2011 Set B Mock Paper 1 Solutions
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8/4/2019 2011 Set B Mock Paper 1 Solutions
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2011 JC2 H2 Mathematics Set B, Mock Paper 1
Specimen Paper for Examination from 2007 Page 1 of 15
Set B, Paper 1
1. Use the method of mathematical induction to prove that the sum of the first n terms of the
series
...)753()642()531( +××+××+××
is )5)(4)(1(41 +++ nnnn . [5]
- - - - - -
Let Pn be the statement
...)753()642()531( +××+××+×× + )4)(2( ++ nnn = )5)(4)(1(4
1+++ nnnn , ∀
+∈ Z n .
When n = 1,
L.H.S. = 15)531( =××
R.H.S. = 15)6)(5)(2)(1(41 = = L.H.S.
Therefore, P1 is true.
Assume that Pk is true for some+
∈ Z k .
That is, ...)753()642()531( +××+××+×× + )4)(2( ++ k k k = )5)(4)(1(4
1+++ k k k k .
We want to show Pk +1 is true.
That is,
...)753()642()531( +××+××+×× + )5)(3)(1( +++ k k k = )6)(5)(2)(1(4
1++++ k k k k .
L.H.S. = ...)753()642()531( +××+××+×× + )4)(2( ++ k k k + )5)(3)(1( +++ k k k
= )5)(4)(1(4
1+++ k k k k + )5)(3)(1( +++ k k k
= )5)(1(4
1++ k k )]3(4)4([ +++ k k k
= )5)(1(4
1++ k k ]128[
2++ k k = )6)(2)(5)(1(
4
1++++ k k k k = R.H.S.
That is, Pk is true ⇒ Pk +1 is true.
Since, P1 is true and Pk is true ⇒ Pk +1 is true, by mathematical induction, Pn is true for all+
∈ Z n .
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2011 JC2 H2 Mathematics Set B, Mock Paper 1
Specimen Paper for Examination from 2007 Page 2 of 15
2. A piece of wire of length d units is cut into two pieces. One piece is bent into the form of
a square and the other piece is bent into the form of a circle. Prove that, when the
combined area of the two shapes is a minimum, then the radius of the circle is
approximately 0.07d units. [7]
- - - - - -
Let r = radius of circle and A = combined area of the two shapes.
Hence,
A =
22
4
2
−+
r d r
π π = ( )22
216
1r d r π π −+
)2)(2(8
12
d
dπ π π −−+= r d r
r
A= r
d r
2
2
1
42 π
π π +− =
42
12
2 d r
π π π −
+
Let 0dd =
r A .
⇒ 42
12
2 d r
π π π −
+ = 0
⇒
+÷
= π
2
12
4
d r ≈ 0.07 d
Since =r
A
d
d
42
12
2 d r
π π π −
+ ⇒ =
2
2
d
d
r
A
+
2
2
12 π π > 0.
Hence, r ≈ 0.07 d gives minimum A.
r
Circumference = 2 π r Perimeter = d − 2 π r
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2011 JC2 H2 Mathematics Set B, Mock Paper 1
Specimen Paper for Examination from 2007 Page 3 of 15
3. (i) Show that)32)(12)(12(
1616
32
1
12
2
12
3
++−
+=
+−
+−
− r r r
r
r r r . [2]
(ii) Hence find ∑=
++−
+n
r r r r
r
1 )32)(12)(12(
1,
giving your answer in the form k – f(n), where k is a constant. [4]
(iii) State the sum to infinity of the series in which the r th term is
)32)(12)(12(
1
++−
+
r r r
r . [1]
- - - - -
(i)
)32)(12)(12(
)12)(12()32)(12(2)32)(12(3
32
1
12
2
12
3
++−
+−−+−−++=
+−
+−
− r r r
r r r r r r
r r r
=)32)(12)(12(
)14()344(2)384(3 222
++−
−−−+−++
r r r
r r r r r
= )32)(12)(12(
1616
++−
+
r r r
r
(ii) ∑=
++−
+n
r r r r
r
1 )32)(12)(12(
1
= ∑=
++−
+n
r r r r
r
1 )32)(12)(12(
1616
16
1= ∑
=
+
−+
−−
n
r r r r 1 32
1
12
2
12
3
16
1
= +
−−
5
1
3
2
1
3
16
1
+
−−
7
1
5
2
3
3
+
−−
9
1
7
2
5
3…
+
−−
−−
− 12
1
32
2
52
3
nnn…
+
+−
−−
− 12
1
12
2
32
3
nnn…
+−
+−
− 32
1
12
2
12
3
nnn =
−−
+−
+−+−
32
1
12
2
12
1
3
3
3
2
1
3
16
1
nnn
=
++
+−
32
1
12
3
16
1
24
5
nn.
