Download - Nota Matematik Tambahan Ting 4 Dan 5 SPM
1
Paper 1 1 Based on the given ordered pairs
{(2, 1), (4, 3), (6, 5), (6, 7)}, an arrow diagram can be drawn as shown below.
(a) The image of 2 is 1. (b) The object of 7 is 6. 2 (a) Let g−1(7) = y
Thus, g(y) = 7 4y − 1 = 7 4y = 8 y = 2 ∴ g−1(7) = 2
(b) hg(x)
= h(4x − 1) = (4x − 1)2 − 3(4x − 1) + 5 = 16x2 − 8x + 1 − 12x + 3 + 5 = 16x2 − 20x + 9
3
(a) The range is {3, 7}. (b) The above relation is a many-to-one
relation.
Form 4: Chapter 1 (Functions) SPM Flashback
Fully-Worked Solutions
'3' and '7' are linked to object(s) but '5' and '11' are not linked to any object(s). Therefore, the range is {3, 7}.
Element '7' in the codomain is linked to two elements, i.e. '28' and '49' in the domain. Therefore, it is a many-to-one relation.
2
4 h : x → 2x + m h(x) = 2x + m Let h−1(x) = y h(y) = x 2y + m = x
y =
x − m2
∴ h−1(x) =
x − m2
=x2
−m2
But it is given that h−1(x) = 3kx +
32
.
Hence, by comparison,
3k =
12
⇒ k =16
and
−
m2
=32
⇒ m = −3 .
5 (a) hg(x) = 6x − 2
h[g(x)] = 6x − 2 3g(x) + 1 = 6x − 2 3g(x) = 6x − 3 g(x) = 2x − 1
(b) gh(x) = g(3x + 1) = 2(3x +1) −1 = 6x +1
When gh(x) =13
,
6x +1 =13
18x + 3 = 118x = −2
x = −19
6 (a) From the given arrow diagram,
f (−2) = −5. Hence, f−1(−5) = −2 .
(b) gf (−2) = 3
7 (a) Let w−1(x) = z
w(z) = x6
3− 2z= x
6 = x(3− 2z)6 = 3x − 2xz
2xz = 3x − 6
z =3x − 6
2x
∴ w−1( x) =3x − 6
2x, x ≠ 0
(b) w−1h −
52
⎛
⎝ ⎜
⎞
⎠ ⎟
= w−1 2 −52
⎛
⎝ ⎜
⎞
⎠ ⎟ + 3
⎡
⎣ ⎢ ⎤
⎦ ⎥
= w−1(−2)
=3(−2) − 6
2(−2)= 3
8 Let
n−1(x) = yn( y) = x
4 y − 1 = x4 y = x +1
y =x +1
4
n−1(x) =x +1
4
mn−1(x) = m x +14
⎛
⎝ ⎜
⎞
⎠ ⎟
=3
8 x + 14
⎛
⎝ ⎜
⎞
⎠ ⎟ − 5
=3
2x − 3, x ≠
32
It is given that h(x) = 3x + 1. Hence, h[g(x)] = 3g(x) + 1.
This is a composite function gf(x) which maps x directly onto z.
Change the subject of the formula to z.
3
9 (a) The relation between set P and set Q is a many-to-one relation.
(b) The relation can be represented by
f ( x) = x4 .
10
m(2) = 72 − h
h= 7
2 − h = 7h8h = 2
h =14
Function notation
4
Paper 2 1 (a)
f : x → 2x − 3f (x) = 2x − 3
Let f −1(x) = yf ( y) = x
2y − 3 = x
y =x + 3
2
∴ f −1(x) =x + 3
2
f −1g(x)
= f −1 x2
+ 2⎛
⎝ ⎜
⎞
⎠ ⎟
=
x2
+ 2⎛
⎝ ⎜
⎞
⎠ ⎟ + 3
2
=
x + 4 + 622
=x +10
4
∴ f −1g : x →x + 10
4
(b) hg : x → 2x + 4
hg(x) = 2x + 4
h x2
+ 2⎛
⎝ ⎜
⎞
⎠ ⎟ = 2x + 4
Let x2
+ 2 = u
x2
= u − 2
x = 2u − 4
h(u) = 2(2u − 4) + 4= 4u − 8 + 4= 4u − 4
∴ h : x → 4x − 4
5
Paper 1 1
2x(x − 3) = (2 − x)(x +1)2x2 − 6x = 2x + 2 − x2 − x
3x2 − 7x − 2 = 0
x =−(−7) ± (−7)2 − 4(3)(−2)
2(3)
x =7 ± 49 + 24
6
x =7 ± 8.5440
6x = 2.591 or −0.2573
2 The roots are 23
and −
52
.
Sum of roots =
23
+ −52
⎛
⎝ ⎜
⎞
⎠ ⎟ = −
116
Product of roots =
23
× −52
⎛
⎝ ⎜
⎞
⎠ ⎟ = −
53
The quadratic equation is
x2 +
116
x −53
= 0
6x 2 + 11x − 10 = 0
3 y = 5x − 2 …
y = 3x2 + 3x + k … Substituting into :
3x2 + 3x + k = 5x − 23x2 − 2x + k + 2 = 0
a = 3, b = −2, c = k + 2 In the case where a curve does not meet a straight line, b2 − 4ac < 0 is applied.
b2 − 4ac < 0(−2)2 − 4(3)(k + 2) < 0
4 −12k − 24 < 0−12k − 20 < 0
−12k < 20
k >20
−12
k > −1 23
4
7 − 2(1+ x)2 = x(x + 5)7 − 2(1+ 2x + x2 ) = x2 + 5x
7 − 2 − 4x − 2x2 = x2 + 5x3x2 + 9x − 5 = 0
x =−b ± b2 − 4ac
2a
x =−9 ± 92 − 4(3)(−5)
2(3)
x =−9 ± 141
6x = 0.4791 or −3.479
Form 4: Chapter 2 (Quadratic Equations) SPM Flashback
Fully-Worked Solutions
x2 − (sum of roots)x + (product of roots) = 0
6
5
9x2 + qx + 1 = 4x9x2 + qx − 4x + 1 = 09x2 + (q − 4)x + 1 = 0
a = 9, b = q − 4, c = 1 If the equation has equal roots, then
b2 − 4ac = 0(q − 4)2 − 4(9)(1) = 0q2 − 8q +16 − 36 = 0
q2 − 8q − 20 = 0(q + 2)(q −10) = 0
q = −2 or 10
7
Paper 1 1
x(x + 2) = 3px − 4x2 + 2x = 3px − 4
x2 + 2x − 3px + 4 = 0x2 + (2 − 3p)x + 4 = 0
a = 1, b = 2 − 3p, c = 4 For the case of two distinct roots, b2 − 4ac > 0 is applied.
(2 − 3p)2 − 4(1)(4) > 04 − 12p + 9p2 − 16 > 0
9p2 −12p − 12 > 03p2 − 4p − 4 > 0
(3p + 2)( p − 2) > 0
Hence, the ranges of values of p are
p < −
23
or p > 2 .
2
x(x − 1) > 12x2 − x > 12
x2 − x −12 > 0(x + 3)(x − 4) > 0
The ranges of values of x are x < −3 or x > 4 .
3
(a) The curve passes through the point (0, −4).
y = −(x − k )2 − 3−4 = −(0 − k )2 − 3−1 = −k 2
k 2 = 1k = 1
(b) The equation of the curve is
y = −(x − 1)2 − 3. Hence, the equation of the axis of symmetry is x −1 = 0 ⇒ x = 1.
(c) The coordinates of the maximum point are
(1, −3). 4 f (x) = −2(x + p)2 − 2
The maximum point of the graph of f (x) is
(−p, −2) . But it is given that the maximum point of the graph of f (x) is (−3, q) . Hence, by comparison,
(a) −p = −3 ⇒ p = 3 (b) q = −2 (c) the equation of the axis of symmetry of the
curve is x = −3.
Form 4: Chapter 3 (Quadratic Functions) SPM Flashback
Fully-Worked Solutions
8
5 (a) The equation of the axis of symmetry is
x =−6 + (−2)
2x = −4
(b) The minimum point is (−4, −5). Hence,
f (x) = ( x + 4)2 − 5 .
6
(1− 2x)(3 + x) > x + 33+ x − 6x − 2x2 > x + 3
−2x2 − 6x > 02x2 + 6x < 02x(x + 3) < 0
Hence, the required range of values of x is −3 < x < 0 .
