Dapatkan Kertas Soalan Ramalan MATHS dan ADD MATHS 2010 di http://maths-catch.com/exam
SPM ZOO ZOOM–I M–IN N Form orm 4: Chap Chapte terr 1 Func Functi tion onss Paper 1
4. (a) gf : x → x2 + 6 x + 2 gf ( x x) = x2 + 6 x + 2 g( x x + 4) = x2 + 6 x + 2
1. The relation in the given given graph can be represented using the following arrow diagram. A
Let x + 4 = u x = u – 4
B
1
10
2
20
3
30
4
40
g(u) = (u – 4)2 + 6(u – 4) + 2 = u2 – 8u + 16 + 6u – 24 + 2 = u2 – 2u – 6 ∴ g( x x) = x2 – 2 x – 6
Based on the above arrow diagram, (a)) th (a thee obje object ct of of 40 is 3, (b) the typ typee of the the relatio relation n is many-to-many relation.
(b)
5. Let g–1( x x) = y g( y y) = x 3 y + k = x x – k y = 3 1 k y = x – 3 3 k 1 –1 x) = x – ∴ g ( x 3 3
–4 –2 2 3
16 9 4
4
(a) The The abov abovee relati relation on is a many-to-one relation. (b) The fu functio nction n which which represent representss the above above relation is f is f ( x) = x2.
It is given that g–1( x x) = mx –
5 6
Hence, Hence, by comparison, comparison, 1 5 k 5 m = and – = – ⇒ k = 3 2 3 6
x) = ff ( x x) 3. f 2 ( x px + q) = f ( px px + q) + q = p ( px = p2 x + pq + q x) = 4 x + 9 It is given that f 2 ( x By comparison, p2 = 4 pq + q p = – 2 –2q + q –q q
]
= f (2) (2) =2+4 =6
2. –3
[
fg(4) = f 42 – 2(4) – 6
=9 =9 =9 = –9
The question requires p < 0.
1
Dapatkan Kertas Soalan Ramalan MATHS dan ADD MATHS 2010 di http://maths-catch.com/exam
Paper 2
x) 2. (a) Let f –1( x f ( y y) y –2 2 y 2 y y –1 ∴ f ( x x) –1 ∴ f (3)
hx 1. (a) f : x → x – 3 hx f ( x x) = x – 3 x) Let f –1( x f ( y y) hy y – 3 hy hy 3 x 3 x
= y = x = x
x + 2) = 2( x = 2 x + 4 = 2 x + 4 = 2(3 2(3)) + 4 = 10
Hence, by comparison, comparison, 2k + 4 = –4 2k = –8 k = –4
[
(c) hf ( x x) : x → 9 x – 3 h[ f f ( x x)] = 9 x – 3 x h – 2 = 9 x – 3 2 x Let –2 =u 2 x =u+ 2 2 x = 2u + 4
]
gf –1( x x) = g f f –1( x x)
x3– x2
=g
1 3 x x – 2 x – 2 = 3 x gf –1( xx) = –5 x x – 2 = –5 x 3 x x – 2 = –15 x2 2 15 x + x – 2 = 0 (3 x – 1)(5 x + 2) = 0 1 2 x = or – 3 5
(
= x + 2
But it is given that f –1g : x → 6 x – 4 f –1g ( x x) = 6 x – 4
kx x) = But it is given that f –1( x , x ≠ 2. x – 2 Hence,, by comparison, Hence comparison, h = 2 and k = 3.
=
= x
x) = f –1[g( x x)] (b) f –1g( x –1 = f (3 x + k ) = 2(3 x + k ) + 4 = 6 x + 2k + 4
= x ( y y – 3) = xy – 3 x = xy – hy = y( x x – h) 3 x y = x – h 3 x –1 ∴ f ( x x) = x – h
(b)
= y = x
)
h(u) = 9(2u + 4) – 3 = 18u + 33 ∴
2
www.nitropdf.com
h : x → 18 x + 33
Dapatkan Kertas Soalan Ramalan MATHS dan ADD MATHS 2010 di http://maths-catch.com/exam
Paper 2
x) 2. (a) Let f –1( x f ( y y) y –2 2 y 2 y y –1 ∴ f ( x x) –1 ∴ f (3)
hx 1. (a) f : x → x – 3 hx f ( x x) = x – 3 x) Let f –1( x f ( y y) hy y – 3 hy hy 3 x 3 x
= y = x = x
x + 2) = 2( x = 2 x + 4 = 2 x + 4 = 2(3 2(3)) + 4 = 10
Hence, by comparison, comparison, 2k + 4 = –4 2k = –8 k = –4
[
(c) hf ( x x) : x → 9 x – 3 h[ f f ( x x)] = 9 x – 3 x h – 2 = 9 x – 3 2 x Let –2 =u 2 x =u+ 2 2 x = 2u + 4
]
gf –1( x x) = g f f –1( x x)
x3– x2
=g
1 3 x x – 2 x – 2 = 3 x gf –1( xx) = –5 x x – 2 = –5 x 3 x x – 2 = –15 x2 2 15 x + x – 2 = 0 (3 x – 1)(5 x + 2) = 0 1 2 x = or – 3 5
(
= x + 2
But it is given that f –1g : x → 6 x – 4 f –1g ( x x) = 6 x – 4
kx x) = But it is given that f –1( x , x ≠ 2. x – 2 Hence,, by comparison, Hence comparison, h = 2 and k = 3.
=
= x
x) = f –1[g( x x)] (b) f –1g( x –1 = f (3 x + k ) = 2(3 x + k ) + 4 = 6 x + 2k + 4
= x ( y y – 3) = xy – 3 x = xy – hy = y( x x – h) 3 x y = x – h 3 x –1 ∴ f ( x x) = x – h
(b)
= y = x
)
h(u) = 9(2u + 4) – 3 = 18u + 33 ∴
2
www.nitropdf.com
h : x → 18 x + 33
Dapatkan Kertas Soalan Ramalan MATHS dan ADD MATHS 2010 di http://maths-catch.com/exam
SPM Zoom-In Form Form 4: Chapte Chapterr 2 Quadra Quadratic tic Equa Equatio tions ns Paper 1
1.
4. x2 + 2 x – 1 + k (2 (2 x + k ) = 0 x2 + 2 x – 1 + 2kx + k 2= 0 x2 + 2 x + 2kx + k 2 – 1= 0 x2 + (2 + 2k ) x x + k 2 – 1 = 0
12 x2 – 5 x(2 x – 1) 1) = 2(3 2(3 x + 2) 2 2 12 x – 10 x + 5 x = 6 x + 4 12 x2 – 10 x2 + 5 x – 6 x – 4 = 0 2 x2 – x – 4 = 0 x = x =
–b
a = 1, b = 2 + 2k , c = k 2 – 1
b2 – 4ac 2a
If a quadratic equation has two real and distinct roots, roots, then then b2 – 4ac > 0.
(–1)2 – 4(2)( 4(2)(– –4) 2(2)
–(–1)
b2 – 4ac > 0 (2 + 2k )2 – 4(1)(k 2 – 1) > 0 4 + 8k + 4k 2 – 4k 2 + 4 > 0 8k + 8 > 0 8k > –8 k > –1
1 33 4 x = 1.6861 or –1.1861 x =
2 3 2 Product of roots = – – = 3 5 5
2. Sum of roots = –
2 3 19 + – =– 3 5 15
5.
3( x x2 + 4) = 2mx 3 x2 + 12 = 2mx 3 x2 – 2mx + 12 = 0 a = 3, b = –2m, c = 12
The required quadratic equation is 19 2 x2 + x + = 0 15 5
If a quadratic equation has equal roots, then b2 – 4ac = 0. b2 – 4ac = 0 (–2m)2 – 4(3)(1 4(3)(12) 2) = 0 2 4m – 144 = 0 4m2 = 144 m2 = 36 m = ±6
x2 – (sum of roots) x x + (product of roots) = 0
15 x2 + 19 x + 6 = 0 3. 3 x2 + 4 p + 2 x = 0 3 x2 + 2 x + 4 p = 0 a = 3, b = 2, c = 4 p
If a quadratic equation does not have real roots, then b2 – 4ac < 0. b2 – 4ac 2 – 4(3)(4 p) 4 – 48 p –48 p 2
<0 <0 <0 < –4
p
>
–4 –48
p
>
1 12
3
www.nitropdf.com
Dapatkan Kertas Soalan Ramalan MATHS dan ADD MATHS 2010 di http://maths-catch.com/exam
6. x2 + 2 x – 8 = 0 a = 1, b = 2, c = –8
x + 2k = 0 7. x2 – (k + 2) x a = 1, b = –(k + 2), c = 2k
The roots are p and q. Sum of roots = – b a
If one of the roots is α, then the other other root is 2 α. Sum of roots = –
p + q = – 2 1 p + q = –2
α
+ 2 α = –
b a
–(k 1+ 2)
3 α = k + 2 k + 2 α = … 1 3
Product of roots = c a 8 pq = – 1
Product of roots =
pq = –8
2 α2 =
The new roots are 2 p and 2q. Sum of new roots = 2 p + 2q = 2( p p + q) = 2(– 2(–2) = –4
2
α
c a
2k 1
= k … 2
Substituting 1 into 2 :
Product of new roots = (2 p)(2q) = 4 pq = 4(– 4(–8) = –32
k + 2 3
2
(k + 2)2 9 (k + 2)2 2 k + 4k + 4 k 2 – 5k + 4 (k – 1)(k – 4) k
The quadratic equation that has the roots 2 p and 2q is x2 + 4 x – 32 = 0.