(iii) As ∞→n , 012
3→
+n, 0
32
1→
+n.
Hence ∑∞
=∞→ ++−
+
1 )32)(12)(12(
1lim
r n r r r
r =
24
5.
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2011 JC2 H2 Mathematics Set B, Mock Paper 1
Specimen Paper for Examination from 2007 Page 4 of 15
4. On the same Argand diagram, sketch the loci
(i) ,5i68 =−− z [2]
(ii) arg( z – 4 – 3i) = 2α , where α =
−
4
3tan
1. [2]
The complex number w is represented by the point of intersection of the loci in parts (i)
and (ii). Find w, in the form x + i y, giving the exact values of x and y. [5]- - - - - -
Note that
1. ∆ BCD is an isosceles triangle with =∠=∠ DBC BCD and α π 2−=∠ BDC
2. α π 2cos)2cos( −=−
3. 4
3tan =α ⇒
5
4cos =α and
5
3sin =α
Applying cosine rule on ∆ BCD : 2)( BC = )2cos()5)(5(255
22α π −−+
2)( BC = 2cos5050 +
2)( BC = )1cos2(5050
2−+ α
2)( BC = 6415
425050
2
=
−
+ ⇒ BC = 8
BC
AB BCA =∠sin ⇒
82sin
AB=α ⇒
25
192
5
4
5
316cossin162sin8 =
=== α α α AB .
CA =
22
25
1928
− =
625
3136=
25
56.
Hence x = 4 +25
56=
25
156and = y
25
1923+ =
25
267. Implies w =
25
156+
25
267i .
5
D(8,6)•
α
2α
C (4,3)
•
• A
B
Re( z)
Im( z)
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2011 JC2 H2 Mathematics Set B, Mock Paper 1
Specimen Paper for Examination from 2007 Page 5 of 15
5. A curve has parametric equations2
1
t x = , t y 2= , where t is a non-zero parameter.
(i) Sketch the curve. [2]
(ii) Find the equation of the tangent to the curve at the point
t
t 2,
12
and show that
the equation of the normal to the curve at the same point can be written in the form
12 625−=− t xt yt . [6]
(iii) The tangent at the point P
4,
4
1meets the y-axis at T and the normal at P meets
the y-axis at N . Prove that the area of the triangle PTN is256
65. [4]
- - - - - -
(i)
(ii) 2
1
t x = ⇒
3
2
d
d
t t
x −= ; t y 2= ⇒ 2
d
d=
t
y.
Hence,3
3
22
d
d
d
d
d
dt
t
t
x
t
y
x
y−=
−×=÷= .
Equation of tangent at the point
t
t 2,
12
is
−−=−
2
3 12
t
xt t y
⇒ 033=+− xt t y
x
y
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2011 JC2 H2 Mathematics Set B, Mock Paper 1
Specimen Paper for Examination from 2007 Page 6 of 15
Equation of normal at the point
t
t 2,
12
is
−=−
23
112
t x
t t y
⇒
−=−
243 12
t xt yt
⇒ 12265−=− xt t yt
⇒ 12625−=− t xt yt
(iii) At the point P
4,
4
1, t = 2.
Hence equation of tangent at P is 68 +−= x y … (1)
And equation of normal at P is . 127432 =− x y . … (2)
Substitute x = 0 into (1) : 6= y .
Hence T is )6,0(
Substitute x = 0 into (2) :32
127= y .
Hence N is )32
127,0(
Area of ∆PTN =256
65
4
1
32
1276
2
1=
− .
x
y
P
T
N
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2011 JC2 H2 Mathematics Set B, Mock Paper 1
Specimen Paper for Examination from 2007 Page 7 of 15
6. (i) Show that ⌡
⌠ +1
dln
n
n
x x = 1ln)1ln()1( −−++ nnnn . [4]
(ii)
The diagram shows the graph of x y ln= between x = n and x = 1+n . The area of
the shaded region represents the error when the value of the integral in part (i) is
approximated by using a single trapezium. Show that the area of the shaded region
is
11
1ln2
1−
+
+
nn . [4]
(iii) Use a series expansion to show that if n is large enough for3
1
nand higher
powers of n
1to be neglected, then the area in part (ii) is approximately equal to
2n
k ,
where k is a constant to be determined. [3]
- - - - - -
(i) xu ln= ⇒ xdx
du 1= ; 1=
dx
dv ⇒ xv =
⌡
⌠ +1
dln
n
n
x x = x x
x x x
n
n
n
n
d1
))((ln
11
⌡
⌠
−
++
=
1
ln)1ln()1(
+
−−++
n
n
xnnnn
= ( )nnnnnn −+−−++ )1(ln)1ln()1(
= 1ln)1ln()1( −−++ nnnn .