9
Paper 2 1 (a)
f (x) = −x2 + 4kx − 5k 2 − 1= −(x2 − 4kx + 5k 2 +1)= −(x2 − 4kx + 4k 2 − 4k 2 + 5k 2 +1)
= −[(x − 2k )2 + k 2 + 1]= −(x − 2k )2 − k 2 −1
Maximum value = −k 2 −1 But it is given that the maximum value = −r2 − 2k . By comparison,
−r2 − 2k = −k 2 − 1r2 = k 2 − 2k +1r2 = (k −1)2
r = k − 1 (shown)
(b) From f (x) = −(x − 2k )2 − k 2 − 1, the axis
of symmetry is x = 2k . But it is given that the axis of symmetry is x = r2 − 1. By comparison, r2 − 1 = 2k Solve the following simultaneous equations: r = k − 1 … r2 − 1 = 2k … Substitute into :
(k −1)2 −1 = 2kk 2 − 2k +1− 1− 2k = 0
k 2 − 4k = 0k(k − 4) = 0
k = 0 or k = 4k = 0 is not accepted.∴ k = 4When k = 4, r = 4 − 1 = 3
Add and subtract
12
× (−4k )⎡
⎣ ⎢ ⎤
⎦ ⎥ 2
= 4k 2
10
Paper 2 1 4x + y = x2 + x − y = −3
4x + y = −3 …
x2 + x − y = −3 …
From : y = −3− 4x …
Substitute into :
x2 + x − (−3− 4x) = −3x2 + x + 3+ 4x + 3 = 0
x2 + 5x + 6 = 0(x + 2)(x + 3) = 0
x = −2 or −3
From : When x = −2, y = −3 − 4(−2) = 5 When x = −3, y = −3− 4(−3) = 9
Hence, the solutions are x = −2, y = 5
or x = −3, y = 9 . 2 x − y = 1 …
x2 + 3y = 6 …
From : x = 1+ y …
Substituting into , we have:
(1+ y)2 + 3y = 61+ 2y + y2 + 3y − 6 = 0
y2 + 5y − 5 = 0
y =−b ± b2 − 4ac
2a
y =−5 ± 52 − 4(1)(−5)
2(1)
y =−5 ± 45
2(1)y = 0.854 or −5.854
From : When y = 0.854, x = 1+ 0.854 = 1.854 When y = −5.854, x = 1+ (−5.854) = −4.854
Hence, the solutions are x = 1.854, y = 0.854 or x = −4.854, y = −5.854 (correct to 3 decimal places).
3 x +
y2
= 1 …
3xy − 7 y = 2 … From :
2x + y = 2 y = 2 − 2x … Substituting into :
3x(2 − 2x) − 7(2 − 2x) − 2 = 06x − 6x2 − 14 + 14x − 2 = 0
−6x2 + 20x −16 = 03x2 −10x + 8 = 0
(3x − 4)(x − 2) = 0
x =43
or 2
From :
When x =
43
, y = 2 − 2 4
3⎛
⎝ ⎜
⎞
⎠ ⎟ = −
23
When x = 2, y = 2 − 2(2) = −2
Hence, the solutions are x = 1 1
3, y = −
23
or
x = 2, y = −2 .
Form 4: Chapter 4 (Simultaneous Equations) SPM Flashback
Fully-Worked Solutions
11
4 3x + y = 2 …
x2 + 2 y2 + xy = 4 …
From : y = 2 − 3x … Substituting into :
x2 + 2(2 − 3x)2 + x(2 − 3x) = 4x2 + 2(4 −12x + 9x2 ) + 2x − 3x2 − 4 = 0
x2 + 8 − 24x + 18x2 + 2x − 3x2 − 4 = 016x2 − 22x + 4 = 0
8x2 − 11x + 2 = 0
x =−(−11) ± (−11)2 − 4(8)(2)
2(8)
x =11± 57
16x = 1.159 or 0.216
From : When x = 1.159, y = 2 − 3(1.159) = −1.477 When x = 0.216, y = 2 − 3(0.216) = 1.352 Hence, the solutions are x = 1.159, y = −1.477 or x = 0.216, y = 1.352 (correct to 3 decimal places).
12
Paper 1 1 log2 R − log4 Q = 2
log2 R −log2 Qlog2 4
= 2
log2 R −log2 Qlog2 22
= 2
log2 R −log2 Q
2= 2
2log2 R − log2 Q = 4log2 R2 − log2 Q = 4
log2R2
Q= 4
R2
Q= 24
R2 = 24QR = 24QR = 22 QR = 4 Q
2 42x− 3 = 5x
log 42x− 3 = log 5x
(2x − 3) log 4 = x log 52x log 4 − 3log 4 = x log 52x log 4 − x log 5 = 3log 4x(2log 4 − log 5) = 3log 4
x =3log 4
2log 4 − log 5x = 3.576
3
2434x = 98 x+6
(35 )4x = (32 )8x+6
320x = 32(8x+6)
Equating the powers, we have:
20x = 2(8x + 6)20x = 16x +12
4x = 12x = 3
4 log5 2.7
= log5 2 710
= log52710
= log533
2 × 5⎛
⎝ ⎜
⎞
⎠ ⎟
= log5 33 − (log5 2 + log5 5)= 3log5 3 − log5 2 − log5 5= 3 p − m − 1
5 3s + 2 − 3s + 1 =
23
3s(32) − 3s(31) =23
9(3s) − 3(3s) =23
(9 − 3)(3s) =23
6(3s) =23
3s =23
×16
3s =19
3s = 3−2
∴ s = −2
6 log5 (2y − 1) = 1 + log5 ( y − 8)
log5 (2 y − 1) − log5 ( y − 8) = 1
log52 y − 1y − 8
⎛
⎝ ⎜
⎞
⎠ ⎟ = 1
2y − 1y − 8
= 51
2y − 1 = 5( y − 8)2y − 1 = 5y − 40
3y = 39y = 13
Form 4: Chapter 5 (Indices and Logarithms) SPM Flashback
Fully-Worked Solutions
13
7
2 = ms 3 = mt
logm 2 = s logm 3 = t
logmm6
⎛
⎝ ⎜
⎞
⎠ ⎟
= logmm
12
2 × 3
⎛
⎝
⎜ ⎜ ⎜
⎞
⎠
⎟ ⎟ ⎟
= logm m12 − logm 2 − logm 3
=12
− s − t
8 272x−5 =
19x+1
33(2x−5) =1
32( x+1)
33(2x−5) =1
[32( x+1) ]12
33(2x−5) =1
3( x+1)
33(2x−5) = 3−( x+1)
Equating the powers, 3(2x − 5) = −(x +1)
6x −15 = −x −17x = 14
x = 2
9 log3
x4
y⎛
⎝ ⎜
⎞
⎠ ⎟ = 2 + 2 log3 x + log3 y
log3x4
y⎛
⎝ ⎜
⎞
⎠ ⎟ − 2 log3 x − log3 y = 2
log3x4
y⎛
⎝ ⎜
⎞
⎠ ⎟ − log3 x2 − log3 y = 2
log3
x4
yx2 y
⎛
⎝
⎜ ⎜ ⎜ ⎜
⎞
⎠
⎟ ⎟ ⎟ ⎟
= 2
log3x4
y⎛
⎝ ⎜
⎞
⎠ ⎟
1x2 y
⎛
⎝ ⎜
⎞
⎠ ⎟ = 2
log3x2
y2
⎛
⎝ ⎜
⎞
⎠ ⎟ = 2
x2
y2= 32
xy
⎛
⎝ ⎜
⎞
⎠ ⎟
2
= 32
xy
= 3
y =x3
10 2 + 2log4 ( p − 6) = log2 p
2 log4 ( p − 6) − log2 p = −2
2 log2 ( p − 6)log2 4
⎡
⎣ ⎢ ⎤
⎦ ⎥ − log2 p = −2
2 log2 ( p − 6)2
⎡
⎣ ⎢ ⎤
⎦ ⎥ − log2 p = −2
log2 ( p − 6) − log2 p = −2
log2p − 6
p⎛
⎝ ⎜
⎞
⎠ ⎟ = −2
p − 6p
= 2−2
p − 6p
=14
4p − 24 = p3p = 24
p = 8
log2 4 = log2 22 = 2log2 2 = 2(1) = 2
14
Paper 1 1
B = (b, c)4(2a) + 3(2b)
3 + 4, 4a + 3(3c)
3 + 4⎛
⎝ ⎜
⎞
⎠ ⎟ = (b, c)
8a + 6b7
, 4a + 9c7
⎛
⎝ ⎜
⎞
⎠ ⎟ = (b, c)
Equating the x-coordinates:
8a + 6b7
= b
8a + 6b = 7b8a = b
a =
b8
…
Equating the y-coordinates:
4a + 9c7
= c
4a + 9c = 7c4a + 2c = 0
2a + c = 0 … Substitute into :
2 b8
⎛
⎝ ⎜
⎞
⎠ ⎟ + c = 0
b4
+ c = 0
b + 4c = 0b = −4c
2 yk
+x3
= 1
3y + kx3k
= 1
3y + kx = 3k3y = −kx + 3k
y = −k3
x + k
∴ m1 = −
k3
5y = 3x + 25
y =35
x + 5
∴ m2 =
35
m1m2 = −1
−k3
⎛
⎝ ⎜
⎞
⎠ ⎟
35
⎛
⎝ ⎜
⎞
⎠ ⎟ = −1
−k5
= −1
k = 5
Form 4: Chapter 6 (Coordinate Geometry) SPM Flashback
Fully-Worked Solutions
15
3 x3
+y4
= 1 is an equation in the intercept form
where the x-intercept is 3 and the y-intercept is 4.
∴ Q = (0, 4)∴ P = (3, 0)
mQP =
4 − 00 − 3
= −43
Gradient of the perpendicular line
= −1
−43
⎛
⎝ ⎜
⎞
⎠ ⎟
=34
Hence, the equation of the straight line that is perpendicular to PQ and passes through the point Q(0, 4) is
y − y1 = m(x − x1 )
y − 4 =34
(x − 0)
4( y − 4) = 3x4y − 16 = 3x
4y = 3x + 16
4 A = (−2, −2), B = (3, 5)
Let P = (x, y)
PA : PB = 2 : 3PAPB
=23
3PA = 2PB3 [x − (−2)]2 + [ y − (−2)]2 = 2 (x − 3)2 + ( y − 5)2
Squaring both sides, we have:
32[(x + 2)2 + ( y + 2)2 ] =22[(x − 3)2 + ( y − 5)2 ]
9(x2 + 4x + 4 + y2 + 4y + 4) =4(x2 − 6x + 9 + y2 − 10y + 25)
9x2 + 36x + 36 + 9y2 + 36y + 36 =4x2 − 24x + 36 + 4y2 − 40y + 100
5x2 + 5y2 + 60x + 76y − 64 = 0 Hence, the equation of the locus of P is 5x2 + 5y 2 + 60x + 76y − 64 = 0 .
