Anderson Junior College 2015 Preliminary Examination H2 Mathematics Paper 1 (9740/01)
1
5
x+2 < 2x +1. 2x −1
2x2 + 1 2 + x2 . < 2 − x2 x2
Do not use a graphic calculator in answering this question. 2x
It is given that sin x >
(i)
[6]
Explain why
π
∫
2
e
− sin x
dx <
0
π < arg ( z − 4) 2 < 2π
(iii)
Sketch clearly the locus of z on an Argand diagram.
[3]
(ii)
Find the range of values of | z − 8 | .
[2]
(iii) Find maximum value of arg( z − 8) .
4
[3]
7
−
2
e
2x
π
dx .
[2]
π
Hence show that
e − sin x dx =
∫
π 2
e − sin u du .
[2]
( e − 1)
[3]
0
2
∫
π
e− sin x dx <
π e
.
A curve C1 has the equation p 2 x 2 − y 2 = p 2 where p > 1 . (i)
Sketch C1, stating the coordinates of any points of intersection with the axes, [3]
(ii)
the coordinates of any stationary points and the equations of any asymptotes in terms of p. C1 undergoes a single transformation to become C2. Given that C2 has a line of symmetry x = 2 and the point (4,3) lies on C2, find p.
[2]
(iii)
The graph of y = f ( x) is given below. It has a maximum point at x = −1 and a
−1
The diagram shows the curve C with equation y = 2sin x and the line L with 8π 1 equation y = x − π . C and L intersect at the point where x = . 3 2
.
0
0
(i)
∫
2
π
By making the substitution u = π − x , show that π
The complex number z satisfies
π
for 0 < x <
π
∫ and
[7]
[3]
(ii)
z − 4 − 3i ≤ 2
radians and AC = 3.
2 3 ≈ a + bθ + cθ 2 , 2 + 2 3θ − θ 2 where a, b and c are constants to be determined in exact form.
dy x + 5 It is given that = 2 and when x = 0, y = 5. Find the Maclaurin’s series dx y expansion for y, up to and including the term in x3 , leaving the coefficients in exact form.
3
6
AB ≈
[3]
6 2
π
Given that θ is sufficiently small, show that
Without using a graphic calculator, solve the inequality Hence find the exact solution of the inequality
In the triangle ABC, angle BAC = θ radians, angle ACB =
y
horizontal asymptote y = 0 . y
x C L
x −1
The region S is defined by y ≤ 2sin −1 x , y ≥
8π x −π 3
and x ≥ 0 .
Sketch the graph of y = f ′( x) on the same diagram as C1.
Find the exact volume of the solid obtained when S is rotated through 2π radians
2
about the y-axis.
[6] Page 1 of 4 AJC / 2015 Preliminary Examination / 9740 / P1
Hence, state the number of roots of the equation p 2 x 2 − [ f ′( x)] = p 2 .
Page 2 of 4 AJC / 2015 Preliminary Examination / 9740 / P1
[4]
8 (i) Prove by the method of mathematical induction that
11
n
2 11 2n + 9 ∑ (r + 3)(r + 5) = 30 − ( n + 4 )( n + 5) . r =2 n+4
(ii) Hence find
A
[5]
2
∑ r (r + 2) .
[3]
r =4
F
n+4
(iii) Deduce that
9
1 9 ∑ (r + 1)2 < 40 . r =4
(
)
E 3 2
.
[2]
2
t 2 −1
y = ln t
,
where t > 1.
R is the finite region bounded by the curve C, the x-axis and the lines x = x = 3 . Show that the area of R is given by
∫
t
2 2 3
(t
3
2
− 1) 2
( ln t )
[6]
1 , x ∈ ℝ, x ≤ 0, x ≠ − 2 2 − x2
(i)
Define the inverse function f
(ii)
Sketch the graphs of f and f
−1
in a similar form.
−1
( x) .
12
The line l passes through the points P(0, 0, 1) and Q(0, 6, 2). The plane p1 is perpendicular to the vector 2i − j + k and contains the point Q. (i)
Find the acute angle between line l and plane p1 .
[3]
(ii)
The point R lies on the x-y plane such that its distance from the mid-point M of the line segment PQ is 2 units. If MR is perpendicular to the line l, find the coordinates of R.