x
y
n n + 1O
y = ln x
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2011 JC2 H2 Mathematics Set B, Mock Paper 1
Specimen Paper for Examination from 2007 Page 8 of 15
(ii) Area of the shaded region = ⌡
⌠ +1
dln
n
n
x x − ( ))1ln(ln)1(2
1++ nn
=
−−++ 1ln)1ln()1( nnnn − )1ln(
2
1ln
2
1+− nn
= 1ln)21()1ln()
21( −+−++ nnnn
= 11
ln)2
1( −
++
n
nn
= 11
1ln2
1−
+
+
nn
(iii) Note that
+
n
11ln = ...
1
3
11
2
1132
+
+
−
nnn
= ...3
1
2
1132++−
nnn
Hence, 111ln21 −
+
+
nn = 1...
31
211
21
32−
++−
+
nnnn
= 1...4
1
3
1
2
1
2
11
22−
+−++−
nnnn
≈ 212
1
n(for
3
1
nand higher powers of
n
1to be neglected)
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2011 JC2 H2 Mathematics Set B, Mock Paper 1
Specimen Paper for Examination from 2007 Page 9 of 15
7. (a) The first four terms 1u , 2u , 3u , 4u of an arithmetic progression are such that
1524 =− uu and 13 94 uu = .
Find the value of 1u . [3]
(b) The positive numbers n x satisfy the relation
2
1
1 )5( +=+ nn x x , for n = 1, 2, 3, … .
As ∞→n , n x → .
(i) Find (in either order) the value of to 3 decimal places and the exact value
of . [3]
(ii) Prove that ( ) −=−+ nn x x 221 . [2]
(iii) Hence show that, if n x > , then n x > 1+n x > . [3]
- - - - - - (a) Let a be the first term and d be the common difference of the A.P.
Given 1524 =− uu ⇒ 15)()3( =+−+ d ad a ⇒ 2
15=d
Given 13 94 uu = ⇒ ad a 9)2(4 =+ ⇒ d a 85 = ⇒ 1u = 122
15
5
8=
=a .
(b)(i) Given : When ∞→n , n x → . Then 1+n x → .
Since 21
1 )5( +=+ nn x x , therefore 21
)5( += .
⇒ 52
+= …. (*)
⇒ 052
=−−
⇒ 2
211±=
Since n x > 0, 0> . Therefore 791.22
211=
+= (to 3 d.p.)
(b)(ii) Given 21
1 )5( +=+ nn x x
⇒ ( ) 52
1 +=+ nn x x
⇒ ( ) 2221 5 −+=−+ nn x x
⇒ ( ) )5(5221 +−+=−+ nn x x using (*)
⇒ ( ) −=−+ nn x x22
1
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2011 JC2 H2 Mathematics Set B, Mock Paper 1
Specimen Paper for Examination from 2007 Page 10 of 15
(b)(iii) From (b)(ii) : ( ) −=−+ nn x x 221
⇒ ( )( ) −=+− ++ nnn x x x 11 … (1)
Given n x > ⇒ 0>− n x .
Hence (1) becomes ( )( ) 011 >+− ++ nn x x
⇒ −<+1n x or >+1n x …. (2)
(rejected)
From (1) : ( )( ) −=+− ++ nnn x x x 11
⇒ ( )( )
( )
−+
=−
+
+ nn
n x x
x1
1
1
Note that 791.21 ≈>++ n x .
Hence, ( ) 11
01
<+<+ n x .
⇒ ( ) ( ) −<−+ nn x x 1
⇒ nn x x <+1 …. (3)
Combining results (2) and (3), we have n x > 1+n x > .
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2011 JC2 H2 Mathematics Set B, Mock Paper 1
Specimen Paper for Examination from 2007 Page 11 of 15
8. (a) Find ⌡⌠
+
+ x
x
xd
9
122
. [3]
(b) Using the substitution θ tan4 += x , find ⌡⌠
+−
x x x
d178
12
. [5]
(c) Find the volume of revolution when the region bounded by the part of the curve
x y 2cos= between 0= x and π 21= x and by the x- and y-axes is rotated
completely about the x-axis. Give your answer correct to 3 significant figures. [3]
- - - - -
(a) ⌡⌠
+
+ x
x
xd
9
122
= ⌡⌠
+
+⌡⌠
+
x x
x x
xd
3
1d
9
2222
= ( ) C x
x +++−
3tan
3
19ln
12
(b) θ tan4 += x ⇒ θ
θ
2sec=
d
dx.