4
www.nitropdf.com
= k = k = 9k = 9k =0 =0 = 1 or 4
Dapatkan Kertas Soalan Ramalan MATHS dan ADD MATHS 2010 di http://maths-catch.com/exam
Paper 2
From 1 : When m = 2, 6(2) = k – 3 k = 12 + 3 k = 15
1. (2 x – 1)( x + 3) = 2 x – 3 – k 2 x2 + 6 x – x – 3 = 2 x – 3 – k 2 x2 + 3 x + k = 0 a = 2, b = 3, c = k
3. (a) 2 x2 + px + q = 0 a = 2, b = p, c = q
The roots are –2 and p. Sum of roots = –
b a
–2 + p = –
3 2
– p = – p =
The roots are – 3 and 2. 2 Sum of roots = – b a
3 +2 2
p – 3 + 2=– 2 2 1 = – p 2 2 p = –1
1 2
Product of roots =
c a
–2 p =
k 2
Product of roots = c a
12 = 2k
–2
–3 2
k = –2
2. 2 x2 + (3 – k ) x + 8m = 0 a = 2, b = 3 – k , c = 8m
(b)
The roots are m and 2m. Sum of roots = – m + 2m = –
b a
2m2 m2 m2 – 2m m(m – 2) m m ∴m
2 x2 – x – 6 = k 2 x2 – x – 6 – k = 0 a = 2, b = –1, c = –6 – k
When b2 – 4ac < 0, (–1)2 – 4(2)(–6 – k ) < 0 1 + 48 + 8k < 0 8k < –49 49 k < – 8 1 k < –6 8
6m = k – 3 … 1
m(2m) =
2=
If the quadratic equation does not have real roots, then b2 – 4ac < 0.
3 – k 2
Product of roots =
q 2 q = –6
c a
8m 2
= 4m = 2m =0 =0 = 0 or 2 = 0 is not accepted. =2
5
www.nitropdf.com
Dapatkan Kertas Soalan Ramalan MATHS dan ADD MATHS 2010 di http://maths-catch.com/exam
SPM ZOOM–IN Form 4: Chapter 3 Quadratic Functions Paper 1
1. f ( x) = 2 x2 + 8 x + 6 = 2( x2 + 4 x + 3) 4 2 4 = 2 x2 + 4 x + – 2 2 2 2 2 = 2( x + 4 x + 2 – 2 + 3) = 2[( x + 2)2 – 1] = 2( x + 2)2 – 2 ∴ a = 2, p = 2, q = –2
–1
+ 3]
[
2
p
5
Hence, the required range of values of p is p < –1 or p > 5. 5. 3 x2 + hx + 27 = 0 a = 3, b = h, c = 27 If a quadratic equation does not have real roots, b2 – 4ac < 0 h2 – 4(3) (27) < 0 h2 – 324 < 0 (h + 18)(h – 18) < 0
2. From f ( x) = – ( x – 4)2 + h, we can state that the coordinates of the maximum point are (4, h). But it is given that the coordinates of the maximum point are (k , 9). Hence, by comparison, (a) k = 4 (b) h = 9 (c) The equation of the tangent to the curve at its maximum point is y = 9.
h –18
2
3. (a) y = ( x + m) + n The axis of symmetry is x = –m. But it is given that the axis of symmetry is x = 1. ∴ m = –1
Hence, the required range of values of h is –18 < h < 18. 6. g( x) = (2 – 3k ) x2 + (4 – k ) x + 2 a = 2 – 3k , b = 4 – k , c = 2
When m = –1, y = ( x – 1)2 + n Since the y-intercept is 3, the point is (0, 3). 2 ∴ 3 = (0 – 1) + n n =2
If a quadratic curve intersects the x-axis at two distinct points, then b2 – 4ac > 0 (4 – k )2 – 4(2 – 3k ) (2) > 0 16 – 8k + k 2 – 16 + 24k > 0 k 2 + 16k > 0 k (k + 16) > 0
(b) When m = –1 and n = 2, y = ( x – 1)2 + 2 Hence, the minimum point is (1, 2). 4.
18
(2 + p)(6 – p) < 7 12 + 4 p – p2 – 7 < 0 – p2 + 4 p + 5 < 0 p2 – 4 p – 5 > 0 ( p + 1)( p – 5) > 0
–16
0
k
Hence, the required range of values of k is k < –16 or k > 0.
6
www.nitropdf.com
Dapatkan Kertas Soalan Ramalan MATHS dan ADD MATHS 2010 di http://maths-catch.com/exam
Paper 2
(b) g( x) = –2 x2 + 8 x – 12 = –2( x – 2)2 – 4
(a) f ( x) = 2 x2 + 10 x + k k = 2 x2 + 5 x + 2
1 2
25 25 k = 2 x + 5 x + – + 4 4 2 k 5 25 = 2 [ x + – + 2 4 2 ] 5 25 = 2 x + – + k 2 2
2
5
=
25 4
The maximum point is (2, –4). When x = 0, y = –12 ∴ (0, –12) The graph of the function g( x) is as shown below.
2
y
2
O (2, –4)
2
(b) (i) Minimum value = 32 25 – + k = 32 2 89 k = 2 (ii)
–12
3. y = h – 2 x… 1 y2 + xy + 8 = 0 … 2
2
b – 4ac < 0 10 – 4(2)(k ) < 0 100 – 8k < 0 – 8k < –100 –100 k > –8 25 k > 2 2
(c)
Substituting 1 into 2 : (h – 2 x)2 + x(h – 2 x) + 8 = 0 h2 – 4hx + 4 x2 + hx – 2 x2 + 8 = 0 2 x2 – 3hx + h2 + 8 = 0 a = 2, b = –3h, c = h2 + 8
If a straight line does not meet a curve, then b2 – 4ac < 0 (–3h)2 – 4(2) (h2 + 8) < 0 9h2 –8h2 – 64 < 0 h2 – 64 < 0 (h + 8)(h – 8) < 0
1 Minimum point is –2 , 32 . 2
2. (a) g( x) = –2 x2 + px – 12 = –2( x + q)2 – 4 –2 x2 + px – 12 = –2( x2 + 2qx + q2) – 4 = –2 x2 – 4qx – 2q2 – 4 By comparison, p = – 4q … 1 and –12 –2q2 q2 q
x
–8
8
h
2
= –2q – 4 = –8 =4 = ±2
Hence, the required range of values of h is –8 < h < 8.
From 1 : When q = 2, p = –4(2) = –8 (Not accepted) When q = –2, p = –4(–2) = 8 (Accepted) because p > 0 and q < 0)
7
www.nitropdf.com
Dapatkan Kertas Soalan Ramalan MATHS dan ADD MATHS 2010 di http://maths-catch.com/exam
SPM ZOOM–IN Form 4: Chapter 4 Simultaneous Equations Paper 2
From 3 , When x = 0.70156, y = 2 – 4(0.70156) = –0.80624
… 1
1. 2 x – 3 y = 2
… 2
x2 – xy + y2 = 4
When x = –5.70156, y = 2 – 4(–5.70156) = 24.80624
From 1 : 2 + 3 y x = … 3 2
Hence, the solutions are x = 0.70156, y = –0.80624 or x = –5.70156, y = 24.80624 (correct to five decimal places).
Substituting 3 into 2 :
2 + 3 y 2
– y 2 +23 y + y – 4 = 0 2
2
(2 + 3 y)2 y(2 + 3 y) – + y2 – 4 4 2 (2 + 3 y)2 – 2 y(2 + 3 y) + 4 y2 – 16 4 + 12 y + 9 y2 – 4 y – 6 y2 + 4 y2 – 16 7 y2 + 8 y – 12 (7 y – 6)( y + 2)
3. (a) Since (16, m) is a point of intersection of
=0
1 x – 2 and y2 + ky – x – 4 = 0, then 4 x = 16 and y = m satisfy both the equations. y =
=0 =0 =0 =0 6 y = or –2 7
Therefore, 1 (16) – 2 = 2 and m= 4 m2 + km – 16 – 4 = 0 22 + k (2) – 16 – 4 = 0 2k = 16 k = 8
From 3 : When y =
6 , x = 7
When y = –2, x =
2+3
( 67 ) = 16
2
(b) When k = 8, 1 y = x – 2 … 1 4
7
2 + 3(–2) = –2 2
2 6 Hence, the points of intersection are 2 , 7 7 and (–2, –2).
y2 + 8 y – x – 4 = 0 … 2
From 1 : 4 y = x – 8 x = 4 y + 8 … 3
2. 4 x + y = 2 … 1
Substituting 3 into y2 + 8 y – (4 y + 8) – 4 y2 + 8 y – 4 y – 8 – 4 y2 + 4 y – 12 ( y – 2)( y + 6) y
x2 + x – y = 2 … 2
From 1 :
y = 2 – 4 x… 3
Substituting 3 into 2 , x2 + x – (2 – 4 x) = 2 x2 + 5 x – 4 = 0 x =
=
, =0 =0 =0 =0 = 2 or –6 2
–5 ± 52 – 4(1)(–4) 2(1)
From 3 : When y = 2, x = 4(2) + 8 = 16 When y = –6, x = 4(–6) + 8 = –16
–5 ± 41 2
Hence, the other point of intersection, other than (16, 2), is (–16, –6).
