.01 0.01 0.01 0.02 0.01 0.1112 2 0.01 11 0.012
0.12 0.66
k k k
k
k
= + + + + + + +
= +
= +
Since interest owed to both banks is the same,
0.12 0.66 0.214
0.14240.142 (3 s.f.)
kk
+ ===
( )( )
( )
Interest owed under special loan 12 0.01 1000120
120 0.214 10001.78
pp
pp
=
=
=
=
9
Let n be the number of months in the loan tenure with Bank A. Interest owed on special loan 0.015n=
( ) ( )1.1 1Interest owed on regular loan 0.01 0.1 1.1 1
1.1 1
nn
−= = −
−
( )( )
0.1 1.1 1 0.015
0.1 1.1 1 0.015 0
n
n
n
n
− <
− − <
Using GC, ( )( )
8,0.1 1.1 1 0.015 0.00564 0
9,0.1 1.1 1 0.015 0.000795 0
n
n
n n
n n
= − − = − <
= − − = >
Therefore max n = 8.
Qn Solution Q9 Mathematical Induction + Method of Difference
(i)
Let Pn denotes the statement ( ) ( )1
2 4 1 23 ! 3 3 !
n
r
rr n
=
+= −
+ +∑ for all 1n ≥ .
When 1n = ,
LHS ( )
( )( )
1
1
2 1 42 4 13 ! 1 3 ! 4r
rr
=
++= = =
+ +∑
RHS ( )
1 2 13 1 3 ! 4
= − =+
LHS.
Therefore 1P is true. Assume Pk
is true for some value of k where 1k ≥
i.e
( ) ( )1
2 4 1 23 ! 3 3 !
k
r
rr k
=
+= −
+ +∑ Need to prove that 1Pk+
is also true
i.e ( ) ( )
1
1
2 4 1 23 ! 3 4 !
k
r
rr k
+
=
+= −
+ +∑
(1)
10
LHS ( ) ( )
( )( )
1
1 1
2 1 42 4 2 43 ! 3 ! 1 3 !
k k
r r
kr rr r k
+
= =
+ ++ += = +
+ + + +∑ ∑
( ) ( )1 2 2 63 3 ! 4 !
kk k
+= − +
+ + [from (1)]
( ) ( )1 2 2 63 3 ! 4 !
kk k
+= − − + +
( ) ( )
( )2 4 2 61
3 4 !k k
k + − +
= − +
( )1 2 RHS3 4 !k
= − = +
Therefore Pk
is true 1Pk+⇒
is also true.
Hence since 1P is true and 1P is true P is truek k+⇒ , by mathematical induction, Pn is true for
all 1n ≥ .
(ii)
( ) ( ) ( )0
2 6 f 1 f4 !
n
r
r r rr
=
++ + − +
∑
( ) ( ) ( )0 0
2 6 f 1 f4 !
n n
r r
r r rr
= =
+= + + − +∑ ∑
Method 1
( ) ( ) ( )0
2 6 6 8 2 4 2 64 ! 4! 5! 3 ! 4 !
n
r
r n nr n n
=
+ + += + + +
+ + +∑
( ) ( )1
2 4 2 63 ! 4 !
n
r
r nr n
=
+ += +
+ +∑
( )
1
1
2 43 !
n
r
rr
+
=
+=
+∑
( ) ( )
1 2 1 23 1 3 ! 3 4 !n n
= − = −+ + +
Method 2
(2)
(3)
11
( ) ( ) ( )0
2 6 6 8 2 4 2 64 ! 4! 5! 3 ! 4 !
n
r
r n nr n n
=
+ + += + + +
+ + +∑
( ) ( )1
2 4 2 63 ! 4 !
n
r
r nr n
=
+ += +
+ +∑
( ) ( )
1 2 2 63 3 ! 4 !
nn n
+= − +
+ +
( ) ( ) ( ) ( )
0 0
f 1 f f 1 f 0n n
r r
r r= =
+ − = − ∑ ∑
( ) ( )f 2 f 1+ −
( ) ( )f 3 f 2+ −
( ) ( )( ) ( )
f f 1
f 1 f
n n
n n
+
+ − −
+ + −
( ) ( )f 1 f 0n= + −
( ) ( )11 e 0nn − + = + −
Sub (3) and (4) into (2)
( ) ( ) ( ) ( ) ( ) ( )1
0
2 6 1 2f 1 f 1 e4 ! 3 4 !