⌡⌠
+−
x x x
d178
12
= ⌡
⌠
+−
x x
d1)4(
12
= ⌡
⌠
+
θ θ
θ
dsec1)(tan
1 2
2
= ⌡
⌠ θ d1
= C +θ = C x +−−
)4(tan1
(c) Volume required
= ( ) x x dcos2
0
22⌡
⌠ π
π
= x x dcos2
0
4⌡
⌠ π
π
= 1.85 unit3.
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2011 JC2 H2 Mathematics Set B, Mock Paper 1
Specimen Paper for Examination from 2007 Page 12 of 15
9. Consider the curve)6(
63
+
−=
x x
x y .
(i) State the coordinates of any points of intersection with the axes. [1]
(ii) State the equations of the asymptotes. [3]
(iii) Prove, using an algebraic method, that
)6(
63
+
−
x x
xcannot lie between two certain
values (to be determined). [5]
(iv) Draw a sketch of the curve
(a) )6(
63
+
−=
x x
x y ,
(b) )6(
63
+
−=
x x
x y ,
making clear the main relevant features of each curve. [4]
- - - - - -
(i) )6(
63
+
−
= x x
x y .
When y = 0, 063 =− x ⇒ 2= x
(2,0) is the only point of intersection with the x-axis.
(ii) The asymptotes are 0= x , 6−= x and 0= y .
(iii) )6(
63
+
−=
x x
x y ⇒ 636
2−=+ x yx yx
⇒ 06)36(2=+−+ x y yx … (1)
For any valid values of y substituted into (1), there must correspond some values of x.Hence, equation (1) has solution for x.
Hence, we consider discriminant 0≥ .
That is, 0)6(4)36( 2≥−− y y
⇒ 096036 2≥+− y y
⇒ 0320122
≥+− y y
⇒ 0)16)(32( ≥−− y y
⇒ 6
1≤ y or
2
3≥ y
Hence,)6(
63
+
−
x x
x cannot lie between6
1 and2
3 .
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2011 JC2 H2 Mathematics Set B, Mock Paper 1
Specimen Paper for Examination from 2007 Page 13 of 15
(iv)(a)
(iv)(b)
(6,1/6)−2 1/2
2•
•
x = −6
x
y
(6,1/6)−2 1/2
2
•
•
x = −6
y
x
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2011 JC2 H2 Mathematics Set B, Mock Paper 1
Specimen Paper for Examination from 2007 Page 14 of 15
10. (i) Expand 31
)8( x+ in ascending powers of x up to and including the term in2
x , and
state the values of x for which this expansion is valid. [5]
(ii) The function f( x) = 31
)8(3 x+− is defined for 190 ≤≤ x . Find the inverse
function )(f 1 x− . [3]
(iii) Sketch the graphs of y = f( x) and y = )(f 1 x−
on the same axes, and show the
geometrical relation between the two graphs and the line y = x. [3]
(iv) Show that the x-coordinate of the point of intersection of the graphs of y = f( x) and
y = )(f 1 x−
is given by
31
)8( x+ = x−3 . [2]
(v) Use the first two terms of expansion of 31
)8( x+ to obtain an approximation to the
x-coordinate of the point of intersection. Give your answer as a simple fraction. [1]
- - - - - -
(i) 31
)8( x+ =31
31
818
+
x=
( )( )
+
−+
+ ...
8!283
112
232
31
x x
=
+−+ ...
576
1
24
112
2 x x
= ...288
1
12
12
2+−+ x x
Expansion is valid when 18<
x ⇒ 8< x .
(ii) Let y = f( x) = 31
)8(3 x+−
⇒ y x −=+ 3)8( 31
⇒ ( )338 y x −=+
⇒ ( ) 833−−= y x
⇒ ( ) 83)(f 31−−=
− y y
⇒ ( ) 83)(f 31−−=
− x x
Hence, ( ) 83:f 31−−→
− x x , ]1,0[∈ x
y = f( x)
1
19
R f = [0,1]
x
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2011 JC2 H2 Mathematics Set B, Mock Paper 1
Specimen Paper for Examination from 2007 Page 15 of 15
(iii)
The graphs of y = f( x) and y = )(f 1 x− are reflection of each other about the line
y = x.
(iv) The point of intersection of the graphs of y = f( x) and y = )(f 1 x−
is the same as the point
of intersection between the graphs of y = f( x) and y = x .
Hence, x = 31
)8(3 x+−
⇒ 31
)8( x+ = x−3 … (1)
(v) From (i) 31
)8( x+ = ...288
1
12
12
2+−+ x x . … (2)
Substitute the first two terms of (2) into (1) :
x x −≈+ 312
12
⇒ 13
12≈ x
x
y
y = x
y = f( x)
y = f
−1
( x)
1
1
19
19