= 0.70156 or –5.70156
8
www.nitropdf.com
Dapatkan Kertas Soalan Ramalan MATHS dan ADD MATHS 2010 di http://maths-catch.com/exam
SPM ZOOM–IN Form 4: Chapter 5 Indices and Logarithms Paper 1
5 x lg 5 x x lg 5 x lg 5 x lg 5 – 2 x lg 3 x(lg 5 – 2lg 3)
5.
log10 ( p + 3) = 1 + log10 p log10 ( p + 3) – log10 p = 1
1. 2 x + 3 + 2 x + 16 (2 x – 1) 2 x = 2 x.23 + 2 x + 16 2 = 8(2 x) + 2 x + 8(2 x) = (8 + 1 + 8)( 2 x) = 17(2 x)
2.
3 x + 3 – 3 x + 2 3 x (33) – 3 x (32) 27(3 x) – 9(3 x) (27 – 9)(3 x) 18(3 x)
=6 =6 =6 =6 =6 6 3 x = 18
= 32 x – 1 = lg 32 x –1 = (2 x – 1) lg 3 = 2 x lg 3 – lg 3 = – lg 3 = –lg 3 –lg 3 x = lg 5 – 2 lg 3 x = 1.87
4.
log10
p + 3 = 101 p p + 3 = 10 p 9 p = 3 1 p = 3
1 3 x 3 = 3–1 x = –1 3 x =
3.
m = 3a log3 m = a
n = 3b log3 n = b
6.
mn4 27 = log3 m + log3 n4 – log3 27 = log3 m + 4 log3 n – log3 33 = a + 4 b – 3
log3
p p+ 3 = 1
log2 y – log8 x = 1 log2 x log2 y – =1 log2 8 log2 x 3 3 log2 y – log2 x log2 y3 – log2 x y3 log2 x y3 x log2 y –
=1 =3 =3
=3 = 23
y3 = 8 x y3 x = 8
9
www.nitropdf.com
log2 8 = log2 23 = 3
Dapatkan Kertas Soalan Ramalan MATHS dan ADD MATHS 2010 di http://maths-catch.com/exam
SPM ZOOM–IN Form 4: Chapter 6 Coordinate Geometry Paper 1
Hence, the area of ∆PQR 1 4 0 2 4 2 0 –3 5 0 1 = |–12 – (–6 + 20)| 2 =
1. Let point A be (0, k ). AB = 10 (0 – 8)2 + (k – 7)2 = 10 64 + k 2 – 14k + 49 = 102 k 2 – 14k + 13 = 0 (k – 1)(k – 13) = 0 k = 1 or 13
=
1 26 2 = 13 units2 =
Based on the diagram, k < 7. ∴ k = 1 ∴ A(0, 1)
4. (a) 2 y = 3 x – 12 At point L (on the x-axis), y = 0 2(0) = 3 x – 12 x = 4 ∴ L (4, 0)
2. (a) x + 2 y + 6 = 0 x + 2 y = –6 x 2 y –6 + = (–6) (–6) –6
–3–6 = – 12
At point N (on the y-axis), x = 0. 2 y = 3(0) – 12 y = –6 ∴ N (0, –6)
Intercept form:
x y + =1 (–6) (–3)
(b) m MN = –
x y + =1 a b
m=–
∴ M =
y-intercept x-intercept
(b) m LN =
Therefore, the gradient of the perpendicular line is 2.
4 2+ 0 , 0 +2(–6) = (2, –3) –6 – 0 3 = 0–4 2
∴ Gradient of perpendicular line = –
Hence, the equation of the straight line which passes through the point N and is perpendicular to the straight line MN is y = 2 x – 3. 3.
1 |–26| 2
2 3
Hence, the equation of the perpendicular line is y – y1 = m( x – x1) 2 y – (–3) = – ( x – 2) 3 3( y + 3) = –2( x – 2) 3 y + 9 = –2 x + 4 3 y = –2 x – 5
x y – =1 4 3 At point P (on the x-axis), y = 0. x 0 – = 1 ⇒ x = 4 4 3 ∴ P is point (4, 0).
PA = PB
5.
At point Q (on the y-axis), x = 0. 0 y – = 1 ⇒ y = –3 4 3 ∴ Q is point (0, –3).
1)2
( x – + ( y – 2)2 = ( x – 0)2 + ( y – 3)2 ( x – 1)2 + ( y – 2)2 = ( x – 0)2 + ( y – 3)2 2 x – 2 x + 1 + y2 – 4 y + 4 = x2 + y2 – 6 y + 9 –2 x – 4 y + 5 = –6 y + 9 –2 x + 2 y – 4 = 0 – x + y – 2 = 0 y = x + 2
10
www.nitropdf.com
Dapatkan Kertas Soalan Ramalan MATHS dan ADD MATHS 2010 di http://maths-catch.com/exam
Paper 2
(c) A(–18, 0), B(2, 0), C (0, –6), D(–20, –6) Area of ABCD
1. (a) y – 3 x + 6 = 0 At point B ( x-axis), y = 0. 0 – 3 x + 6 = 0 ⇒ x = 2 ∴ B is point (2, 0).
1 –18 2 0 –20 –18 2 0 0 –6 –6 0 1 = | –12 – (120 + 108)| 2 =
y – 3 x + 6 = 0 At point C ( y-axis), x = 0. y – 3(0) + 6 = 0 ⇒ y = –6 ∴ C is point (0, –6).
1 240 2 = 120 units2 =
2. (a) (i) y – 3 x + 6 = 0 At point P (on the y-axis), x = 0. y – 3(0) + 6 = 0 ⇒ y = – 6 ∴ P is point (0, –6). (ii) The coordinates of point S are 4(0) + 3(7) 4(–6) + 3(15) , = (3, 3) 3 +4 3 +4
y = 3 x – 6 m BC = 3 ∴m AC = –
1 3
Let A(k , 0). ∴
1 m AC = – 3
(b) Area of ∆QRS = 48 units2
0 – (–6) = – 1 k – 0 3 –k = 18 k = –18 ∴ A is point (–18, 0). (b) Let D ( p, q). Midpoint of BD = Midpoint of AC 2 + p 0 + q –18 + 0 0 + (–6) , = , 2 2 2 2
2 2+ p , q2 = (–9, –3)
1 k 7 3 k = 48 2 0 15 3 0 15k + 21 – (45 + 3k ) = 96 12k – 24 = 96 12k = 120 k = 10 (c) S(3, 3), Q(10, 0), T ( x, y) TS : TQ = 2 : 3 2 TS = TQ 3 3TS = 2TQ 9(TS)2 = 4(TQ)2 9[( x – 3)2 + ( y – 3)2] = 4[( x – 10)2 + ( y – 0)2] 9( x2 – 6 x + 9 + y2 – 6 y + 9) = 4( x2 – 20 x + 100 + y2) 9 x2 – 54 x + 81 + 9 y2 – 54 y + 81 = 4 x2 – 80 x + 400 + 4 y2 5 x2 + 26 x + 5 y2 – 54 y – 238 = 0
Equating the x-coordinates, 2 + p = –9 2 p = –20 Equating the y-coordinates, q = –3 2 q = –6 ∴ D is point (–20, –6).
11
www.nitropdf.com
Dapatkan Kertas Soalan Ramalan MATHS dan ADD MATHS 2010 di http://maths-catch.com/exam
SPM ZOOM–IN Form 4: Chapter 7 Statistics Paper 1 2
=
1. After the given score are arranged in ascending order, we have 6
6
k
6
k
=
9
For 7 to be the median, k = 8, as shown below. 6
6
8
8
9
=
n
1
k
6
k + 3
11
13
2
2
1
3
1
1
6 < k + 3 < 11 3 < k < 8 k = 4, 5, 6, 7
7, 7, 7, 11, 13 M
Q3
∴ Q3 = 7
Paper 2
1.