5 AB : y =
h2
x +k2
PQ : y =k + 1
3x +
h3
mAB =h2
mPQ =k + 1
3
Since the straight lines AB and PQ are perpendicular to each other,
(mAB ) (mPQ ) = −1h2
k + 13
⎛
⎝ ⎜
⎞
⎠ ⎟ = −1
h =−6
k + 1
6 The equation of PQ is
2y = −x + 4
y = −12
x + 2
∴ mPQ = −12
∴ mQR = −1
mPQ= 2
Hence, the equation of QR is y = 2x − 3. Equation of PQ: 2y = −x + 4 … Equation of QR: y = 2x − 3 …
× 2 + : 5y = 5 y = 1 From : 2(1) = −x + 4 x = 2 Hence, the coordinates of point Q are (2, 1) .
16
Paper 2 1
(a) (i) Radius of the circle
= MA= (−3 − 1)2 + (0 − 3)2
= 16 + 9= 25= 5
MR = MA(x − 1)2 + ( y − 3)2 = 5(x − 1)2 + ( y − 3)2 = 52
x2 − 2x +1+ y2 − 6y + 9 = 25x2 − 2x + y2 − 6y −15 = 0
Hence, the equation of the locus of the point R(x, y) is x
2 − 2x + y2 − 6y − 15 = 0 .
(ii) Point B(4, k) lies on the circumference of the circle.
42 − 2(4) + k 2 − 6k − 15 = 0k 2 − 6k − 7 = 0
(k − 7)(k + 1) = 0k = 7 or −1
(b) mMA =
0 − 3−3− 1
=34
∴ mAC = −1
mMA= −
134
= −43
Equation of the straight line AC is
y − 0 = −43
[x − (−3)]
3y = −4(x + 3)3y = −4x − 12
At point C (y-axis), 0=x .
3y = −4(0) −12 ⇒ y = −4∴ C = (0, − 4)
Area of ∆OAC
=12
0 −3 0 00 0 −4 0
=12
12
= 6 units2
17
2 (a)
The equation of AB in the intercept form is
x6
+y
(−3)= 1 .
(b)
2AD = DBADDB
=12
D =
2(0) + 1(6)1+ 2
, 2(−3) + 1(0)1+ 2
⎛
⎝ ⎜
⎞
⎠ ⎟ = (2, −2)
(c)
mAB =0 − (−3)
6 − 0=
12
mCD = −1
mAB= −
112
⎛
⎝ ⎜
⎞
⎠ ⎟
= −2
Hence, the equation of CD is
y − y1 = m(x − x1 )y − (−2) = −2(x − 2)
y + 2 = −2x + 4y = −2x + 2
∴ y-intercept = 2
18
3 (a) (i) The gradient of the straight line 2x − y − 5 = 0 ⇒ y = 2x − 5 is 2.
∴ mBC = 2
∴ mAB = −1
mBC= −
12
Hence, the equation of the straight line AB is:
y − y1 = m(x − x1)
y − (−3) = −12
[x − (−9)]
2( y + 3) = −(x + 9)2y + 6 = −x − 9
x + 2y + 15 = 0 (general form)
(ii) Equation of BC: 2x − y − 5 = 0 …
Equation of AB: x + 2y +15 = 0 …
+
4x − 2y −10 = 0x + 2y + 15 = 0
5x + 5 = 0∴ x = −1
K × 2K
From :
2(−1) − y − 5 = 0∴ y = −7
∴ B is point (−1, −7) .
(b)
B = (−1, −7)3(−9) + 2h
2 + 3, 3(−3) + 2k
2 + 3⎛
⎝ ⎜
⎞
⎠ ⎟ = (−1, −7)
−27 + 2h5
, −9 + 2k5
⎛
⎝ ⎜
⎞
⎠ ⎟ = (−1, −7)
Equating the x-coordinates:
−27 + 2h5
= −1
2h = 22h = 11
Equating the y-coordinates:
−9 + 2k5
= −7
2k = −26k = −13
∴D is point (11, −13). Area of ∆ADO
=12
−9 11 0 −9−3 −13 0 −3
=12
117 − (−33)
=12
150
= 75 units2
(c) Let P be point (x, y).
PA = 2[x − (−9)]2 + [y − (−3)]2 = 2
(x + 9)2 + ( y + 3)2 = 22
x2 + 18x + 81+ y2 + 6y + 9 = 4x 2 + 18x + y2 + 6y + 86 = 0
At the point A(−9, −3), x1 = −9, y1 = −3.
19
4 (a) (i) Area of ∆OAB
=12
0 −3 6 00 −5 1 0
=12
−3− (−30)
=12
27
= 13.5 units2
(ii) AB = [6 − (−3)]2 + [1− (−5)]2
= 92 + 62
= 117= 10.82 units
Area of ∆OAB = 13.512
× AB × h = 13.5
12
×10.82 × h = 13.5
h = 2.50
Hence, the perpendicular distance from O to AB is 2.50 units.
(b) (i) Let P = (x, y)
PA = 2PB(PA)2 = (2PB)2
PA2 = 4PB2
[x − (−3)]2 + [y − (−5)]2 =4[(x − 6)2 + ( y − 1)2 ]
(x + 3)2 + ( y + 5)2 =4[(x − 6)2 + ( y − 1)2 ]
x2 + 6x + 9 + y2 +10y + 25 =4(x2 −12x + 36 + y2 − 2y + 1)
x2 + 6x + 9 + y2 +10y + 25 =4x2 − 48x +144 + 4y2 − 8y + 4
0 = 3x2 − 54x + 3y2 −18y +114x 2 + y2 − 18x − 6y + 38 = 0
(ii) At the y-axis, x = 0.
∴02 − 18(0) + y2 − 6y + 38 = 0y2 − 6y + 38 = 0
b2 − 4ac= (−6)2 − 4(1)(38)= 36 − 152= −116
Since b2 − 4ac < 0 , there are no real roots and hence, the locus will not cut the y-axis.
20
Paper 1 1 σ = k
2255 km −=
−=
=−
=⎟⎟
⎠
⎞
⎜⎜
⎝
⎛−
=⎟⎟
⎠
⎞
⎜⎜
⎝
⎛−
=⎟⎟
⎠
⎞
⎜⎜
⎝
⎛−
∑∑
∑
∑
∑
∑∑
x
kmx
xkm
kxm
kxm
kn
xn
x
22
2
2
2
2
2
22
255
255
55
55
)(
)(
2 Variance = 14
x2∑n
−x∑
n
⎛
⎝
⎜ ⎜
⎞
⎠
⎟ ⎟
2
= 14
12 + 42 + k 2
3−
1 + 4 + k3
⎛
⎝ ⎜
⎞
⎠ ⎟
2
= 14
17 + k 2
3−
5 + k3
⎛
⎝ ⎜
⎞
⎠ ⎟
2
= 14
17 + k 2
3−
(5 + k )2
9= 14
3(17 + k 2 ) − (5 + k )2 = 12651 + 3k 2 − (25 + 10k + k 2 ) = 126
51 + 3k 2 − 25 − 10k − k 2 − 126 = 02k 2 − 10k − 100 = 0
k 2 − 5k − 50 = 0(k + 5)(k − 10) = 0
k = −5 or 10k = −5 is not accepted.
∴ k = 10
Form 4: Chapter 7 (Statistics) SPM Flashback
Fully-Worked Solutions
21
Paper 2 1 (a) (i) Given that x = 10,
x∑6
= 10
x = 60∑
(ii) Given that σ = 3,
∑
∑
∑
=
=−
=−
=
6542
22
22
2
9106
9)(
9
x
x
xn
x
σ
(b) If each mark is multiplied by 4 and then 5
is added to it: (i) New mean
= (4 × original mean) + 5= (4 ×10) + 5= 45
(ii) New standard deviation
= (4 × original standard deviation)= 4 × 3= 12
Hence, the new variance
= 122
= 144
2 (a) Mean, x =
x∑n
=20010
= 20
Variance, σ 2 =x2∑
n− (x)2
=440010
− 202
= 40
(b) Assume that a number k is added to the
first set of data. (i)
New mean = 20 + 2
x∑ + k
11= 22
200 + k11
= 22
k = 42
(ii) New standard deviation
=x2∑ + 422
11− 222
=4400 + 422
11− 222
= 8.739
22
3 (a) The frequency table which represents the given histogram is as follows.
Marks Frequency Cumulative frequency
10 − 24 2 2 25 − 39 6 8 40 − 54 8 16 55 − 69 13 29 70 − 84 5 34 85 − 99 2 36
The median class is given by
Tn
2
= T362
= T18 .