[5]
The plane p2 is given by the equation x + 5 y − 10 z − 10 = 0 . (iii) The plane p3 passes through the point (1, 1, 1) and contains all the common
[6]
points of p1 and p2. Find a vector equation of p3, giving your answer in the scalar product form.
Another function g is defined by g : x ֏ 1 − e λ − x , x ∈ ℝ, x ≥ 0 where λ is a constant.
(iii) Given that the composite function f −1g exists, find the greatest value of λ. −1
With this value of λ , find the range of f g .
Page 3 of 4 AJC / 2015 Preliminary Examination / 9740 / P1
[4]
[3]
on the same diagram, giving the exact equation
of any asymptote(s) and showing clearly the relationship between the two graphs. Hence find the set of values of x, in exact form, for which f ( x) ≤ f
[4]
dt .
The function f is defined by
−1
3 Show that the volume V of the box in cm3 is given by V = 2 3 h a − h . 2
(ii) Use differentiation to find, in terms of a, the value of h which would result in the volume of the box being maximum.
1 and 3
By using the results in (i) & (ii), find the exact area of R.
f:x ֏
Fig. 2
It is then folded to form the open hexagonal box of height h cm, as shown in Fig. 2.
(i)
1
D
A piece of vanguard sheet, ABCDEF, is in the form of a regular hexagon of side a cm. A kite shape is cut out from each corner to form the shaded shape, as shown in Fig. 1.
A curve C has parametric equations
10
h cm
Fig. 1
[1]
d −1 1 1 sin =− dx x x x2 − 1
x=
C
[2]
Show that for x > 1, d 1 x (i) =− dx x 2 − 1 x2 − 1
(ii)
B
a cm
[2]
END OF PAPER
[1]
Page 4 of 4 AJC / 2015 Preliminary Examination / 9740 / P1
[4]
Anderson Junior College Preliminary Examination 2015 H2 Mathematics Paper 1 (9740/01)
Qn 1
3(i)
Solution x+2 < 2x + 1 2x − 1 x + 2 − ( 2 x + 1)( 2 x − 1)
2x −1 −4 x 2 + x + 3 <0 2x −1
π < arg( z − 4)2 < 2π π < 2 arg( z − 4) < 2π π < arg( z − 4) < π 2
2
(4,3)
<0 Re(z) (4,0)
( 2 x − 1) ( 4 x 2 − x − 3) > 0
3(ii)
( 2 x − 1)( 4 x + 3)( x − 1) > 0 −
Im(z)
3 1 or x > 1 (ans)
C -3/4
1
1/2
2
5
B
2 x2 + 1 2 + x2 < 2 − x2 x2
A(8,0)
From diagram, AB < z − 8 ≤ AC
1 x2 < 2 + 1 2 2 −1 x 2 x
2+
3 4
1
⇒− < ⇒0<
x 1 x2
2
<
<
12 + 42 < z − 8 ≤ 32 + 42 + 2 17 < z − 8 ≤ 7
1 1 or 2 > 1 2 x
3(iii)
1 1 or 2 > 1 2 x 2
⇒ x 2 > 2 or x2 < 1 ⇒ x > 2 or x < − 2 or −1 < x < 1, x ≠ 0 2
(8,0)
maximum arg( z − 8) 3 2 = π − tan −1 + sin −1 4 5 = 2.9096