_ ∑ x 2 —— – ( x )2 n 5278 ——— – 27 2 7
=
10
(b) 1, 1, 4, 4, 6,
= 189 27 =7
=
2
Taking into consideration both cases, k = 4 or 5
∑ x n
(b)
318
(a) 1 < k < 6 k = 2, 3, 4, 5
27 = 189 n n
–
10
2
∑ f
Frequency
After two new scores, 7 and 10, are added to the original scores, the mean of the eight scores = 6 + 6 + 6 + 8 + 8 + 9 + 7 + 10 8 = 7.5 – x
10 132
∑ fx
4. Number
Median = 7
2. (a)
∑ f
–
= 1.96
Since the mode is 6, then k ≠ 9.
6
∑ fx 2
= 25 = 5
(a)
Mass (kg)
Frequency
Cumulative frequency
1.1 – 2.0
5
5
2.1 – 3.0
9
14
3.1 – 4.0
12
26
4.1 – 5.0
8
34
5.1 – 6.0
6
40
Frequency 12
3.
x
f
fx
fx2
10
30
3
90
2700
8
32
5
160
5120
34
2
68
2312
Sum
10
318
10 132
6 4 2
0
1.05
2.05
Mode = 3.5 kg 12
www.nitropdf.com
3.05 4.05 5.05 3.5 (Mode)
6.05 Mass (kg)
Dapatkan Kertas Soalan Ramalan MATHS dan ADD MATHS 2010 di http://maths-catch.com/exam
(b) The Q1 class is given by
(a)
Median = 46.5
T 40 = T 10 = 2.1 – 3.0
Q1 = 2.05 +
40 –5 4 (1) = 2.61 kg 9
39.5 +
The Q3 class is given by
4
26 + k – –11 2 (10) = 46.5 k 26 + k – –11 2 (10) = 7 k
T 3 (40) = T 30 = 4.1 – 5.0 Q3 = 4.05 +
L +
4
n – F 2 c = 46.5 f m
3 (40) – 26 4 (1) = 4.55 kg 8
26 + k – 11 = 0.7k 2 26 + k – 22 = 1.4k
Hence, the interquartile range = Q3 – Q1 = 4.55 – 2.61 = 1.94 kg
0.4k = 4 k = 10
(c) New interquartile range = Original interquartile range = 1.94 kg 2.
(b)
Cumulative frequency
Marks
f
20 – 29
4
4
30 – 39
7
11
40 – 49
k
11 + k
50 – 59
8
19 + k
60 – 69
5
24 + k
70 – 79
2
26 + k
Mid point ( x)
fx
fx2
4
24.5
98.0
2401.00
30 – 39
7
34.5
241.5
8331.75
40 – 49
10
44.5
445.0
19802.50
50 – 59
8
54.5
436.0
23762.00
60 – 69
5
64.5
322.5
20801.25
70 – 79
2
74.5
149.0
11100.50
1692
86199
Marks
f
20 – 29
36
Variance =
=
∑ fx 2 ∑ f
–
86 199 36
∑ fx
2
∑ f
–
1692
2
36
= 185.42 (c) (i) New median = Original median + 10 = 46.5 + 10 = 56.5 (ii) New variance = Original variance = 185.42
13
www.nitropdf.com
Dapatkan Kertas Soalan Ramalan MATHS dan ADD MATHS 2010 di http://maths-catch.com/exam
SPM ZOOM–IN Form 4: Chapter 8 Circular Measure Paper 1 1.
B
(
20
π–
1.8
0.9 rad 8 cm
0.9 rad O
8 cm
= 44 cm
s θ
3. Area of the shaded region = Area of sector OAB – Area of sector OXY
= 1.342 rad.
= Area of sector BOC 2
rad r =
= 3.142 – 0.9 – 0.9
=
)
3.142 180
D
∠ BOC = π – ∠ AOB – ∠COD
1
180
15.36
C
OB =
A
3.142
2. ∠ BOC = 20º = 20
1
2
82
1.2 –
2
8 1.342
1 2 r θ
2
= 42.94
= 38.4 – 10 = 28.4 cm2
14
www.nitropdf.com
1
2
54
1
2
rs
Dapatkan Kertas Soalan Ramalan MATHS dan ADD MATHS 2010 di http://maths-catch.com/exam
Paper 2
(c) Perimeter of the shaded region
1.
= 2r sin
C
θ
2
+ r θ
= 2(10) sin 1.982 2 = 16.73 + 19.82 = 36.55 cm
6 cm 8 cm
8 cm
M
+ 10(1.982) r
B
A
(r – 6) cm
3.
r cm
D
O
4 c m
(a) MO = r – 6 In ∆OMB, using Pythagoras’ theorem, MO2 + MB2 = OB2 (r – 6)2 + 82 = r 2 MB =
1 1 AB = 2 2
16
C 4 c m
3 cm
5 cm
1
In ∆ ADB,
3
cos ∠ ABD = 8 10 ∠ ABD = 0.6435 rad
(b) In ∆ BOM , sin ∠ BOM =
8 8
sin ∠ BOM =
1 3
(b) ∠ AOD = 2 ∠ ABD = 2 0.6435 = 1.2870 rad
24
25
= 5 1.2870 = 6.435 cm
= 2 1.287 = 2.574 rad.
(c) Area of the shaded region
2.574
=1 81 2 3 = 70.71 cm2
2
The angle at the centre is twice the angle at circumference.
∴ Length of the arc AD
∠ BOM = 1.287 rad. ∴ ∠ AOB
5 cm
(a) Since ∆ ADB is inscribed in a semicircle, it is a right-angled triangle.
r 2 – 12r + 36 + 64 – r 2 = 0 –12r + 100 = 0 r = 8
B
O
A
= 8 cm
(c) Area of ∆ODB
r
– sin 2.574
= 1 83 2 = 12 cm2
2. (a) ∠ BOA = π – 0.822 = 1.160 rad. 2 ∴ ∠ BOQ = π – 1.160 = 1.982 rad.
Area of sector BOC = 1 52 0.6435 2 = 8.04375 cm2
(b) Area of the shaded region = 1 r 2 (θ – sin θ) 2
Hence, the area of the shaded region = Area of ∆ODB – Area of sector BOC = 12 – 8.04375 = 3.956 cm2
= 1 (10)2 (1.982 – sin 1.982r ) 2 = 53.27 cm2 15
www.nitropdf.com
Dapatkan Kertas Soalan Ramalan MATHS dan ADD MATHS 2010 di http://maths-catch.com/exam
SPM ZOOM–IN Form 4: Chapter 9 Differentiation Paper 1
4. z = xy z = x(30 – x) z = 30 x – x2 dz = 30 – 2 x dx
1 = (5 x – k )–2 (5 x – k )2 f ′ ( x) = –2(5 x – k )–3 (5) –10 = x (5 – k )3
1. f ( x) =
When z has a stationary value, dz =0 dx 30 – 2 x = 0 x = 15
f ′ (1) = 10 –10 = 10 [5(1) – k ]3 (5 – k )3 = –1 5 – k = – 1 k = 6
2.
d 2 z = –2 (negative) dx2
y = ( x + 1) (2 x – 1) dy = ( x + 1) 2 (2 x – 1)1 (2) + (2 x – 1)2 (1) dx = (2 x – 1)[4( x + 1) + (2 x – 1)] = (2 x – 1)(6 x + 3)
[
3.
Hence, the maximum value of z = 30(15) – 152 = 225
2
]
5.
y=
1 = (2 x – 5)–3 (2 x – 5)3
–6 dy = –3 (2 x – 5)–4 (2) = dx (2 x – 5)4
3
y = 2 x – 4 x + 5 dy = 6 x2 – 4 dx
y dy δ ≈ δ x dx δ y ≈
Gradient at the point (–1, 7) = 6 (–1)2 – 4 =2
dy δ y dx
–6 (3.01 – 3) (2 x – 5)4 –6 0.01 = [2(3) – 5]4 = –0.06 =
Equation of the tangent is y – 7 = 2[ x – (–1)] y – 7 = 2( x + 1) y – 7 = 2 x + 2 y = 2 x + 9
6. A = 2 πr 2 + 2 πrh = 2 πr 2 + 2 πr (3r ) = 8 πr 2 dr dA dA = dt dt dr
= 16 πr 0.1 = 16 π (5) 0.1 = 8 π cm2 s–1
16
www.nitropdf.com
Dapatkan Kertas Soalan Ramalan MATHS dan ADD MATHS 2010 di http://maths-catch.com/exam
Paper 2
1.