Thus, the median class is 55 − 69. Median
= L +
n2
− F
fm
⎛
⎝
⎜ ⎜ ⎜
⎞
⎠
⎟ ⎟ ⎟ c
= 54.5 +
362
− 16
13
⎛
⎝
⎜ ⎜ ⎜
⎞
⎠
⎟ ⎟ ⎟ (15)
= 56.81
(b) Marks f Midpoint
(x) fx fx2
10 − 24 2 17 34 578 25 − 39 6 32 192 6144 40 − 54 8 47 376 17 672 55 − 69 13 62 806 49 972 70 − 84 5 77 385 29 645 85 − 99 2 92 184 16 928
f∑= 36
fx∑
= 1977
fx2∑
= 120 939 Standard deviation, σ
=fx2∑f∑
−fx∑f∑
⎛
⎝
⎜ ⎜
⎞
⎠
⎟ ⎟
2
=120 939
36−
197736
⎛
⎝ ⎜
⎞
⎠ ⎟
2
= 18.54
Median class
23
4 (a)
Marks Frequency Cumulative frequency
20 − 29 2 2 30 − 39 4 6 40 − 49 6 (12) 50 − 59 (12) 24 60 − 69 k 24 + k 70 − 79 2 26 + k
f = 26 + k∑
If the median is 52, then the median class is 50 − 59.
Median = 52
L +
n2
− F
fm
⎛
⎝
⎜ ⎜ ⎜
⎞
⎠
⎟ ⎟ ⎟ c = 52
49.5 +
26 + k2
− 12
12
⎛
⎝
⎜ ⎜ ⎜
⎞
⎠
⎟ ⎟ ⎟ (10) = 52
26 + k − 242
12
⎛
⎝
⎜ ⎜ ⎜
⎞
⎠
⎟ ⎟ ⎟ (10) = 52 − 49.5
2 + k2
12
⎛
⎝
⎜ ⎜ ⎜
⎞
⎠
⎟ ⎟ ⎟ (10) = 2.5
2 + k24
(10) = 2.5
2 + k = 6k = 4
(b)
Modal mark = 53.75 (c) New mode = Original mode − 3
= 53.75 − 3= 50.75
Median class
24
Paper 1 1 Let OR = r cm
Perimeter of the sector ROS = 40 cm
r + r + 8 = 402r = 32r = 16
θ =
sr
=8
16= 0.5 rad.
2
Length of the major arc AB = 46.64 cmr(2π − 0.454) = 46.64
r[(2 × 3.142) − 0.454] = 46.64r(5.83) = 46.64
r = 8∴ Radius = 8 cm
3 (a) Area of sector = 20 cm2
12
r2(0.4) = 20
r2 = 100r = 10
Length of the arc AB
= 10 × 0.4= 4 cm
(b) Reflex angle AOB
= 2π − 0.4= (2 × 3.142) − 0.4= 5.884 rad.
= 5.884 ×180
3.142= 337° ′ 5
4 ∠AOB =
812
=23
rad.
Area of the shaded region
= Area of sector OAB − Area of sector AMN
=12
(12)2 23
⎛
⎝ ⎜
⎞
⎠ ⎟ −
12
(6)2 (0.7)
= 48 −12.6= 35.4 cm2
Form 4: Chapter 8 (Circular Measures) SPM Flashback
Fully-Worked Solutions
25
Paper 2 1
(a) From ∆OPB,
cos θ =9
15=
35
θ = 0.9273 rad.
(b) Area of the shaded region
= Area of sector OAB − Area of ∆OPB
=12
(15)2(0.9273) −12
× 9 ×12
= 104.32125 − 54= 50.32 cm2
OP =35
×15
= 9 cm
Make sure that your calculator is in the radian mode.
26
2 (a)
Insert a line OQ in the diagram. • Since OPQR is a rhombus,
OP = PQ = QR = RO . • Since the radius of a circle is a constant,
OP = OQ = OR.
It can now be concluded that ∆OPQ is an equilateral triangle because OP = PQ = OQ .
It can also be concluded that ∆ORQ is an equilateral triangle because OR = RQ = OQ . Therefore, ∠POQ = ∠ROQ = 60°.
Hence, ∠POR = α= 120°
= 120 ×π
180
=23
π rad.
(b) Based on ∆JOQ ,
cos ∠JOQ =OQOJ
cos 60° =8
OJ
OJ =8
cos 60°OJ = 16 cm
Length of the arc JLK
= OJ × ∠JOK
= 16 ×23
π
= 33.51 cm
(c) Area of the shaded segment
=12
r2 (α − sin α)
=12
(16)2 23
π − sin 120°⎛
⎝ ⎜
⎞
⎠ ⎟
= 157.23 cm2
27
3 (a) In ∆OQR ,
cos π3
⎛
⎝ ⎜
⎞
⎠ ⎟
r
=7
OR
cos 60° =7
OR
OR =7
cos 60°OR = 14 cm
In ∆OQR ,
tan 60° =QR7
QR = 7 × tan 60°QR = 12.1244 cm
PR = OR − OP= 14 − 7= 7 cm
Length of arc PQ
= 7 ×π3
= 7 ×3.142
3= 7.3313 cm
Hence, the perimeter of the shaded region A = PR + QR + Length of arc PQ = 7 + 12.1244 + 7.3313 = 26.46 cm
(b) Area of ∆OQR
=12
× 7 ×12.1244
= 42.4354 cm2
Area of the sector ORS
=12
×142 ×3.142
3= 102.6387 cm2
Hence, the area of the shaded region B = Area of sector ORS − Area of ∆OQR = 102.6387 − 42.4354 = 60.20 cm2
28
4 (a)
∆CAO is an isosceles triangle.
∴∠CAO = ∠COA= π −1.982= 3.142 −1.982= 1.16 rad.
(b)
Perimeter of the shaded region
= CR + Length of arc CQ + Length of arc RQ
= 8 + (10 ×1.982) + (18 ×1.16)= 48.7 cm
(c) MC = CO2 − MO2
= 102 − 42
= 84 cm
Area of shaded region
= Area of sector ARQ − Area of ∆ACO− Area of sector COQ
=12
(18)2(1.16) −12
× 8 × 84 −12
(10)2(1.982)
= 52.16 cm2
29
Paper 1 1 (a) y = 15x(3 − x) = 45x − 15x2
dydx
= 45 − 30x
When dydx
= 0,
45 − 30x = 0,x = 1.5
(b) When x = 1.5, y = 15(1.5)(3−1.5) = 33.75
d 2 ydx2
= −30 (negative)
Therefore, y is a maximum.
2 δyδx
≈dydx
δy ≈dydx
× δx
= (2x + 3) × (2.01− 2)= (2 × 2 + 3) × 0.01= 0.07
3 y = 3x3(2x −1)3
Let u = 3x3 and v = (2x −1)3
dudx
= 9x2 dvdx
= 3(2x − 1)2(2)
= 6(2x −1)2
dydx
= u dvdx
+ v dudx
= 3x3[6(2x −1)2 ]+ (2x − 1)3(9x2 )= 18x3(2x −1)2 + 9x2 (2x − 1)3
= 9x2 (2x −1)2[2x + (2x −1)]= 9x 2(2x − 1)2(4x − 1)
4 y = 4x +
4x
= 4x + 4x−1
dydx
= 4 − 4x−2 = 4 −4x2
∴dxdy
=1dydx
=1
4 − 4x2
⎛
⎝ ⎜
⎞
⎠ ⎟
Rate of change of x:
dxdt
=dxdy
×dydt
=1
4 −4x2
⎛
⎝ ⎜
⎞
⎠ ⎟
× 2
=1
4 −432
⎛
⎝ ⎜
⎞
⎠ ⎟
× 2
=9
16units s−1
5 g(x) =
16(3x − 4)2
=
16
(3x − 4)−2
′ g (x) =−26
(3x − 4)−3(3)
= −(3x − 4)−3
′ ′ g (x) = 3(3x − 4)−4 (3)= 9(3x − 4)−4
=9
(3x − 4)4
∴ ′ ′ g 13
⎛
⎝ ⎜
⎞
⎠ ⎟ =
9
3× 13
− 4⎛
⎝ ⎜
⎞
⎠ ⎟
4=
19
Form 4: Chapter 9 (Differentiation) SPM Flashback
Fully-Worked Solutions
Rate of change of y = 2 units
30
6 V =
13
πh2 (21− h)
V = 7πh2 −13
πh3
dVdh
= 14πh − πh2
dhdV
=1
14πh − πh2
Rate of change of depth of water
=dhdt
=dhdV
×dVdt
=1
14πh − πh2× 9
=1
14π (3) − π (3)2× 9
=3
11πcm s−1
7 y = (x + 3)2 = x2 + 6x + 9
dydx
= 2x + 6
If the gradient of the normal is −
16
, then the
gradient of the tangent is 6.
∴dydx
= 6
2x + 6 = 6x = 0
When x = 0, y = (0 + 3)2 = 9 Hence, the coordinates of point Q are (0, 9) .