dy x + 5 dy = 2 ⇒ y2 = x+5 dx y dx 2
d2 y dy 2 y + y2 2 = 1 dx dx dy 3 dy d 2 y dy d 2 y d3 y 2 + 2 y ⋅ 2 + 2 y ⋅ 2 + y 2 3 = 0 dx dx dx dx dx dx 3
dy d 2 y d3 y dy ⇒ 2 + 6 y ⋅ 2 + y2 3 = 0 dx d x dx dx dy 1 = When x = 0, y = 5 ⇒ dx 5 d2 y d2 y 3 1 10 + 25 2 = 1 ⇒ 2 = dx dx 125 25 3
d y d y 4 1 1 3 2 + 30 + 25 3 = 0 ⇒ 3 = − dx dx 625 5 5 125 1 3 4 ∴ y = 5+ x + x2 − x3 + … 5 125(2!) 625(3!) 1 3 2 2 3 = 5+ x + x − x +… 5 250 1875 3
5
3
= 2.91rad (3sf) 4
C: y = 2sin −1 x 8π L: y = x −π 3 1 π C & L intersect at , 2 3 And y-intercept of L is -π. Volume obtained when S is rotated 2π radians about the y-axis π
=
2 y 2 1 1 π π + π - π∫03 sin dy 3 2 3 2
π 1− cos y π 4π dy = - π∫03 2 12 3
y
x C L
π
6(iii)
π2 π - [ y − sin y ]03 9 2 π π2 π π = - − sin 9 2 3 3 2 π2 π 3 π + = 9 6 2 2
5
=
1 2
π
0
2
2 = 2 ∫ e− sin x dx
from the result in (ii)
0 π 2
−
2x
< 2 ∫ e π dx
from the result in (i)
0 π
2x 2 −π − = 2 e π 2 0 = -π [ e-1 – e0 ] π ( e − 1) = e
B
A
θ
C
3
cos θ +
π
2
dx = ∫ e− sin x dx + ∫ e− sin x dx π
1 3 2 sin 56π cos θ − cos 56π sin θ 1 2
π
− sin x
0
π 3 π2 − = 4 18 By sine rule, AB 3 = sin π6 sin( 56π − θ ) AB =
π
∫e
=
7(i)
sin θ
3 2
≈
2 3 2 (1 − θ 2 ) + 2 3 (θ )
=
2 3 2 + 2 3θ − θ 2
y2 =1 p2 y x → ∞, → ± x, ie. y → ± px p
y C1
x2 −
since θ is small
1 2
p 2 x 2 − y 2 = p 2 where p > 1
(shown)
( -1, 0) y = px
( 1, 0)
x
y = - px
Applying binomial expansion,
( 3 1 − (
AB ≈ 3 1 + ≈
) )+(
3θ − 12 θ 2 3θ − 12 θ 2
−1
7(ii)
)
3θ − 12 θ 2 2
≈ 3 1 − 3θ + 12 θ 2 + 3θ 2 = 3 − 3θ + 6(i)
2x
esin x > e π 1 1 ⇒ 0< < 2x , esin x e π π 2
⇒
π
∫e
− sin x
π
∫e
− sin x
π
p2 = 3
2
p = 3 (rej − 3 ∵ p > 1)
since y = e x is increasing for 0 < x < −
7(iii)
π
2x
0 0
dx = ∫ e
− sin (π − u )
( -1, 0)
( −du )
π
2
y
2
dx < ∫ e π dx
0
6(ii)
3 p2 = 9
2x π
Given that sin x > ⇒
4 p2 − 9 = p2
7 3 a = 3, b = −3, c = 2
7 3 2 θ 2
The transformation is that of a translation of 2 units in the direction of the positive xaxis. The equation of C2: p 2 ( x − 2) 2 − y 2 = p 2 Sub (4,3) into C2: p 2 (4 − 2)2 − 32 = p 2
du = −1 Let u = π − x ⇒ dx
=
x
(1, 0) C1
2
π 2
= ∫e 0
− sin ( u )
du
since sin (π-u) = sin u
No. of roots = no. of intersection points between both graphs = 3
8(i)
n
Let Pn be the proposition:
2
2n + 9
11
∑ (r + 3)(r + 5) = 30 − ( n + 4 )( n + 5) , n ∈ Z + , n ≥ 2 .
8(iii)
r =2
n+4 r =4
Since LHS = RHS, P2 is true.