(b) When p = –3 and k = 4, y = –3 x3 + 4 x dy = –9 x 2 + 4 dx
1 1 – = x–2 – x–3 x 2 x 3 2 3 dy = –2 x – 3 + 3 x – 4 = – 3 + 4 dx x x y=
d 2 y = –18 x dx2
d 2 y = 6 x– 4 – 12 x– 5 = 6 – 12 x4 x5 dx2
At turning points, dy =0 dx –9 x2 + 4 = 0 4 x2 = 9 2 x = ± 3
2 3 6 12 + + – + x – x x x x x4
dy d 2 y + 2 + x 2 y + 5 = 0 dx dx
4
3
4
4
5
x 2
–2 x + 3 + 6 –
x1 – x1 + 5 = 0 2
3
12 1 +1– +5 =0 x x
+ 4 23 = 1 79 d y = – 18 2 = –12 (< 0) 3 dx 23 , 1 79 is a turning point which is
y = –3
2 3
3
2
2
∴
3
2. (a) y = px + kx dy = 3 px2 + k dx At (1, 1), x = 1 and m =
2 , 3
When x =
13 –2 x + 15 – =0 x –2 x2 + 15 x – 13 = 0 2 x2 – 15 x + 13 = 0 (2 x – 13)( x – 1) = 0 13 or 1 x = 2
a maximum. When x = – dy = –5. dx
2 , 3
+ 4– 23 = –1 79 d y = –18 – 2 = 12 (> 0) 3 dx – 23 , –1 79 is a turning point which is
y = –3 –
2
∴ 3 px + k = –5
2 3
3
2
3 p(1)2 + k = –5 3 p + k = –5 … 1
2
∴
The curve passes through point (1, 1). ∴ 1 = p(1)3 + k (1) p + k = 1 … 2 1
a minimum.
– 2 : 2 p = –6 ⇒ p = – 3
From 2 : –3 + k = 1 ⇒ k = 4
17
www.nitropdf.com
Dapatkan Kertas Soalan Ramalan MATHS dan ADD MATHS 2010 di http://maths-catch.com/exam
3.
3888 = 96 x2 + 3888 x–1 x 3888 dL = 192 x – 3888 x–2 = 192 x – dx x2 L = 96 x2 +
(b)
V
4 x m 5 x m H
E
F
G
3 x m y m
D
At stationary point, dL =0 dx 3888 192 x – =0 x 2 3888 192 x = x 2 3888 x 3 = 192
C
6 x m A
(a)
6 x m
B
Volume of the cuboid = 5832 cm3 (6 x)(6 x)( y) = 5832 36 x2 y = 5832 x2 y = 162 162 y = 2 x
x 3 = 20.25 x = 2.73
L = Area of ABCD + 4 (Area of GBCH ) + 4 (Area of VGH ) 1 L = (6 x)2 + 4(6 xy) + 4 (6 x)(5 x) 2 2 2 L = 36 x + 24 xy + 60 x L = 96 x2 + 24 xy 162 L = 96 x2 + 24 x x2 3888 (shown) L = 96 x2 +
d 2 L = 192 + 7776 x–3 = 192 + 7776 (> 0) x3 dx 2 ∴ L is a minimum.
4.
y =
h = h(1 + 2 x)–2 (1 + 2 x)2
4h dy = –2h(1 + 2 x)–3 (2) = – dx (1 + 2 x)3 dy δ x δ y = dx
x
8c 4h c =– 3 (1 + 2 x)3 8c 4h c – =– 3 [1 + 2(1)]3 8c 4hc – =– 3 27 8 27 h = 3 4 h = 18 –
18
www.nitropdf.com
Dapatkan Kertas Soalan Ramalan MATHS dan ADD MATHS 2010 di http://maths-catch.com/exam
SPM ZOOM–IN Form 4: Chapter 10 Solution of Triangles 2. (a) In ∆PQS, using the sine rule, sin ∠QSP sin 35º = 8 7 sin 35º 8 sin ∠QSP = 7 sin ∠QSP = 0.65552 ∠QSP = 40.96º
Paper 2
1. (a) ∠UST = 180º – 65º = 115º ∠SUT = 180º – 43º – 115º = 22º In ∆UST , using the sine rule, US 9 = sin 43º sin 22º 9 sin 43º US = sin 22º = 16.385 cm
∴ ∠PQS = 180º – 35º – 40.96º
= 104.04º Hence, the area of ∆PQS 1 8 7 sin 104.04º = 2
U
22°
= 27.16 cm2 (b) This problem involves the ambiguous case of sine rule. The sketch of ∆QRS1 is as shown below.
1 6 . 3 8 5 c m
Q
115° 65° R
43° S 9 cm
7 cm
T
10 cm
(b) In ∆USR, using the cosine rule, UR2 = 72 + 16.3852 – 2(7)(16.385)cos 65º UR2 = 220.5238 UR = 14.85 cm
1 2
7 cm
43° R
S1
S
In ∆QRS, using the sine rule, sin ∠QSR sin 43º = 10 7 sin 43º 10 sin ∠QSR = 7
Area of ∆ RSV = 41.36 cm2
(c)
7 cm
7 12 sin ∠ RSV = 41.36 sin ∠ RSV = 0.98476 Basic ∠ = 79.98º ∠ RSV = 180º– 79.98º = 100.02º
sin ∠QSR = 0.974283 Basic ∠ = 76.98º
R
7 cm
∴ ∠QSR = 76.98º or ∠QS1 R = 103.02º
S
100.02°
In ∆QS1 R, ∠ RQS1 = 180º – 43º – 103.02º = 33.98º
12 cm
In ∆QS1 R, using the sine rule, 10 RS1 = sin ∠ RQS1 sin ∠ RS1Q RS1 10 = sin 33.98º sin 103.02º 10 sin 33.98º RS1 = sin 103.02º = 5.737 cm
V
In ∆ RSV , using the cosine rule, RV 2 = 72 + 122 – 2(7)(12)cos 100.02º RV 2 = 222.23064 RV = 14.91 cm 19
www.nitropdf.com
Dapatkan Kertas Soalan Ramalan MATHS dan ADD MATHS 2010 di http://maths-catch.com/exam
SPM ZOOM–IN Form 4: Chapter 11 Index Numbers Paper 2
(a) Supplement A x 400
1. (a)
100 = 120 P2004 I = P 2002
x = 480 y =
I 2004 (based on 2002) = 115 P2004 100 = 115 P2002 69 100 = 115 P2002
525 500
100 = 105
660 z
100 = 110
100
P2002 = 69 100 115 P2002 = RM60.00
z = 600
(b) Supplement B I 2006 (based on 2002) P = 2006 100 P2002 P P = 2006 2004 100 P2004 P2002 = 130 120 100 100 100 = 156
– I = 115
(b)
(120 20) + 130m + (105 80) + (110 40) = 115 20 + m + 80 + 40 15 200 + 130m = 115 140 + m 15 200 + 130m = 16 100 + 115m 15m = 900 m = 60
(c)
– (c) I 2006 (based on 2002) 100 + 25 – I = 2004 100 125 115 = 100
(115 3) + (120 2) + 105 x 3 + 2 + x 585 + 105 x 5 + x 585 + 105 x 30 x – (d) I 2006 (based on 2004)
= 143.75 (d) Total yearly cost in 2006 143.75 5 500 000 = 100 = RM7 906 250
I (based on 2002)
2004
I 2006 (based Weightage on 2004)
A
115
150
3
B
120
130
2
C
105
120
x
= 111 = 111 = 555 + 111 x = 6 x =5
= (150 3) + (130 2) + (120 5) 3+2+5 = 1310 10 = 131 P Thus, 2006 100 = 131 P2004 P2006 100 = 131 300
2. Health supplement
– I 2004 (based on 2002) = 111
P2006 = 131 300 100 P2006 = RM393
20
www.nitropdf.com
Dapatkan Kertas Soalan Ramalan MATHS dan ADD MATHS 2010 di http://maths-catch.com/exam
SPM ZOOM–IN Form 5: Chapter 1 Progressions Paper 1
4.
T 3 – T 2
=3 ar – ar = 3 4r 2 – 4r = 3 4r 2 – 4r – 3 = 0 (2r + 1)(2r – 3) = 0 2
1. (a)
T 6
= 38 a + 5d = 38 a + 5(7) = 38 a =3
r = –
(b) S9 – S3 9 3 = [2(3) + 8(7)] – [2(3) + 2(7)] 2 2 = 279 – 30 = 249 2. (a)
1 3 or 2 2
5. 0.242424 … = 0.24 + 0.0024 + 0.000024 + … 0.24 a = S = 1 – 0.01 1 – r 0.24 = 0.99 24 = 99 8 = 33 ∞
T 2 – T 1 = T 3 – T 2 2h – 1 – (h – 2) = 4h – 7 – (2h – 1) h + 1 = 2h – 6 h =7
(b) When h = 7, the arithmetic progression is 5, 13, 21, … with a = 5 and d = 8.
6. The numbers of bacteria form a geometric progression 3, 6, 12, …
S8 – S3
The number of bacteria after 50 seconds = T 11 = ar 10 = 3(210) = 3072
8 3 [2(5) + 7(8)] – [2(5) + 2(8)] 2 2 = 264 – 39 = 225 =
T 2
3.