8 y =
13
u6
y =13
(3x − 6)6
dydx
=63
(3x − 6)5(3)
= 6(3x − 6)5
9 (a) y = 3 + 14x − 2x3
dydx
= 14 − 6x2
When x = 2, dy
dx= 14 − 6(2)2 = −10
(b) δy ≈
dydx
× δx
= (−10) × [(2 + k ) − 2]= −10k
31
Paper 2 1
Using the concept of similar triangles,
r0.5
=h
0.7
r =h
0.7× 0.5
r =57
h
V =13
π r2h
V =13
π 57
h⎛
⎝ ⎜
⎞
⎠ ⎟
2
h
V =25
147πh3
dVdh
=25
147π (3h2 )
dVdh
=2549
π h2
dhdV
=49
25πh2
Rate of change of the height of the water level:
dhdt
=dhdV
×dVdt
dhdt
=49
25πh2× 0.1
dhdt
=49
25π(0.3)2× 0.1
dhdt
=49
25(3.142)(0.3)2× 0.1
= 0.6931 m s−1
2 y = 2x3 − 3x2 − 12x + 11
dydx
= 6x2 − 6x − 12
d 2 ydx2
= 12x − 6
(a) At turning point,
dydx
= 0
6x2 − 6x −12 = 0x2 − x − 2 = 0
(x + 1)(x − 2) = 0x = −1 or 2
When x = −1,
y = 2(−1)3 − 3(−1)2 −12(−1) + 11 = 18 ∴ (−1, 18) is a turning point.
When x = −1,
d 2 ydx2
= 12(−1) − 6 = −18 (negative)
∴ (−1, 18) is a maximum point.
When x = 2,
y = 2(2)3 − 3(2)2 − 12(2) + 11 = −9 ∴ (2, −9) is a turning point.
When x = 2,
d 2 ydx2
= 12(2) − 6 = 18 (positive)
∴ (2, −9) is a minimum point. (b) At point (3, 2),
dydx
= 6(3)2 − 6(3) − 12 = 24
mgradient = 24
∴ mnormal = −124
The equation of normal is
y − 2 = −124
(x − 3)
24( y − 2) = −(x − 3)24y − 48 = −x + 3
24y = −x + 51
At the x-axis, y = 0.
24(0) = −x + 51x = 51
∴ P is point (51, 0).
Rate of increase of the volume of water:
dVdh
= 0.1 m3 s−1
32
Paper 2 1 (a)
Area of ∆PRQ = 4 m2
12
(3)(3)sin∠RPQ = 4
sin∠RPQ =89
∠RPQ = 62.73°
Using the cosine rule,
RQ2 = 32 + 32 − 2(3)(3) cos 62.73°RQ2 = 9.75268RQ = 3.123 m (correct to 4
significant figures)
(b) Step 1 (Find PM where M is the midpoint
of QR)
Area of ∆PRQ = 4 m2
12
× RQ × PM = 4
12
× 3.123× PM = 4
PM = 2.5616 m
Step 2 (Determine ∠VMP )
The angle between the inclined plane VQR and the base PQR is 55°. Therefore, ∠VMP = 55°. Step 3 (Find VM) From ∆VRM , using the Pythagoras' Theorem, VM = 22 −1.56152 = 1.2497 m Step 4 (Calculate PV)
Using the cosine rule,
VP2 = 2.56162 + 1.24972 −2(2.5616)(1.2497) cos 55°
VP = 2.110 m (correct to 4significant figures)
(c) Step 1 (Find ∠VQP)
cos ∠VQP =22 + 32 − 2.1102
2(2)(3)cos ∠VQP = 0.712325cos ∠VQP = 44.58°
Step 2 (Find the area of ∆VPQ ) Area of ∆VPQ
=12
× 2 × 3× sin 44.58°
= 2.106 m2
Form 4: Chapter 10 (Solution of Triangles) SPM Flashback
Fully-Worked Solutions
33
2 (a)
(i) Based on ∆ABC , using the sine rule,
sin ∠ABC15
=sin 30°
9
sin ∠ABC =sin 30°
9×15
sin ∠ABC = 0.83333∠ABC = 56.44°
(ii) Based on ∆ADC , using the cosine
rule,
152 = 102 + 82 − 2(10)(8)cos ∠ADC
cos ∠ADC =102 + 82 −152
2(10)(8)cos ∠ADC = −0.38125
∠ADC = 112.41°
(iii) Based on ∆ABC ,
∠ACB = 180° − 30° − 56.44°= 93.56°
Area of the quadrilateral ABCD
= Area of ∆ADC + Area of ∆ABC
=12
(10)(8) sin 112.41° +
12
(15)(9) sin 93.56°
= 36.9792 + 67.3697= 104.35 cm2
(b) (i) Based on ∆ABC , since the length of BC is shorter than the length of AC and ∠BAC is an acute non-included angle, the ambiguous case will occur. Another triangle (∆A ′ B C) that can be drawn is as shown below.
(ii)
∠A ′ B C = 180° − 56.44°
= 123.56°
3 (a) In ∆PQR , using the cosine rule,
PR2 = 102 + 72 − 2(10)(7) cos 75°= 112.76533
PR = 10.62 cm
(b)
In ∆PSR , using the sine rule,
sin ∠PSR10.62
=sin 40°
8sin ∠PSR = 0.85330
∠PSR = 58.57° or 121.43°
∠PS2 R ∠PS1R
34
(c) (i) The obtuse ∠PSR is represented by
∠PS1R . In ∆PS1R ,
∠PRS1 = 180° − 40° − 121.43°
= 18.57°
In ∆PS1R , using the sine rule,
PS1
sin 18.57°=
10.62sin 121.43°
PS1 =10.62
sin 121.43°× sin 18.57°
PS1 = 3.964 cm
(ii) Area of quadrilateral PQRS
= Area of ∆PQR + Area of ∆PS1R
=12
×10 × 7 × sin 75° +
12
× 3.964 ×10.62 × sin 40°
= 47.34 cm2
4 (a) Area of ∆DBC = 29 cm2 12
× 6 ×10 × sin ∠DCB = 29
sin ∠DCB = 0.96667Basic ∠ = 75.16°∠DCB = 180° − 75.16°∠DCB = 104.84° (obtuse)
(b)
BD2 = 62 +102 − 2(6)(10) cos 104.84°BD2 = 166.73448BD = 12.91 cm
(c)
In ∆ABD,
sin ∠A12.91
=sin 45°
9.5
sin ∠A =sin 45°
9.5×12.91
sin ∠A = 0.96092∠A = 73.93°
∴∠DAB = 73.93°
and ∠D ′ A B = 180° − 73.93°
= 106.07°
Thus, ∠ABD = 180° − 45° − ∠DAB= 180° − 45° − 73.93°= 61.07°
Thus, ∠ ′ A BD = 180° − 45° − ∠D ′ A B= 180° − 45° −106.07°= 28.93°
First quadrant
Second quadrant
35
Paper 2 1
Item Price in 2000
Price in 2002
Price index
for 2002
based on 2000 (I)
Weekly expenses(weigh-tage, w)
Iw
P RMp RM1.75 140 12 1680 Q RM2.00 RM2.30 115 28 3220 R RM4.00 RM4.80 q = 120 20 2400 S RM6.00 RM7.50 125 30 3750 T RM2.50 RMr 110 10 1100
w =∑
100
Iw =∑12 150
(a) (i) For item P,
I2002 =P2002
P2000×100
140 =1.75
p×100
p = 1.25
(ii) For item R,
I2002 =P2002
P2000×100
q =4.804.00
×100
q = 120
(iii) For item T,
I2002 =P2002
P2000×100
110 =r
2.50×100
r = 2.75
(b)
I =Iw∑w∑
=12 150
100= 121.5
(c)
I =P2002
P2000×100
121.5 =P2002
RM500×100
P2002 = RM607.50
Hence, the corresponding total monthly expenses in the year 2002 was RM607.50.
(d)
Hence, I 2004 (based on the year 2000)
=100 + 20
100× I 2002
=120100
×121.5
= 145.8
Form 4: Chapter 11 (Index Numbers) SPM Flashback
Fully-Worked Solutions
+ 21.5% + 20% Year 2000 Year 2002 Year 2004
36
2 (a) (i) For item S,
I2004 =P2004
P2002×100
110 =1.50P2002
×100
P2002 =1.50110
×100
P2002 = RM1.36
Hence, the price of item S in the year 2002 was RM1.30.
(ii) For item P,
It is given that:
I2002 (based on 2000) = 105P2002
P2000×100 = 105
∴ P2000 =
P2002 ×100105
…
From the table, we can see that:
I2004 (based on 2002) = 115P2004
P2002×100 = 115
∴ P2004 =
115 × P2002
100 …
I2004 (based on 2000)
=P2004
P2000×100
=
115 × P2002
100P2002 ×100
105
×100
=115100
×105100
×100
= 120.75
Hence, the price index of item P for the year 2004 based on the year 2000 is 120.75.
(b) (i) I = 110
Iw∑w∑
= 110
(115 × 20) + (10x) + (105 × 40) + (110 × 30)
20 + 10 + 40 + 30= 110
9800 + 10x100
= 110
9800 + 10x = 11 00010x = 1200
x = 120
(ii) I 2004 =
P2004
P2002× 100
110 =22
P2002×100
P2002 =22
110×100
P2002 = 20.00
Hence, the price of a box of icecream in the year 2002 was RM20.00.
From
From
37
3 (a) I2004 (based on 2002)
=
P2004
P2002×100
For material K,
p =
1.751.40
×100 = 125
For material M,
q2
×100 = 140
q =140 × 2
100q = 2.80
For material N,
2.40r
×100 = 80
r =2.40 ×100
80r = 3.00
(b) (i)
Material I2004 (based on the year
2002)
Angle of pie chart (degrees)
w
K 125 75 15 L 150 40 8 M 140 155 31 N 80 90 18
I 2004 (based on 2002)
=Iw∑w∑
=(125 × 15) + (150 × 8) + (140 × 31) + (80 × 18)
15 + 8 + 31 + 18
=885572
= 122.99
(ii) I 2004 (based on 2002) = 122.99
P2004
P2002×100 = 122.99
RM9P2002
×100 = 122.99
P2002 =RM9 ×100
122.99= RM7.32
(c)
I 2002+ 22.99%⎯ → ⎯ ⎯ ⎯ I 2004
+ 20%⎯ → ⎯ ⎯ I 2006
(100) (122.99) (?)
