CONFIDENTIAL*
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Skema Pemarkahan
Mathematics T Paper 1 (954/1) PEPERIKSAAN PERCUBAAN PENGGAL 1 SIJIL TINGGI PERSEKOLAHAN MALAYSIA 2014
2
CONFIDENTIAL*
1.
The functions f and g are defined by : f : x x + 4, x , a x b , x , x 4
g : x x2 – 4 x, (a) (b)
1.(a)
1.(b)
By finding the values of g (1) and g (3) , explain why the inverse function g -1( x) is not defined [2 marks] Given that the function g f ( x) is defined, find the minimum value of a and the maximum value of b. [3 marks]
g (1) 3 , g (3) 3 g (1) g (3) 3 g is not one-to-one
M1
function g 1 ( x) not defined
A1
Get g (1) and g (3) and
conclude
For function g f ( x) defined
R f D g f ( x) D g x D f f ( x) [4 , 4] 4 f ( x) 4 4 x4 4 8 x 0 But a x b minimum a = – 8 , maximum b = 0
100
2.(a) Evaluate
3
r
M1 A1 A1
3r 1 , give your answer in the standard form A 10n .
[3 marks]
r 1
(b) Express
4 (2 x) 1 3 x
in the form of increasing power of x including the term of x3.
Determine the range of values of x so that this expansion is valid.
[6 marks]
2(a)
M1 A1 A1 (b)
=
1
x 21 1 3x (2 x) 1 3 x 2 x (1)(2) x 2 (1)(2)(3) x 21 (1)( ) ( ) ( ) 3 .... 2 2! 2 3! 2 ( 12 )( 32 ) ( 12 )( 32 )( 52 ) 1 2 3 1 ( )( 3 x ) ( 3 x ) ( 3 x ) ... 2 2! 3! 4
1 2
See
M1
M1
1
k (1 ) (1 3x ) x 2
Either one expanded
1 2
3
CONFIDENTIAL* 1 1 1 21 x x 2 x 3 ... 4 8 2 23 2(1 x x 2 7 x 3 ...) 8 23 2 2 x x 2 14x 3 ... 4 The expansion is valid when x 1 and 3 x 1 2 =
correctly
27 2 135 3 3 1 x x x ... 2 8 16
A1
=
=
x 2 and x
3.
x
1 3
or
Either one series correct
A1 M1
1
3 1
1
3
3
x
A1
Matrix P is invertible if P 0 where P is the determinant of P. 3 4 1 3 4 1 The matrix A 1 0 3 has an inverse A1 because 1 0 3 10 . 2 5 4 2 5 4
(a) (b)
Find A 1 by using the method of elementary row operations. [3 marks] Solve the following system of linear equations by method of matrices that involves A and A1 . 3 x 4 y z 1 2 x 5 y 4 z 3
[4 marks]
x 3 z 2
(c)
State the value of 6 4 1 (i)
3(a)
A I 1 = 3 2 1 = 0 0
2
0
3
4
5
4
(ii)
1
3 = 1 2
4
1
1
0
0
3
0
1
5
4
0
0
0
0
1
0
1 1 40
0
4 5 0 4 5
3
3
0
10 1 10 0
1 1
3 2
0
, 0 1
4
0
[2 marks]
3
0
Idea from A I to
0 1
M1
R1
0
0
1 2 5 4 3
get I A
1
R2
3 R1 R2 R2 2 R1 R3 R3
M1
A1
See two ERO carried out correctly
4
CONFIDENTIAL*
= I A
A1
1
3.(b)
B1
M1
A1
A1 3.(c)
(i) – 20 (ii) 10
B1 B1
4.(a) Given p(1 + 5i) – 2q = 3 + 7i, find the values of p and q if p and q are both real numbers. [3 marks] 4.(b) Express the complex number 3 i in the form r (cos i sin ) , where r is the modulus and is the argument of the complex number. Hence, simplify ( 3 i )5. 4.(a)
p – 2q = 3 and 5 p = 7 7 p 5 4 q 5
4.(b)
,
[5 marks] M1 A1 A1 M1
Either one correct
5
CONFIDENTIAL*
r=2
,
M1
6
i sin 6 6 5 5 32 cos i sin 6 6 3 i 2 cos
Both correct
A1 M1
16 3 16i
See k (cos 56
isin 56 )
A1
Show that the two curves 4 x2 + 9 y2 = 36 and 4 x2 – y2 = 4 have the same foci. For the hyperbola, state the equations of the asymptotes. [6 marks] 2 2 2 2 Sketch the curves 4 x + 9 y = 36 and 4 x – y = 4 on the same axes, showing clearly the asymptotes of the hyperbola. [4 marks]
5.