T 1 x +
=
2 = 9 x + 4
Paper 2
1. (a) The volumes of cylinders are 2 2 2 πr h, πr (h + 1), πr (h + 2), …
T 3 T 2 x – x +
4 2
T 2 – T 1
= πr 2 (h + 1) – πr 2h = πr 2h + πr 2 – πr 2h = πr 2
T 3 – T 2
= πr 2 (h + 2) – πr 2 (h + 1) = πr 2h + 2 πr 2 – πr 2h – πr 2 = πr 2
( x = ( x – 4)(9 x + 4) 2 2 x + 4 x + 4 = 9 x – 32 x – 16 8 x2 – 36 x – 20 = 0 2 x2 – 9 x – 5 = 0 (2 x + 1)( x – 5) = 0 1 x = – or 5 2
Since T 2 – T 1 = T 3 – T 2 = πr 2, the volumes of cylinders form an arithmetic progression with a common difference of π r2.
21
www.nitropdf.com
Dapatkan Kertas Soalan Ramalan MATHS dan ADD MATHS 2010 di http://maths-catch.com/exam
(b)
2
2. (a)
2
a = πr h, d = πr
= 32 π a + 3d = 32 π 2 2 πr h + 3 πr = 32 π 2 2 r h + 3r = 32 2 r (h + 3) = 32 … 1 T 4
S4
1
:
2
r
(2h + 3) 3) 2h + 3 h+ 3 16h + 24 3h
2
r (h +
= 45 ar – ar = 45 ar (r – 1) = 45 … 2 2
= 104 π
h
=
= 150 T 1 + T 2 = 150 a + ar = 150 a (1 + r ) = 150 … 1 T 3 – T 2
4 (2a + 3d ) = 104 π 2 4a + 6d = 104 π 2 4 πr h + 6 πr 2 = 104 π 2r 2h + 3r 2 = 52 2 r (2h + 3) = 52 … 2 2
S2
1
:
2
a (1 + r ) ar (r – 1)
1 + r r (r – 1)
=
150 45
=
10 3
3 + 3r = 10r 2 – 10r 10r 2 – 13r – 3 = 0 (2r – 3)(5r + 1) = 0
52 32
13 8 = 13h + 39 = 15 =5
r =
=
3 2
or –
1 5
(b) For the sum to infinity to exist,– 1 < r < 1. 3 Thus, r = is not accepted. 2 1 Therefore, r = – 5
From 1 : 2 r (5 + 3) = 32 2 r = 4 r = 2
From 1 : a
1 – 15 = 150 a
= 187
1 2
1 a 2 = = 1 – r 1 – – 1 5 187
∴
22
www.nitropdf.com
S∞
= 156
1 4
Dapatkan Kertas Soalan Ramalan MATHS dan ADD MATHS 2010 di http://maths-catch.com/exam
SPM ZOOM–IN Form 5: Chapter 2 Linear Law Paper 1 1.
Paper 2
2 + qx x 2 = 2 +q x 1 =2( 2)+q x
y = y x y x
1. (a)
y = hxk
log10 y = log10 (hxk ) 1
log10 y 2 = log10 h + log10 xk 1 log y = log h + k log x 10 10 10 2 y x
1 x2
log10 y = 2log10 h + 2k log10 x log10 y = 2 k log10 x + 2 log10 h
5 = 2(1) + q q=3
(1, 5)
(b) y x
1 x2 (3, p)
2.
q=3 p = 2(3) + 3 p = 9
y = p k x y lg x = lg p k
lg y – lg k
x
x
1.5
2.0
2.5
3.0
3.5
y
142
338
660
1348
1995
log10 x
0.18
0.30
0.40
0.48
0.54
log10 y
2.15
2.53
2.82
3.13
3.30
The graph of log10 y against log10 x is as shown below. log10 y
= lg p
Graph of log10 y against log10 x
3.5
lg y – x lg k = lg p 3.0
lg y = x lg k + lg p
2.0
2 b 3. y – ax = 2 x x 2 3 xy – ax = b xy2 = ax3 + b
(–1, 10):
10 = a(–1) + b … 1
(5, –2):
–2 = a(5) + b … 2
1 – 2 :
From 2 :
0.55 – 0.06 = 0.49
1.55 1.5 1.0 0.5
O
x3
3.35 – 1.75 = 1.6
2.5
∴ Y = lg y , X = x , m = lg k, c = lg p
xy2
12 = –6a a = –2 –2 = –2(5) + b b =8
23
www.nitropdf.com
0.1
0.2
0.3
0.4
0.5
0.6
log10 x
Dapatkan Kertas Soalan Ramalan MATHS dan ADD MATHS 2010 di http://maths-catch.com/exam
(c)
2k = Gradient 3.35 – 1.75 2k = 0.55 – 0.06 2k = 3.2653 k = 1.63 2 log10 h 2 log10 h log10 h h
(b) (i) 1 = y
1 x + p = y2 q p 1 1 x + 2 = y q q
= Y –intercept = 1.55 = 0.775 = 5.96
Gradient =
x
0.1
0.3
0.4
0.5
0.7
0.8
y
0.78
0.60
0.54
0.50
0.44
0.42
1 y2
1.64
2.78
3.43
4.00
5.17
5.67
Graph of
5.5
Y - intercept p q p 0.17 p
1 against x y2
4.6 – 2 0.6 – 0.16
= 1.1 = 1.1 = 1.1 = 0.19
(ii) When x = 0.6, from the graph, 1 = 4.6 y2 y2 = 0.2174 y = 0.47
5.0 4.6 4.5 4.0 3.5 4.6 – 2 = 2.6
3.0 2.5 2.0
0.6 – 0.16 = 0.44
1.5 1.1 1.0 0.5
O
Squaring both sides.
1 = 5.91 q q = 0.17
2. (a)
1 y2
x + p q
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
x
24
www.nitropdf.com
Dapatkan Kertas Soalan Ramalan MATHS dan ADD MATHS 2010 di http://maths-catch.com/exam
SPM ZOOM–IN Form 5: Chapter 3 Integration Paper 1
4. Area of the shaded region k
1.
5
( y – 5) dy = 8
[ 2
k – 5k – 2
k
y2 –5 y 2
]
5
=8
=
2
=
2
y dx
–1
–1
( x2 – 2 x + 1) dx 2
3 = x – x2 + x 3
[
52 – 5(5) = 8 2
]
–1
8 1 –4+2– – –1 – 1 3 3 = 3 units2 =
k 2 – 5k + 25 = 8 2 2 k 2 – 10k + 25
= 16
Paper 2
k 2 – 10k + 9 = 0 (k – 1)(k – 9) = 0 k = 1 or 9
y
1. y = – x3 – x
2.
2
–1
4
3g ( x) dx +
[ = 3[ =3
2
–1 4
2
3g ( x) dx
g( x) dx +
4
2
]
g( x) dx
1 O
]
g( x) dx
–1
x
2
P
Q
= 3(20) = 60 3.
dy = x4 – 8 x3 + 6 x2 dx
Area P 0
dx y = x – 8 x + 6 x + c 5 4 3
4
3
2
4
3
1, – 1 45 , 9 1 = –2 + 2 + c 5 5 c = –2
5
–1
(– x3 – x) dx 0
]
Area Q 2
y dx
0
=
Hence, the equation of the curve is 5
0
[
Since the curve passes through the point
x
[
5
y =
=
4 2 = – x – x 4 2 –1 4 2 = 0 – – (–1) – (–1) 4 2 1 1 = + 4 2 3 = 4
y = x – 2 x4 + 2 x3 + c 5
–
y dx
–1
y = x – 8 x + 6 x 5
– 2 x4 + 2 x3 – 2.
2
0
(– x3 – x) dx 2
4 2 = – x – x 4 2 0 4 2 =– 2 – 2 –0 4 2 = –4 – 2 = –6
[
25
www.nitropdf.com
]
]
Dapatkan Kertas Soalan Ramalan MATHS dan ADD MATHS 2010 di http://maths-catch.com/exam
(b) When h = 1 and k = 4, y = x2 + 4
Hence, the total area of the shaded region = Area P + |Area Q| 3 = + |–6| 4 3 = 6 units2 4 2. (a)
y y = x2 + 4 P
y = hx2 + k dy = 2hx dx
x = 3
4
At the point (–2, 8), the gradient of the curve is – 4. dy = –4 ∴ dx 2hx = – 4 2h (–2) = – 4 – 4h = – 4 h =1
x
O
2
3 Q
Volume generated, V x = Volume generated by the curve– Volume generated by the straight line PQ (from x = 0 to x = 2) 3 1 = π y2 dx – πr 2h 0 3 3 2 2 1 = π ( x + 4) dx – π(4)2 (2) 0 3 3 4 2 32 = π ( x + 8 x + 16) dx – π 0 3
The curve y = hx2 + k passes through the point (–2, 8). 2 ∴ 8 = h(–2) + k 8 = 4h + k 8 = 4(1) + k k = 4
3 5 3 32 = π x + 8 x + 16 x – π 0 3 5 3 5 8 32 = π 3 + (3)3 + 16(3) – 0] – π 3 3 5 14 3 = 157 π units 15
[ [
26
www.nitropdf.com
]
Dapatkan Kertas Soalan Ramalan MATHS dan ADD MATHS 2010 di http://maths-catch.com/exam
SPM ZOOM–IN Form 5: Chapter 4 Vectors Paper 1 1. (a)
4. (a) If the vectors _a and _b are parallel, then a _ = hb _ (h is a constant).