I 2006 (based on 2002)
=100 + 20
100×122.99
= 147.59
38
4 (a) For component U,
I2006 = 120P2006
P2004×100 = 120
h50
×100 = 120
h = 60
(b) For component S,
I2006 = 125P2006
P2004×100 = 125
mk
×100 = 125
100m = 125k
4m = 5k … P2006 = 20 + P2004
m = 20 + k … Substituting into :
4(20 + k ) = 5k80 + 4k = 5k
k = 80
From : m = 20 + 80 = 100
(c) (i) I = 132
P2006
P2004×100 = 132
1716P2004
×100 = 132
P2004 = RM1300
(ii)
Component I2006 w U 120 1
R 4025
×100 = 160 3
S 125 4
T 4440
×100 = 110 p
I = 132(120 × 1) + (160 × 3) + (125 × 4) + 110p
1 + 3 + 4 + p= 132
1100 + 110p8 + p
= 132
1100 + 110p = 1056 + 132p44 = 22pp = 2
39
Paper 1 1 (a) k + 1, k + 5, 2k + 6 , … arithmetic progression
d = T2 − T1 = k + 5 − (k + 1) = 4d = T3 − T2 = 2k + 6 − (k + 5) = k + 1
Since common difference is always the same, k +1 = 4 ⇒ k = 3
(b) When k = 3, we have 4, 8, 12, …
S8 =
82
[2(4) + 7(4)] = 144
2 (a) T4 = 24
ar3 = 2481r3 = 24
r3 =2481
=827
r =23
(b)
S∞ =a
1− r=
81
1− 23
= 243
3 Since k, 3, 9
k, m are four consecutive terms
of a geometric progression,
3k
=m9k
3k
=mk9
mk 2 = 27
m =27k 2
4 The arithmetic progression 5, 9, 13, … has a common difference of 4.
S3 = 5732
[2a + (3 − 1)(4)] = 57
32
(2a + 8) = 57
3a +12 = 57a = 15
Hence, the three consecutive terms which sum up to 57 are 15, 19 and 23.
+4 +4 5 Volumes of water in litres:
410, 425, 440, … Volume of water at the end of the 8th day
= T8
= a + 7d= 410 + 7(15)= 515 litres
6 0.848484K
= 0.84 + 0.0084 + 0.000084 +K
=0.84
1− 0.01
=0.840.99
=8499
=2833
Form 5: Chapter 12 (Progressions) SPM Flashback
Fully-Worked Solutions
S∞ =
a1− r
These three terms are not the first three terms but any three consecutive terms with a common difference of 4. Therefore, a new value of a (first term) has to be determined.
40
7 (a) If 3, k, 48 are in an arithmetic progression, then
k − 3 = 48 − k2k = 51
k = 25.5
(b) If 3, k, 48 are in a geometric progression,
then
k3
=48k
k 2 = 144k = 12
8 (a) The common difference of the arithmetic
progression 2, 5, 8, … is 5 − 2 = 3 . (b) The sum of all the terms from the 4th term
to the 23rd term
= S23 − S3
=232
[2(2) + (23− 1)(3)]−32
[2(2) + (3− 1)(3)]
= 805 −15= 790
9 (a) The common ratio of the geometric
progression 2, 6, 18, K is 62
= 3 .
(b) Sn = 6560
a(r n −1)r − 1
= 6560
2(3n −1)3 − 1
= 6560
3n −1 = 65603n = 65613n = 38
∴ n = 8
10 138 += kT
a + 7d = 3k + 1a + 7(4) = 3k + 1
a = 3k − 27 …
S8 = 13k + 682
(2a + 7d ) = 13k + 6
8a + 28d = 13k + 68a + 28(4) = 13k + 6
8a = 13k −106 … Substituting into ,
8(3k − 27) = 13k − 10624k − 216 = 13k − 106
11k = 110k = 10
11 (a) T3 = 10
ar2 = 10 …
T3 + T4 = 1510 + T4 = 15
T4 = 5
ar3 = 5 …
: ar3
ar2=
510
r =12
From :
a 12
⎛
⎝ ⎜
⎞
⎠ ⎟
2
= 10
a 14
⎛
⎝ ⎜
⎞
⎠ ⎟ = 10
a = 40
(b)
S∞ =40
1− 12
= 80
41
Paper 2 1 (a)
T1 = Area of the first triangle
=12
bh
T2 = Area of the second triangle
=12
b2
⎛
⎝ ⎜
⎞
⎠ ⎟
h2
⎛
⎝ ⎜
⎞
⎠ ⎟ =
18
bh
T3 = Area of the third triangle
=12
b4
⎛
⎝ ⎜
⎞
⎠ ⎟
h4
⎛
⎝ ⎜
⎞
⎠ ⎟ =
132
bh
T2
T1=
18
bh
12
bh=
14
T3
T2=
132
bh
18
bh=
14
Since
T2
T1=
T3
T2=
14
(a constant), the
areas of the triangles form a geometric
progression with a common ratio of 14
.
(b) (i)
T1 =
12
(160)(80) = 6400
T2 =18
(160)(80) = 1600
T3 =1
32(160)(80) = 400
Tn = 25ar n−1 = 25
6400 14
⎛
⎝ ⎜
⎞
⎠ ⎟
n−1
= 25
14
⎛
⎝ ⎜
⎞
⎠ ⎟
n−1
=1
256
14
⎛
⎝ ⎜
⎞
⎠ ⎟
n−1
=14
⎛
⎝ ⎜
⎞
⎠ ⎟
4
n − 1 = 4n = 5
Hence, the 5th triangle has an area of 25 cm2.
(ii) Sum to infinity,
S∞ =a
1− r
=6400
1− 14
= 8533 13
cm2
42
2 (a) The number of cubes in each storey forms a geometric progression 1, 4, 16, 64, …, where a = 1 and r = 4.
Tn = 4096ar n−1 = 4096
1(4)n−1 = 40964n−1 = 46
n − 1 = 6n = 7
Hence, the height of the model is 7 × 4 = 28 cm .
(b)
S7 =1(47 −1)
4 − 1= 5461
Hence, the total price of cubes used
= 5461× 80= 436 880 sen= RM4368.80
3 (a) T5 = 320 a + 4d = 320
h + 4k = 320 …
S8 = 244082
(2a + 7d ) = 2440
4(2h + 7k) = 2440
2h + 7k = 610 …
× 2 − : k = 30 From : h + 4(30) = 320 h = 200
(b) Tn (of Epsilon) = Tn (of Sigma)
200 + (n − 1)(30) = 160 + (n − 1)(35)200 + 30n − 30 = 160 + 35n − 35
45 = 5nn = 9
43
Paper 1 1
y = px2 + qxyx
= px + q
For the point (2, 6), x = 2 and yx
= 6.
∴ 6 = p(2) + q …
For the point (10, 2), x = 10 and yx
= 2 .
∴ 2 = p(10) + q …
− : −4 = 8 p ⇒ p = −
12
From : 6 = −
12
⎛
⎝ ⎜
⎞
⎠ ⎟ (2) + q ⇒ q = 7
2
y = 7x2 − x3
yx2
= 7 − x
The straight line passes through the
point (2, h). Thus, x = 2 and y
x2= h .
yx2
= 7 − x
h = 7 − 2h = 5
The straight line passes through the
point (k, 3). Thus, x = k and y
x2= 3.
yx2
= 7 − x
3 = 7 − kk = 4
3 (a) 2mxy =
log10 y = log10 mx2
log10 y = log10 m + log10 x2
log10 y = log10 m + 2 log10 xlog10 y = 2 log10 x + log10 m
(b) (i) Y-intercept = −1
log10 m = −1m = 10−1
m =1
10
(ii) Gradient = 2
k − (−1)2 − 0
= 2
k +1 = 4k = 3
4 y = −3x3 + 4
yx3
= −3 +4x3
yx3
= 4 1x3
⎛
⎝ ⎜
⎞
⎠ ⎟ − 3
(Y = 4X + c)
By comparison, Y =
yx 3
and X =
1x 3
.
Form 5: Chapter 13 (Linear Law) SPM Flashback
Fully-Worked Solutions
Gradient Y-intercept
Divide throughout by x3.
Rearrange
44
Paper 2 1 (a)
x 1 2 3 4 5 y 1.32 1.76 2.83 5.51 13.00x2 1 4 9 16 25
log10 y 0.121 0.246 0.452 0.741 1.114 The graph of log10 y against x2 is as shown below.
(b)
y = abx2
log10 y = log10 a + x2 log10 b
(i) Y-intercept = 0.08
log10 a = 0.08a = antilog 0.08a = 1.2
(ii) Gradient =
0.74 − 0.1216 − 1
log10 b =0.6215
= 0.04133
b = antilog 0.04133b = 1.1
→↑
Non-linear
Linear
45
2 (a) x 2 4 6 8 10 12 y 5.18 11.64 26.20 58.95 132.63 298.42
log10 y 0.71 1.07 1.42 1.77 2.12 2.47
y = pk x
log10 y = log10 p + x log10 k
The graph of log10 y against x is a straight-line graph, as shown below:
(b) (i) log10 p = Y-intercept
log10 p = 0.36p = 2.29
(ii) log10 k = gradient
log10 k =2.12 − 1.42
10 − 6
log10 k =0.74
log10 k = 0.175k = 1.5
→
↑
46
3 (a) x 2.5 3.0 3.5 4.0 4.5 5.0 y 7.0 7.7 8.4 9.9 10.1 11.0
xy 17.5 23.1 29.4 39.6 45.5 55.0x2 6.3 9.0 12.3 16.0 20.3 25.0
The graph of xy against x2 is as shown below.