nn
r
r r r nr n
− +
=
++ + − = − + + + +
∑
or
( ) ( ) ( ) ( ) ( ) ( ) ( )1
0
2 6 1 2 2 6f 1 f 1 e4 ! 3 3 ! 4 !
nn
r
r nr r nr n n
− +
=
+ ++ + − = − + + + + + +
∑
(iii) Method 1
As n →∞ , ( ) ( )11 e 0,nn − ++ → ( )
2 04 !n
→+
( ) ( ) ( )0
2 6 1f 1 f4 ! 3
n
r
r r rr
=
+⇒ + + − → + ∑
(3)
12
The limit of convergence is 13
.
Method 2
As n →∞ , ( ) ( )11 e 0,nn − ++ → ( )
2 0,3 !n
→+
( )2 6 0
4 !n
n+
→+
( ) ( ) ( )0
2 6 1f 1 f4 ! 3
n
r
r r rr
=
+⇒ + + − → + ∑
The limit of convergence is 13
.
Qn Solution Q10 Functions (i) ( )
( )[ )1
2
f ff
Let 1 2
1 2
Since 1, 1 2
Hence , f 1 2.
Since D R ,D 2,
f : 1 2, , 2
y x
x y
x x y
x x
x x x x
−
= − +
∴ = ± −
≥ = + −
= + −
= = ∞
∴ + − ∈ ≥
(ii) ( )1f f x x− = , ( )1ff x x− =
-1 -11 1
ff f ff f is not equal to ff because D [1, ) [2, ) D− − = ∞ ≠ ∞ =
(iii) Method 1:
( )1 3h2
xxa
− −=
( ) 3g ln 2 1 12
3 ln 1 1
xxa
xa
− = − + − = − +
Method 2: ( ) ( )Let g ln 1x Ax B= + +
13
( ) ( )( ) ( )gh ln 2 3 1 ln 2 1 1
By comparison,12 2
33 1 1
x A ax B x
aA Aa
A B Ba
= + + + = − +
= ⇒ =
+ = − ⇒ = − −
( ) 3g ln 1 1xxa
− ∴ = − +
(iv) For fh to exist, h fR D⊂ .
[ )fD 2,= ∞ ,
( )hR 3,a= + ∞ Since h fR D , fh exists.⊂ 1 1a+ +
( )fhR 1 1,a∴ = + + ∞
( )f 1 2x x= + −
1
2 fD
( )fgR 1 1,a= + + ∞
f ( )gR 3,a= + ∞
h h
1D ,2
= ∞
y
x O 3a +
14
Qn Solution
Q11 Differential Equations (a)
d dyd d
u xyu y xx x
=
= +
Sub. d 1dy y yx x
−= − into d dy
d du y xx x= +
d 1d
11
d 1d
u yy x yx x
y y xyxy
u ux
− = + −
= + − −= −
= −
1 d 1 dx
1ln 1
ln 1
1 e e
1 e e1 e where e
1 e1 e
C x
C x
x C
x
x
uu
u x C
u x C
u
uu A A
xy AAyx
− −
− −
− −
−
−
=−
− − = +
− = − −
− =
− = ±
− = = ±
− =
−=
∫ ∫
(b)
22
2
2
2
d edd 2ed
4e
t
t
t
xtx Ct
x Ct D
−
−
−
=
= − +
= + +
when 0, 7t x= =
( )027 4e 0
7 43
C DD
D
−= + +
= +=
15
24e 34, 3
t
x CtB D
−∴ = + += =
2C =
x
t
C = 0
C = 1
C = 1−
3x =
7
O
16
Qn 12 Solution (i) ( )3 12
5 423 51 4
2 13 5 ,1 4
y zx
xyz
− +− = = =
⇒ = −⇒ = +⇒ = − +
− = + ∈ −
r
λ
λλλ
λ λ
(ii) Method 1 2 13 5 ,1 4
− = + ∈ −
r λ λ
1 1
5 1 1 5 4 0 is parallel to 4 1
l− = − + − = ⇒ Π −
⋅
2 13 1 2 3 1 6 A point on lies in 1 1
l = + + = ⇒ Π − −
⋅
Therefore l lies in Π . Method 2
2 13 5 1 2 3 5 1 4 61 4 1
−λ + λ = −λ + + λ + − λ = − + λ −
⋅
Therefore l lies in Π .