2
11
2k + 9
r =2
<
9 40
(since
k +1
2
11
2 ( k + 1) + 9
11
2k + 11
9(ii)
r =2
k +1
LHS = ∑ r =2
d −1 1 sin = dx x
2
=
(r + 3)(r + 5)
2 2 + (k + 4)(k + 6) r =2 ( r + 3) ( r + 5)
11 2k + 21k + 54 − 2k − 10 − 30 ( k + 4 )( k + 5 )( k + 6 )
9(iii)
11 2k + 19k + 44 − 30 ( k + 4 )( k + 5 )( k + 6 )
=
( k + 4 )( 2k + 11) 11 − 30 ( k + 4 )( k + 5 )( k + 6 )
11 2k + 11 = − 30 ( k + 5 )( k + 6 ) Since P2 is true, and Pk is true ⇒ Pk +1 is true, by mathematical induction, Pn is true for 8(ii)
all n ∈ Z + , n ≥ 2 . n+4 n +1 2 2 =∑ ∑ r = 4 r ( r + 2) r =1 ( r + 3)( r + 5) n +1 2 2 =∑ + ( 4 )( 6 ) r = 2 ( r + 3)( r + 5)
= =
11 2n + 11 2 − + 30 ( n + 5 )( n + 6 ) 24 9 2n + 11 − 20 ( n + 5)( n + 6 )
1
1
/ 45 6. /
7
*− 7 / 0 = √7 /
1
-1
*− 7 / 0 + √" & = " 9: " > 1,
1 x x2 −1 2 3
3
∫ y dx
Area =
dx
∫ ln t dt dt
=
1 3
2
2
∫
=
−t
3
ln(t )
2
2
=
1 − 2 2 1 x 1− x 1
= −
2
=
> 0 for all n ∈ Ζ + )
5
k
=∑
11 2k + 9 2 = − + 30 k + 4 k + 5 ( )( ) ( k + 4 )( k + 6 ) 11 ( 2k + 9 )( k + 6 ) − 2 ( k + 5 ) = − 30 ( k + 4 )( k + 5 )( k + 6 )
2n + 11
( n + 5 )( n + 6 )
1 1 ) > r ( r + 2 ) ( r + 1) 2
! 1 ! . 2 −" 1 # (= *+" & − 1,-/ 0 = *−&0 +" & − 1,-/ +2", = 2 !" √" & − 1 !" +" & − 1,/
9(i)
Need to show that Pk +1 is also true. i.e.
∑ (r + 3)(r + 5) = 30 − ( k + 1 + 4)( k + 1 + 5) = 30 − ( k + 5)( k + 6 ) .
2
1 9 2n + 11 − 2 20 ( n + 5 )( n + 6 )
Assume Pk is true for some k ∈ Z , k ≥ 2
∑ (r + 3)(r + 5) = 30 − ( k + 4 )( k + 5) .
(Since ( r + 1) = r 2 + 2r + 1 > r 2 + 2r = r ( r + 2 ) ⇒
= +
k
2
1 n+ 4 2 = ∑ 2 r =4 (r + 1)2 1 n+ 4 2 < ∑ 2 r =4 r (r + 2)
2 2 When n = 2, LHS = = , (5)(7) 35 11 2 ( 2 ) + 9 2 − = . RHS = 30 ( 6 )( 7 ) 35
i.e.
1
∑ (r + 1)
(
)
t 2 −1
2
∫ ln t.
=
2 3
3
t
(t
2
)
−1
−1 = ln t t 2 − 1
from (i)
dt 2
3 2 2
∫
2 3
2 − ln 2 − ln 3 = − 3 1 3
+
2
∫t 2 3
1 t 2 −1
2 −1 1 − ln 2 = + 3 ln − sin t 3 3 =−
1
&√@ @
dt
2
(by part (ii)) 2 3
1
√@
√@
AB2 + √3AB2 − √3 lnG√3H − [:JB-1 *&0 − :JB-1 * & 0]
= +√3 − =
?
−1 1 dt t 2 −1 t
2
− 2 3
>
+9: =+",!",
dt
1
,AB2 −
√@
AB2 −
√@ AB3 &
√@ AB3 & L
+M
L
L
M
@
−+ − ,
10(i)
Let y = f ( x ) ⇒ y =
⇒ x2 = 2 −
1 y
⇒ x = − 2− f
Therefore
10(ii)
11(i)
1 2 − x2
1 since x ≤ 0 y
-1
R 120º
F
T C
S
E
D
2
3 2 3 = 3h a − h × 2 3 2 2
x y = f -1 (x)
2
3 = 2 3 h a − h (shown) 2
y =-√2
y = f -1 (x)
Alternative method: find the height RS of triangle PST
( x)
From triangle PSR, x x 1 2 2 tan 30° = ⇒ = RS 3 RS
⇒ f ( x) = x 1 =x 2 − x2 ⇒ x3 − 2 x + 1 = 0 ⇒
⇒ RS =
⇒ ( x − 1) ( x 2 + x − 1) = 0
3 2 a− h 2 3
Volume, V = base area × height