(b) (i) From the graph, the value of y which is incorrectly recorded is 9.9. The actual value of y is given by:
xyactual = 374(yactual) = 37
yactual = 9.25
(ii)
y = qx +p
qx
xy = qx2 +pq
q = Gradient
q =55 − 525 − 0
q = 2
pq
= Y-intercept
pq
= 5
p2
= 5
p = 10
→↑
47
4 (a) x 1 3 5 7 9 11 y 5 20 80 318 1270 5050
x + 1 2 4 6 8 10 12 log10 y 0.70 1.30 1.90 2.50 3.10 3.70
The graph of log10 y against (x + 1) is as shown below.
(b) y = hq x +1
log10 y = log10 h + (x +1) log10 qlog10 y = (x +1) log10 q + log10 h
Gradient Y-intercept
Y-intercept = 0.1log10 h = 0.1
h = 1.26
Gradient =3.7 − 0.712 − 2
log10 q = 0.3q = 2
→↑
48
Paper 1
1
3(1+ 2x)4
dx∫
= 3(1+ 2x)−4 dx∫= 3 (1+ 2x)−3
−3(2)⎛
⎝ ⎜
⎞
⎠ ⎟ + c
= −12
(1+ 2x)−3 + c
But it is given that
3(1+ 2x)4
dx∫ = k(1+ 2x)n + c , thus by
comparison, k = −
12
and n = −3.
2 Area of the shaded region = 54 units2
y dx0
k
∫ = 54
6x2 dx0
k
∫ = 54
6 x3
3⎛
⎝ ⎜
⎞
⎠ ⎟
⎡
⎣ ⎢ ⎤
⎦ ⎥ 0
k
= 54
[2x3 ]0k = 54
2(k 3 − 03 ) = 54k 3 = 27k = 3
3
(4x −1) dx−1
k
∫ = 3
4x2
2− x
⎡
⎣ ⎢ ⎤
⎦ ⎥ −1
k
= 3
[2x2 − x]−1k = 3
2k 2 − k − [2(−1)2 − (−1)] = 32k 2 − k − 3 = 32k 2 − k − 6 = 0
(2k + 3)(k − 2) = 0
k = −32
or 2
k = −32
is not accepted
∴ k = 2
4
f (x) dx1
2
∫ + [ f (x) + cx] dx2
3
∫ = 30
f (x) dx1
2
∫ + f (x) dx2
3
∫ + cx dx2
3
∫ = 30
f (x) dx1
3
∫ + c x2
2⎡
⎣ ⎢ ⎤
⎦ ⎥ 2
3
= 30
5 + c 32 − 22
2⎛
⎝ ⎜
⎞
⎠ ⎟ = 30
52
c = 25
c = 10
5
12
f (x) dxa
b
∫ =12
f (x) dxa
b
∫ =12
(−6) = −3
Form 5: Chapter 14 (Integration) SPM Flashback
Fully-Worked Solutions
The area of the region below the x-axis has a negative sign.
49
6
[nx − g(x)] dx = 91
4
∫
nx dx1
4
∫ − g(x) dx1
4
∫ = 9
n x2
2⎡
⎣ ⎢ ⎤
⎦ ⎥ 1
4
+ g(x) dx4
1
∫ = 9
n 42
2−
12
2⎛
⎝ ⎜
⎞
⎠ ⎟ − 6 = 9
152
n − 6 = 9
152
n = 15
n = 2
g(x) dxa
b
∫ = − g(x) dxb
a
∫
50
Paper 2
1 (a) dydx
= 2x + 4
y = (2x + 4) dx∫y =
2x2
2+ 4x + c
y = x2 + 4x + c
y = 7 when x = 1 :7 = 12 + 4(1) + cc = 2
∴ y = x 2 + 4x + 2
(b) x2 d 2 y
dx2+ (x − 1) dy
dx+ y + 3 = 0
x2 (2) + (x −1)(2x + 4) + (x2 + 4x + 2) + 3 = 02x2 + 2x2 + 2x − 4 + x2 + 4x + 2 + 3 = 0
5x2 + 6x +1 = 0(5x + 1)(x +1) = 0
x = −15
or −1
2 8y = 3x
x =
83
y …
x = y2 −1 … Substitute into :
83
y = y2 − 1
8y = 3y2 − 33y2 − 8y − 3 = 0
(3y +1)( y − 3) = 0
y = −13
or 3
It is obvious that for point A, y = 3. From , when y = 3,
x =83
y =83
(3) = 8
∴ A = (8, 3)
Volume of the solid generated when the shaded region is revolved through 360° about the y-axis = Volume of the cone generated by the line
segment OA − Volume of the solid generated by the curve x = y2 − 1 from y = 1 to y = 3
=13
πr2h − π x2 dy1
3
∫=
13
π (8)2 (3) − π ( y2 − 1)2 dy1
3
∫= 64π − π ( y4 − 2y2 +1) dy
1
3
∫= 64π − π y5
5−
2 y3
3+ y
⎡
⎣ ⎢ ⎤
⎦ ⎥ 1
3
= 64π − π 35
5−
2(3)3
3+ 3−
15
−23
+ 1⎛
⎝ ⎜
⎞
⎠ ⎟
⎡
⎣ ⎢ ⎤
⎦ ⎥
= 64π − π 2435
−18 + 3−15
+23
− 1⎛
⎝ ⎜
⎞
⎠ ⎟
= 64π −49615
π
= 30 1415
π units3
51
3 (a) dydx
= 3x2 − 4x
y = (3x2 − 4x) dx∫y =
3x3
3−
4x2
2+ c
y = x3 − 2x2 + c
The curve passes through the point A(1, −9).
−9 = 13 − 2(1)2 + cc = −8
Hence, the equation of the curve is y = x3 − 2x2 − 8.
(b) At turning points, dydx
= 0.
dydx
= 0
3x2 − 4x = 0x(3x − 4) = 0
x = 0 or 1 13
d 2 ydx2
= 6x − 4
When x = 0, y = (0)3 − 2(0)2 − 8 = −8 ∴ (0, −8) is a turning point.
d 2 ydx2
= 6(0) − 4 = −4 (negative)
Hence, (0, −8) is a maximum point.
When x = 1 1
3,
y =
43
⎛
⎝ ⎜
⎞
⎠ ⎟
3
− 2 43
⎛
⎝ ⎜
⎞
⎠ ⎟
2
− 8 = −9 527
∴ 1 1
3, −9 5
27⎛
⎝ ⎜
⎞
⎠ ⎟ is a turning point.
d 2 ydx2
= 6 43
⎛
⎝ ⎜
⎞
⎠ ⎟ − 4 = 4 (positive)
Hence, 1 1
3, −9 5
27⎛
⎝ ⎜
⎞
⎠ ⎟ is a minimum point.
52
4 (a) y =
4(2x − 1)2
= 4(2x −1)−2
dydx
= −8(2x −1)−3(2)
= −16
(2x −1)3
At the point A(1, 4),
m =dydx
= −16
[2(1) − 1]3= −16
Hence, the equation of the tangent at the point A(1, 4) is
y − y1 = m(x − x1 )y − 4 = −16(x − 1)y − 4 = −16x + 16
y = −16x + 20
(b)
(i) Area of the shaded region
= y dx2
3
∫=
4(2x − 1)22
3
∫ dx
= 4(2x −1)−22
3
∫ dx
=4(2x −1)−1
−1(2)⎡
⎣ ⎢ ⎤
⎦ ⎥ 2
3
= −2
2x − 1⎡
⎣ ⎢ ⎤
⎦ ⎥ 2
3
= −2
2(3) − 1− −
22(2) −1
⎛
⎝ ⎜
⎞
⎠ ⎟
= −25
− −23
⎛
⎝ ⎜
⎞
⎠ ⎟
=4
15 units2
(ii) Volume generated, Vx
= π y2 dx2
3
∫= π 4
(2x − 1)2
⎛
⎝ ⎜
⎞
⎠ ⎟
2
dx2
3
∫= π 16
(2x − 1)4dx
2
3
∫= π 16(2x − 1)−4 dx
2
3
∫= π 16(2x −1)−3
−3(2)⎡
⎣ ⎢ ⎤
⎦ ⎥ 2
3
= π 8−3(2x − 1)3
⎡
⎣ ⎢ ⎤
⎦ ⎥ 2
3
= −83
π 1(2x − 1)3
⎡
⎣ ⎢ ⎤
⎦ ⎥ 2
3
= −83
π 1[2(3) − 1]3
−1
[2(2) − 1]3
⎛
⎝ ⎜
⎞
⎠ ⎟
= −83
π 1125
−127
⎛
⎝ ⎜
⎞
⎠ ⎟
=784
10125π units3
53
5 (a) The gradient of the straight line x + y − 8 = 0 ⇒ y = −x + 8 is −1.