5
4 x + 9 y = 36
x
2
x
2
y
Centre is (0, 0)
Foci is ( 5 , 0) and ( 5 , 0)
The asymptotes are y = 2 x
1
M1
2
or
9 4 2 c =9-4 =5
=4
Centre is (0, 0)
36
– y
4 x
1 4 2 c =1+4 =5 Foci is ( 5 , 0) and ( 5 , 0)
y
or
2
M1 A1
either one correct both correct
A1
and y = - 2 x
B1 B1
y ●
● (-3, 0)
●
5, 0
(0, 2)
● (-1, 0)
● (1, 0)
5, 0 ●
● (3, 0)
x
●(0, - 2)
6.
D1 D1
Shape of ellipse Vertices and foci shown
D1 D1
Shape of hyperbola Vertices and asymptotes shown
The points A, B and C have position vectors a i j 2k , b 3i 2 j 4k and c i 4 j 4k respectively. Find :
(a) a b
[1 mark]
(b) (b a) (c a)
[3 marks]
6.(a)
a b = 13
B1
6.(b)
b a 2i j 2k ,
B1 B1
c a 2i 3 j 6k
6
CONFIDENTIAL*
(b a) (c a) =
i
j
k
2
1
2
M1
Determinant shown or any two components in answer correct
2 3 6 = 12i 8 j 8k
A1
7.(a) The polynomial Q( x) is defined by Q( x) = x³ + m x² + 5 x – n . If Q( x) has a quadratic factor ( x² + 5) and a zero of – 2 , find the values of m and n. [3 marks] 7.(b) Express cos x 2 sin x in the form R sin( x ) where R is positive and 0
. 2 Hence, state the maximum value of cos x 2 sin x as well as the corresponding value of x in the range 0 x
correct to two decimal places. 2 7.(c) Prove that cot tan 2 cot 2 .
[5 marks] [3 marks]
Hence or otherwise, solve the equation cot tan 12 for 0 360 .
7.(a)
x³ + mx² + 5x – n = (x² + 5)(x + 2) = x³ + 2x² + 5x + 10 By comparison, m = 2 , n = – 10
[4 marks]
M1 M1 A1
7.(b) Let R sin( x + α) cos x + 2 sin x R sin x cos α + R cos x sin α cos x + 2 sin x
R sin
α
= 1 -------- (1) = 2 -------- (2)
M1
Find R or α
A1
R or α correct
5 sin( x + 0.4636) Since the maximum value of sin( x + 0.4636) = 1
A1
CAO
B1
R cos α (1)2 + (2)2,
5 , R > 0
R = (1) ( 2)
R2 = 12 + 22
tan α =
,
α
1 2
= 0.4636
0<α<
2
cos x + 2 sin x
The maximum value of cos x + 2 sin x = and it occurs when x + 0.4636 = 2 x = 1.1072 1.11
7.(c)
cot θ – tan θ = = =
cos
B1
sin
sin cos cos 2 sin 2 sin cos cos 2 1 2
5
M1
Using cot θ and tan θ
M1
Using sin2θ
sin 2
= 2 cot 2θ
A1
7
CONFIDENTIAL*
cot θ – tan θ =
12
2 cot 2θ =
12
M1
2 cot 2θ = 2 3 cot 2θ = tan 2θ =
3 A1
1 3
For 0 360 , 0 2 720 2θ = 30 , 210 , 390 , 570
θ
8.
See equation in one variable term
A1
= 15 , 105 , 195 , 285
A1
5 2 r = 0 1 , 5 0
The line l has equation
.
1 (a) Show that l lies in the plane whose equation is r . 2 5 . 0
[3 marks]
(b) Find the position vector of A, the foot of the perpendicular from the origin O to l. [4 marks] (c) Find an equation of the plane containing O and l . [4 marks] (d) Find the position vector of the point P where l meets the plane whose equation 1
is r . 2 11. 2
[4 marks]
8(a) + λ
=
r
and
r •
= -5
For any point R on l with position vector r r
= M1 •
= -5 – 2λ + 2λ
= -5 r satisfies vector equation of plane, R is a point on the plane ∴ line l lies on the plane.
Get r and try checking if r satisfies equation of plane
A1
A1
(b) =
+ λ 1
•
=0
,
λ 1 ∈ ℝ
M1
Use idea of perpendicular
8
CONFIDENTIAL*
=0
•
M1
10 + 4 λ 1 + λ 1 = 0 λ 1 = -2
=
Scalar product
A1
-2
=
A1
(c) =
×
n
∴
=
M1
=
A1
equation of plane is
or (d)
r •
=
r •
=0
– 5 x
•
Find normal vector
M1
A1
Accept r
•
=0
+ 10 y + 5 z = 0
Since P is on line l , =
+ λ 2
,
λ 2 ∈ ℝ
1 Given that r . 2 11. 2 1 . 2 11. 2 5 + 2 λ 2 + 2 λ 2 + 10 = 11 λ 2 = - 1 = =
M1 M1 A1
-1
A1
Get position vector of point and substitute to equation of plane Scalar product to get λ 2