−1 ± 1 + 4 −1 ± 5 = 2 2 f ( x ) ≤ f -1 ( x )
⇒ x = 1, x =
2 3 2 1 = 6 ( x ) RS × h = 3h a − h a − h 2 3 3 2 = 11(ii)
y = g(x) y =1
2
3 3 3 2 3 h a − h =2 3 h a − h 2 3 2 2 2
−1 − 5 ≤x<− 2 2
10(iii) For f -1 g to exist, range of g ⊆ Domain of f-1 1 ⇒ 1 − e λ ,1) ⊆ , ∞ 2 1 λ ⇒ 1− e ≥ 2 1 ⇒ λ ≤ ln 2 1 Greatest value of λ = ln 2 1 f −1 g [0, ∞ ) ⎯⎯→ ,1 ⎯⎯→ ( −1, 0] . 2 Range of f -1 g = = ( −1, 0]
x
2
0.5 0.5
⇒
h
P
2 3 = 3h a − h × 2 3
y=x
-1
30º
B
1 = 6 x 2 sin 60o × h 2
y = f(x)
f ( x ) =f
a
Q
= 6 ( area of △PST ) × h
y
y = f(x)
A
Volume, V = base area × height
1 1 : x → − 2 − , x < 0 or x ≥ x 2
x = -√2
From triangle APQ, a−x 1 a−x tan 30° = 2 ⇒ = h 2h 3 2 ⇒ x=a− h 3
3 V = 2 3 h a − h 2
2
3 3 dV = 2 3 2 a − h (−1)h + a − h dh 2 2 3 3 = 2 3 a − h −2h + a − h 2 2 3 3 = 2 3 a − h a − 3h 2 2
dV = 0. dh 3 3 ⇒h= a or h = a 2 6
For stationary value of V,
2
2
(shown)
3 a, base area of the box is zero (or the volume is zero). Hence this value of 2 h does not give a maximum volume of the box. 3 To check for maximum at h = a, 6 dV When h = ( 63 a) − , 3 a − h > 0 and 3 a − 3h > 0 , and so, > 0. dh 2 2 dV When h = ( 63 a) + , 3 a − h > 0 and 3 a − 3h < 0 , and so, < 0. dh 2 2 3 So V is maximum when h = a. 6 Alternatively, 2nd Derivative Test When h =
3 3 d 2V = 2 3 − a − 3h − 3 a − h = 12 3h − 12a dh 2 2 2
12(ii)
62 + 12 22 + 12 + 12
3 Let M be the point of PQ = 0,3, . 2
R lies on x-y plane ⇒ R = ( a, b,0 ) .
&&&' MR ⊥ line l
2
2 3 ⇒ a 2 + ( b − 3) + − = 2 2 2
2
13 3 ⇒ a2 + − 3 + − = 4 4 2 3 3 4 3 3 13 3 3 13 ∴ R = , , 0 or − , ,0 4 4 4 4 ⇒a=±
10 − 11 5 24 r = + β 21 , β ∈ ℝ 11 11 0
30 11 5 16 [ OR r = + β 21 , β ∈ ℝ ] 11 11 0
10 21 − 11 − 11 1 21 Another vector parallel to p3 = 24 − 1 = 13 = − 1 −13 11 11 11 1 11 0 −1
−1
[Note that ( 0, 6, 2 ) lies on p3 since it lies on both p1 and p2 ⇒ it lies on p3. Accept 5 1
=
−6 + 1 37 6
& & &' Thus M R
%
⇒ θ = 19.6
a = b − 3 3 − 2
a 0 3 13 ⇒ b − 3 • 6 = 0 ⇒ 6b − 18 − = 0 ⇒ b = 2 4 3 1 − 2
Since Length of MR = 2,
Using GC, line of intersection of the two planes is
11
0 0 Line l : r = 0 + λ 6 , λ ∈ ℝ 1 1 Let θ be the angle between the line l and the plane p1.
sin θ =
p2 : x + 5 y − 10 z = 0
5
2 3 a, d V2 = −6a < 0 . 6 dh 3 So V is maximum when h = a. 6
0 2 6 • −1 1 1
2 2 0 p1 : r • −1 = −1 • 6 = −5 ⇒ 2 x − y + z = −4 1 1 2
A vector parallel to p3 is 21 .
When h =
12(i)
12(iii)
or equivalent as direction vector] 21 − 5 11 −34 17 13 = −16 = −2 8 Normal to p3 = 21 × 11 11 46 −23 −1
Equation of p3 : 17 17 1 17 r • 8 = 8 • 1 ⇒ r • 8 = 2 −23 −23 1 −23