→ 3 → 3 EA = DC = (12 p _ ) = 9 p _ 4 4
2_i – 5 j _ – 3 j _ = h(ki _) 2_i – 5 j _ – 3hj _ = hki _
→ 1 → (b) EQ = ED 2
Equating the coefficients of _j –3h = –5 5 h= 3
→ 1 → → → EA + AB + BC + CD = 2
1 p _r _ q p _ 2 _ 1 – 3 p – 6 r _ – 9q _ 2 _
=
9 – 6 – 9 – 12
=
Equating the coefficients of _i hk = 2 5 k = 2 3 6 k = 5
→ → → 2. XY = XD + DY
1 → 2 → BD + DC 2 3 → 1 → 2 → BA + AD + AB = 2 3 =
=
1 – _ b _a 2 6 +2
+
_a = hb _ 5 _a = _ b 3
(b)
2 (6b) 3 _
|a| 5 = |b| 3
b + _a + 4_ b = –3_
|a| : |b| = 5 : 3
= a _ + b _
→ → → 5. (a) AC = AB + BC
1 a+ _ b + 2_c 3. (a) _ 5 1 = 7 j _ + 5 (10_i – 5 j_ ) + 2(–4_i + j _)
= 9_i – 4 j _ + (–6_i + mj_)
j = 3_i + ( m – 4)_ → (b) If AC is parallel to the x-axis, the coefficient of j_ equals zero. m–4 = 0 m =4
= 7_ j + 2_i – _j – 8_i + 2_ j
j = – 6_i + 8_ 1 _b + 2_c | = (–6)2 + 82 = 10 5 Hence, the unit vector in the direction of
(b) |_ a+
_a + =
1 b + 2_c _ 5
1 –6_i + 8 j _ 10
3 4 = – _i + j 5 5_
27
www.nitropdf.com
Dapatkan Kertas Soalan Ramalan MATHS dan ADD MATHS 2010 di http://maths-catch.com/exam
Paper 2
(c) Since the points O, T and S are collinear, → → then, OT = kOS , where k is a constant.
→ → → 1. (a) OT = OA + AT 1 → _ + AQ = 4 x 3 → 1 → _ + ( AO + OQ ) = 4 x 3 1 = 4 x _ + (– 4 x _ + 6 y _) 3 8 = x _ + 2 y _ 3
→ → OT = kOS
8 x _ + 2 y _] _ = k [(6 – 6h) _y + 16hx 3 8 x _ + 2 y _ _ = k (6 – 6h) _y + 16hkx 3 8 _ x _ + 2 y _ = (6k – 6hk ) _y + 16hkx 3
→ → → (b) OS = OQ + QS → = 6 y_ + hQP → → = 6 y_ + h(QO + OP ) → → = 6 y_ + h(QO + 4OA)
Equating the coefficients of x _, 8 = 16hk 3 1 = 6hk 1 hk = … 1 6
= 6 y_ + h[–6 y _)] _ + 4(4 x
Equating the coefficients of y _, 6k – 6hk = 2 … 2
_ = (6 – 6 h) y _ + 16 hx
Substituting 1 into 2 :
16 = 2
6k – 6
6k = 3 1 k = 2 From 2 :
hk =
1 6
12 = 16
h
h=
28
www.nitropdf.com
1 3
Dapatkan Kertas Soalan Ramalan MATHS dan ADD MATHS 2010 di http://maths-catch.com/exam
→ 5 → 5 2. (a) (i) OM = OB = (14 y _) = 10 y _ 7 7
→ → → (c) AK = AL + LK
→ 1 → (ii) AK = AB 4 → 1 → AO + OB = 4
1 = –2 x _ + 14 y _ 4
1 7 3 7 x _+ qx _ _ + y = – 2 px _ + 10 py _ + qy _ 2 2 2 2
–
1 7 x p+ _ + y = –2 2 2_
Equating the coefficients of y _, 2 20 p + 7q = 7 …
–20 p + 15q = –5 … 1 +
20 p + 7q = 7 … 22q = 2 1 q= 11
= p(–2 x _ + 10 y _) = –2 px + 10 py _ → → (ii) KL = qKO → → = q KA + AO
12 x_ 3 = q _ x 2
–
–
=–
1 7 x _ – _y 2 2
14 11 7 p = 22
– 4 p = –
7 y _ _ – 2 x 2
–
2
From 1 : 1 – 4 p + 3 = –1 11
→ → KA = – AK
= =q
7 y _ 2
3 7 _ – qy qx _ 2 2
29
www.nitropdf.com
3 7 q x _ + 10 p + q _y 2 2
Equating the coefficients of x _, 1 – 4 p + 3q = –1 …
→ → (b) (i) AL = pAM → → = p AO + OM
– x _ + 7 y _ = (– 4 p + 3q) _x + (20 p + 7q) _y
1 7 _ + y = – x 2 2_
–
5
Dapatkan Kertas Soalan Ramalan MATHS dan ADD MATHS 2010 di http://maths-catch.com/ WebsiteZI F505_4th pp
10/15/08
9:40 AM
Page 30
SPM ZOOM–IN Form 5: Chapter 5 Trigonometric Functions Paper 1
3 tan θ = 2 tan (45º – θ)
3.
1tan+ tan45 45 tantan = 2 1 tan 1 + tan
1 + p2 1
θ
3 tan θ
O
–p
θ
º–
3 tan θ = 2
1.
º
–
θ
θ
2
3 tan θ + 3 tan θ = 2 – 2 tan θ 3 tan2 θ + 5 tan θ – 2 = 0 (3 tan θ – 1)(tan θ + 2) = 0
sin (90º – θ) = cos θ
=–
2.
tan θ =
p
1 + p2
1 or tan θ = –2 3 1 , 3 = 18.43º = 18.43º, 198.43º
When tan θ = Basic ∠
3 – 10 tan x = 0 cos2 x 3 sec2 x – 10 tan x = 0 2 3(tan x + 1) – 10 tan x = 0 3 tan2 x + 3 – 10 tan x = 0 3 tan2 x – 10 tan x + 3 = 0 (3 tan x – 1)(tan x – 3) = 0 1 tan x = or tan x = 3 3
θ
When tan θ = –2, Basic ∠ = 63.43º θ = 116.57º, 296.57º ∴ θ = 18.43º, 116.57º, 198.43º, 296.57º
1 , 3 x = 18.43º, 198.43º
When tan x =
When tan x = 3, x = 71.57º, 251.57º ∴ x = 18.43º, 71.57º, 198.43º, 251.57º
30
www.nitropdf.com
θ
Dapatkan Kertas Soalan Ramalan MATHS dan ADD MATHS 2010 di http://maths-catch.com/ WebsiteZI F505_4th pp
10/15/08
9:40 AM
Page 31
Paper 2
2. (a), (b)
1. (a) LHS = =
y
1 – cos 2 x sin 2 x
y = 3 sin
3 2
1 – (1 – 2 sin2 x) 2 sin x cos x
O
x
2
x
2
–2
2 sin2 x = 2 sin x cos x
y = 2 –
sin x cos x = tan x = RHS =
3 sin
(b) (i), (ii) The graph of y = |tan x| is as shown below.
2 x x + =2 2 π 2 x x 3 sin =2– 2 π
Sketch the straight line y = 2 –
y y =
2 x 2
π
x
0
2 π
y
2
–2
(2π, 1) O
π
π
3π 2
π
Number of solution = Number of intersection point =1
x
1 – cos 2 x x – =0 sin 2 x 2 π 1 – cos 2 x x = sin 2 x 2 π x |tan x| = 2 π
Sketch the straight line
y =
x
2 π
.
Number of solutions = Number of points of intersection =4
31
www.nitropdf.com
2 x π
Dapatkan Kertas Soalan Ramalan MATHS dan ADD MATHS 2010 di http://maths-catch.com/ WebsiteZI F506_4th pp
10/15/08
9:40 AM
Page 32
SPM ZOOM–IN Form 5: Chapter 6 Permutations and Combinations 3. Number of different committees that can be formed
Paper 1
1.
Number of arrangements
2
1
3P 2
2
3
3P 2
2
5
3P 2
Choosing a female secretary and a female treasurer from 7 females
= 4C 1 7C 2 8C 3
Hence, the number of 4-digit odd numbers greater than 2000 but less than 3000 that can be formed = 3 3P2 = 18
Choosing a male president from 4 males
= 4704 2. Each group of boys and girls is counted as one item. G1, G2 and G3
B1, B2 and B3
√
√
This gives 2!.
√ √
√
At the same time, B1, B2, and B3 can be arranged among themselves in their group. This gives 3!.
√
√
√
In the same way, G1, G2, and G3 can also be arranged among themselves in their group. This gives another 3!. Using the multiplication principle, the total number of arrangements = 2! 3! 3! = 72
32
www.nitropdf.com
Choosing 3 subcommittee members from 8 (males or females).