(iii) Method 1 Line passing through A and perpendicular toΠ ,
7 1: 6 1 ,
1 1ANl
− = + ∈ − −
r α α
Since N lies on ANl , 7
61
ON→
− + = + − −
ααα
for some α ∈
At point of intersection of ANl and Π ,
17
7 16 1 61 1
7 6 1 63 6
2
− + + = − − − ⇒ − + + + + + =⇒ =⇒ =
αααα α α
αα
7 2 5
6 2 81 2 3
ON→
− + − = + = − − −
(shown)
Method 2
Let 600
OB =
be a point on Π
6 7 130 6 60 1 1
AB−
= − = − −
Π
A ( 7,6, 1)− − ×
N (6,0,0)B ×
18
AN
= Projection of AB→
onto the normal = ( ) AB ⋅
n n
13 1 11 16 1 13 31 1 1
113 6 1 1
31
12 1
1
222
= − − −
− − = −
= −
= −
7 2 5
6 2 81 2 3
ON OA AN→ → →
− − = + + = + = − − −
(shown)
Using ratio theorem,
( )1 '2
ON OA OA= +
' 25 7
2 8 63 1
310
5
OA ON OA= −
− − = − − − − = −
19
(iv) Method 1 to find P using A
The smallest possible area of ANP∆ occurs when P is the foot of perpendicular from A to l .
To find OP→
Let P be the foot of the perpendicular from A to l .
2 13 5 , .1 4
− = + ∈ −
r λ λ
Since P lies on l then23 51 4
OP−λ
= + λ − + λ
for some λ∈
2 7 93 5 6 3 51 4 1 4
AP OP OA−λ − −λ
= − = + λ − = − + λ − + λ − λ
9 1, 3 5 . 5 0
4 49 15 25 16 0
42 2447
AP l− −
⊥ ⇒ − + =
− + − + + ==
=
λλ
λλ λ λ
λ
λ
427 104 13 5 417 7
941 47
OP
−
= + = − +
Π
A
P N
l
20
Method 2 to find P using N The smallest possible area of ANP∆ occurs when P is the foot of perpendicular from N to l .
To find OP→
Let P be the foot of the perpendicular from N to l .
2 13 5 , .1 4
− = + ∈ −
r λ λ
Since P lies on l then23 51 4
OP−λ
= + λ − + λ
for some λ∈
2 5 73 5 8 5 51 4 3 2 4
NP OP ON−λ − −λ
= − = + λ − = − + λ − + λ − + λ
7 1, 5 5 . 5 0
2 4 47 25 25 8 16 0
42 2447
NP l− −
⊥ ⇒ − + = +
− + − + + + ==
=
λλλ
λ λ λλ
λ
Π
A
P N
l
21
427 104 13 5 417 7
941 47
OP
−
= + = − +
Method 3 to find PN using cross-product The smallest possible area of ANP∆ occurs when P is the foot of perpendicular from A to l .
To find PN→
Let P be the foot of the perpendicular from A to l .
Let 231
OC→
= −
be a point on 2 1
: 3 5 , .1 4
l−
= + ∈ −
r λ λ
5 2 7
8 3 53 1 2
CN− − = − = − − −
2 2 2
7 11 5 5
1 5 4 2 4
PN CN= ×
− − = × + + −
d
Π
A
P N
lC
22
301 3042 30
30 342
30 15 14714
= −
=
= =
After finding OP
or PN
,
5 10 31 158 41 17 7
3 9 215 147
5 7 28 6 23 1 2
' 12
PN
PN
AN
AA
− − = − = − −
=
− − = − = − − −
=
2
1Smallest possible area of 21 15 14 122 7
15 42 13.9 units7
ANP PN AN∆ =
=
= =