Dapatkan Kertas Soalan Ramalan MATHS dan ADD MATHS 2010 di http://maths-catch.com/exam
SPM ZOOM–IN Form 5: Chapter 7 Probability Paper 1
3 5 h+ 5 3 = h + k + 5 5 5h + 25 = 3h + 3k + 15 2h = 3k – 10 3 k – 10 h = 2
1.
P(Not a green ball) =
2.
P (not getting any post)
2 3 3 5 8 = 35 =
4 7
–
3. There are 3 ‘ E ’ and 4 ‘ E ’ in the bag 3 2 1 (a) P( EE ) = = 7 6 7 (b)
– P( EE )
=
3 7
4 2 = 6 7
33
www.nitropdf.com
Dapatkan Kertas Soalan Ramalan MATHS dan ADD MATHS 2010 di http://maths-catch.com/exam
SPM ZOOM–IN Form 5: Chapter 8 Probability Distributions Paper 1
3. (a) X – Mass of a crab, in g X ∼ N ( 175, 15)
1. X – Number of penalty goals scored 3 X ∼ B n, 5 16 P( X = 0) = 625
Z = X – µ
n
16 = 625 2 16 (1)(1) = 5 625 25 = 25 3 5
C o
0
2 5
σ
190 – 175 15 =1 =
n
(b) P(175 < X < 190) 175 – 175 190 – 175 =P < Z < 15 15 = P (0 < Z < 1) = 0.5 – 0.1587 = 0.3413
n
n
∴n
4
0.1587
=4 O
2. X ∼ N ( 55, 122) Area of the shaded region P ( X < 37) 37 – 55 = P Z < 12 = P ( Z < –1.5) 0.0668 = 0.0668
4. 0.1841
–0.9
P( Z > – 0.9)= 0.8159 ∴ k = –0.9
–1.5
34
www.nitropdf.com
0.8159
1
Dapatkan Kertas Soalan Ramalan MATHS dan ADD MATHS 2010 di http://maths-catch.com/exam
Paper 2
(ii) P(10 < X < 13) 10 – 12 13 – 12 < Z< =P 3.1201 3.1201 = P(–0.641 < Z < 0.321) = 1 – 0.2608 – 0.3741 0.2608 = 0.3651
1. (a) X – Number of blue beads drawn
186 1 X B 10, 3 X ∼ B 10,
0.3741
∼
–0.641
(i) P( X ≥ 3) = 1 – P( X = 0) – P( X = 1) – P( X = 2)
1 2 C 3 3
– 10
0
1 3
= 1 – 10C 0
2 3
2
2. (a) X – Number of customers requiring a supplementary card
10
– 10C 1
1
1 3
0.321
2 3
280 500 7, 1425
9
X ∼ B 7,
8
X ∼ B
2
= 0.7009
(i) P( X = 3) (ii) Mean = np = 10
1 =3 3
1
= 7C 3
3
14 25
3
11 25
4
= 0.2304 Standard deviation = npq = 10 1 3
(ii) P( X = 3) = P( X = 0) + P( X = 1) + P( X = 2) 14 0 11 7 7 14 1 11 + C 1 = 7C 0 25 25 25 25
2 3
= 1.49
7
C 2
(b) X – Lifespan of a species of dog X ∼ N (12, σ2)
5
(b) X – Time taken to settle invoices X ∼ N ( 30, 52)
= 0.9 –4 P Z > = 0.9
P Z >
2
= 0.1402
P( X > 8) = 90%
(i)
1425 1125
8 – 12 σ
(i) P(28 ≤ X ≤ 36)
28 –5 30
σ
=P
≤ Z ≤
36 – 30 5
= P(–0.4 ≤ Z ≤ 1.2) = 1– 0.3446 – 0.1151 = 0.5403
0.9 0.1
–1.282 0.3446
–
4 σ
= –1.282 –0.4
σ
0.1151
= 3.1201 years
35
www.nitropdf.com
1.2
6
+
Dapatkan Kertas Soalan Ramalan MATHS dan ADD MATHS 2010 di http://maths-catch.com/exam
(ii) P( X < 22)
22 – 30 5 = P( Z < –1.6) = 0.0548 = P Z <
Hence, the expected number of invoices which are given discounts = 0.0548 220 = 12
36
www.nitropdf.com
Dapatkan Kertas Soalan Ramalan MATHS dan ADD MATHS 2010 di http://maths-catch.com/exam
SPM ZOOM–IN Form 5: Chapter 9 Motion Along a Straight Line Paper 2
(c) When particle A reverses its direction, v A = 0 12 + t – t 2 = 0 2 t – t – 12 = 0 (t + 3)(t – 4) = 0 t = –3 or 4 t = –3 is not accepted ∴ t = 4
1. (a) For particle A, at maximum velocity, dv A dt
=0
1 – 2t = 0 1 t = 2 2
d v A 2
dt
= –2 (negative)
1 1 – 2 2
Hence, vmax = 12 + = 12
ds B
v B = 2
dt 2
v B = 6t
1 m s–1 4
– 14t – 15
dv B
a B =
dt
a B =12t – 14
(b)
3
– 7t 2 – 15t When particle B returns to O, s B = 0 3 2 t t 2 – 7 – 15t = 0 2 t (2t – 7t – 15) = 0 t (2t + 3)(t – 5) = 0 3 or 5 t = 0, – 2 3 are not accepted t = 0 and t = – 2 ∴ t = 5 s B = 2t
When t = 4, –2 a B =12(4) – 14 = 34 m s 2 (a)
a = 12 – 6t v= v=
a dt
(12 – 6t ) dt 12t – 3t 2 + c When t = 0, v = 15. Thus, c = 15 2 ∴ v = 12t – 3t + 15 v=
At maximum velocity, s A
= v
s A = s A =
A
dv
dt
dt
2
(12 + t – t ) dt
12t +
2
t
2
–
3
t
3
12 – 6t = 0 t = 2
+c
When t = 0, s A = 0. ∴ c = 0 ∴ s A =
12t +
2
t
2
–
=0
When t = 2, 2 –1 v = 12(2) – 3(2) + 15 = 27 m s
3
t
3
2
d v 2
dt
When t = 5, 2 5 – 5 3 = 30 5 m s A = 12(5) + 6 2 3
= –6 (< 0)
Therefore, v is a maximum.
37
www.nitropdf.com
Dapatkan Kertas Soalan Ramalan MATHS dan ADD MATHS 2010 di http://maths-catch.com/exam
(b)
s= s=
v dt
(c) When the particle travels to the right, v >0 12t – 3t 2 + 15 > 0 3t 2 – 12t – 15 < 0 2 t – 4t – 5 < 0 (t + 1)(t – 5) < 0
2
(12t – 3t + 15) dt 2 – t 3 + 15t + c When t = 0, s = 0. Thus, c = 0. 2 3 ∴ s = 6t – t + 15t s = 6t
At maximum displacement, ds dt
=0
–1
5
t
2
12t – 3t + 15 = 0 3t 2 – 12t – 15 = 0 2 t – 4t – 5= 0 (t – 5)(t + 1) = 0 t = 5 or –1 t = –1 is not accepted ∴ t = 5
–1 < t < 5 Since the values of t cannot be negative, therefore 0 ≤ t < 5.
When t = 5, 2 3 s = 6(5) – 5 + 15(5) = 100 m 2
d s 2
dt
= 12 – 6t
When t = 5,
2
d s 2
dt
= 12 – 6(5) = –18
Therefore, s is a maximum.
38
www.nitropdf.com
Dapatkan Kertas Soalan Ramalan MATHS dan ADD MATHS 2010 di http://maths-catch.com/exam
SPM ZOOM–IN Form 5: Chapter 10 Motion Along a Straight Line Paper 2
(c) (i)
1. (a) I 180 x + 90 y 2 x + y
≤ 5400 ≤ 60
x
0
30
y
60
0
3 The furthest point on the straight line y = x 4 inside the feasible region R is (20,15). ∴ xmax = 20, ymax = 15
II 3 x + 4 y ≤ 120
III
x 4 = y 3 3 x = 4 y 3 y = x 4
x
0
40
y
30
0
x
0
30
200 150 0.1 = 3000
y
0
60
The optimal point is (24, 12).
(ii) Profits = 200 x + 150 y Draw the straight line 200 x + 150 y = 3000
y ≤ 2 x
(b)
Hence, the maximum profit = 200(24) + 150(12) = RM6600
y
60 2 x + y = 60
2. (a) Mixing: 30 x + 10 y ≤ 15 60 3 x + y ≤ 90
y = 2 x
50
Baking:
2 40 x + 40 y ≤ 26 3 x + y ≤ 40
40
30 y =
3 x 4
Max (24, 12) 10
R
3 x + 4 y = 120
O
10 15 20 200 x + 150 y = 3000
30
40
60
Decorating: 10 x + 30 y ≤ 15 60 x + 3 y ≤ 90
(20, 15)
20
x
39
www.nitropdf.com
x
0
30
y
90
0
x
0
40
y
40
0
x
0
90
y
30
0