The gradient function is dydx
= px3 + x2.
dydx
= −1
px3 + x2 = −1p(−1)3 + (−1)2 = −1
−p +1 = −1p = 2
(b) dydx
= px3 + x2 = 2x3 + x2
y = (2x3 + x2 ) dx∫y =
2x4
4+
x3
3+ c
y =x4
2+
x3
3+ c
At the point (−1, 0), x = −1, y = 0
∴ 0 =(−1)4
2+
(−1)3
3+ c
0 =12
−13
+ c
c = −16
Hence, the equation of the curve is
y =
x4
2+
x 3
3−
16
6 (a) Equation of RAQ: 2y = −x + 10 At point Q (on the x-axis), y = 0.
2(0) = −x + 10x = 10
∴ Q = (10, 0)
Area of the shaded region = Area under the curve from x = 0 to
x = 2 + Area under the straight line AQ
=x2
2+ 2
⎛
⎝ ⎜
⎞
⎠ ⎟ dx
0
2
∫ +12
× 8 × 4⎛
⎝ ⎜
⎞
⎠ ⎟
=x3
6+ 2x
⎡
⎣ ⎢ ⎤
⎦ ⎥ 0
2
+ 16
=23
6+ 2(2) − 0 +16
= 21 13
units2
(b)
Volume generated = Volume generated by the curve from
y = 2 to y = 4
= π (2y − 4) dy2
4
∫= π[y2 − 4y]2
4
= π[42 − 4(4) − (22 − 4(2))]= 4π units3
x2
2= y − 2
x2 = 2y − 4
At the point (−1, 0), x = −1.
54
7 (a) y = x − 4 …
x = ( y − 2)2 … Substituting into :
y = ( y − 2)2 − 4y = y2 − 4y + 4 − 4y = y2 − 4y0 = y2 − 5y0 = y( y − 5)y = 0 or 5
From : When y = 0, x = (0 − 2)2 = 4 ∴ B = (4, 0) When y = 5, x = (5 − 2)2 = 9 ∴ A = (9, 5)
(b)
Area of the shaded region P = Area of trapezium − Area under the
curve
=12
(4 + 9)(5) − x dy0
5
∫= 32.5 − ( y − 2)2 dx
0
5
∫= 32.5 −
( y − 2)3
3(1)⎡
⎣ ⎢ ⎤
⎦ ⎥ 0
5
= 32.5 −(5 − 2)3
3−
(0 − 2)3
3⎛
⎝ ⎜
⎞
⎠ ⎟
= 20 56
units2
(c) Vy = π x2 dy
0
2
∫
= π ( y − 2)4 dy0
2
∫= π ( y − 2)5
5(1)⎡
⎣ ⎢ ⎤
⎦ ⎥ 0
2
= π (2 − 2)5
5−
(0 − 2)5
5⎛
⎝ ⎜
⎞
⎠ ⎟
= π 0 − −325
⎛
⎝ ⎜
⎞
⎠ ⎟
⎡
⎣ ⎢ ⎤
⎦ ⎥
= 6 25
π units3
55
Paper 1
1 (a) Since P = (4, 2), OP = 4
2⎛ ⎝ ⎜
⎞ ⎠ ⎟ .
(b) Since Q = (−6, 3), OQ = −6i + 3 j .
Hence, QO = −OQ = 6i − 3 j . 2 r = 2p − 3q
r = 2(3a + 2b) − 3(3a − b)r = 6a + 4b − 9a + 3br = −3a + 7b
But it is given that r = ha + (h + k )b . By comparison,
h = −3 and h + k = 7−3+ k = 7
k = 10
3
(a)
BD = BA + AD= −12 p + 6q
(b) EC = EB + BC
=13
DB + BC
=13
(−BD) + BC
=13
(12 p − 6q) + 6q
= 4p − 2q + 6q
= 4 p + 4q
4 (a) AB = OB − OA
= 3i + 17 j − (−2i + 5 j)
= 5i + 12 j
(b)
AB = 52 +122 = 169 = 13
Unit vector in the direction of AB
=1
AB( AB)
=1
13(5i +12 j)
=5
13i +
1213
j
5 AB + 2BC = 11i −13 j
(OB − OA) + 2(OC − OB) = 11i − 13 j
53
⎛ ⎝ ⎜
⎞ ⎠ ⎟ − −1
7⎛ ⎝ ⎜
⎞ ⎠ ⎟ + 2 m
p⎛ ⎝ ⎜
⎞ ⎠ ⎟ − 5
3⎛ ⎝ ⎜
⎞ ⎠ ⎟
⎡ ⎣ ⎢
⎤ ⎦ ⎥ =
11−13
⎛ ⎝ ⎜
⎞ ⎠ ⎟
6−4
⎛ ⎝ ⎜
⎞ ⎠ ⎟ + 2 m − 5
p − 3⎛ ⎝ ⎜
⎞ ⎠ ⎟ = 11
−13⎛ ⎝ ⎜
⎞ ⎠ ⎟
6 + 2m − 10−4 + 2 p − 6
⎛ ⎝ ⎜
⎞ ⎠ ⎟ = 11
−13⎛ ⎝ ⎜
⎞ ⎠ ⎟
2m − 42 p −10
⎛ ⎝ ⎜
⎞ ⎠ ⎟ = 11
−13⎛ ⎝ ⎜
⎞ ⎠ ⎟
Equating the x-components:
2m − 4 = 11
m = 7 12
Equating the y-components:
2p − 10 = −13
p = −1 12
Form 5: Chapter 15 (Vectors) SPM Flashback
Fully-Worked Solutions
56
6 (a) OQ = 9
−12⎛ ⎝ ⎜
⎞ ⎠ ⎟
(b)
OQ = 92 + (−12)2 = 15
Hence, the unit vector in the direction of
OQ =1
159
−12⎛ ⎝ ⎜
⎞ ⎠ ⎟ =
35
− 45
⎛
⎝
⎜ ⎜ ⎜
⎞
⎠
⎟ ⎟ ⎟
.
7 BM = BA + AM
= CO +12
AO
= (4i − 2 j) +12
(−2i − 6 j)
= 4i − 2 j − i − 3 j
= 3i − 5 j
8 (a) A(6, 3) ⇒ OA = 6
3⎛ ⎝ ⎜
⎞ ⎠ ⎟
B(0, −5) ⇒ OB = 0
−5⎛ ⎝ ⎜
⎞ ⎠ ⎟
AB = OB − OA
= 0−5
⎛ ⎝ ⎜
⎞ ⎠ ⎟ − 6
3⎛ ⎝ ⎜
⎞ ⎠ ⎟
= −6−8
⎛ ⎝ ⎜
⎞ ⎠ ⎟
(b)
AB = (−6)2 + (−8)2 = 100 = 10
Unit vector in the direction of AB
=1
AB( AB)
=1
10−6−8
⎛ ⎝ ⎜
⎞ ⎠ ⎟
=−
35
−45
⎛
⎝
⎜ ⎜ ⎜
⎞
⎠
⎟ ⎟ ⎟
9 (a) Since the point X, Y and Z are collinear,
XY = kYZ2a + 4b = k(3a + mb)2a + 4b = 3ka + kmb
Equating the coefficients of a ,
3k = 2
k =23
Equating the coefficients of b ,
km = 423
m = 4
m = 6
(b) XY =
23
YZ
XY =23
YZ
XY
YZ=
23
XY : YZ = 2 : 3
k =
23
57
Paper 2 1 (a) Using the concept of position vectors:
AB = 1014
⎛ ⎝ ⎜
⎞ ⎠ ⎟
OB − OA = 1014
⎛ ⎝ ⎜
⎞ ⎠ ⎟
46
⎛ ⎝ ⎜
⎞ ⎠ ⎟ − OA = 10
14⎛ ⎝ ⎜
⎞ ⎠ ⎟
OA = 46
⎛ ⎝ ⎜
⎞ ⎠ ⎟ − 10
14⎛ ⎝ ⎜
⎞ ⎠ ⎟
OA = −6−8
⎛ ⎝ ⎜
⎞ ⎠ ⎟
∴ A = (−6, − 8)
(b)
OA = (−6)2 + (−8)2 = 100 = 10
Unit vector in the direction of OA
=1
OA(OA)
=1
10−6−8
⎛ ⎝ ⎜
⎞ ⎠ ⎟
=− 3
5− 4
5
⎛
⎝ ⎜
⎞
⎠ ⎟
(c) Since CD is parallel to AB, CD = k AB (k is a constant).
m7
⎛ ⎝ ⎜
⎞ ⎠ ⎟ = k 10
14⎛ ⎝ ⎜
⎞ ⎠ ⎟
m7
⎛ ⎝ ⎜
⎞ ⎠ ⎟ = 10k
14k⎛ ⎝ ⎜
⎞ ⎠ ⎟
Equating the x-components: m = 10k … Equating the y-components:
7 = 14k
k =12
From , when k =
12
,
m = 10 1
2⎛
⎝ ⎜
⎞
⎠ ⎟ = 5
58
2
(a) (i)
AP = AO + OP
AP = −4a + 4 p
(ii) OQ = OA + AQ
OQ = OA +14
AB
OQ = OA +14
( AO + OB)
OQ = OA +14
( AO + 3OP)
OQ = 4a +14
[−4a + 3(4 p)]
OQ = 4a − a + 3p
OQ = 3a + 3 p
(b) (i)
AR = hAP = h(−4a + 4p)
AR = −4ha + 4h p
(ii) RQ = kOQ = k(3a + 3p)
RQ = 3ka + 3k p
(c) AQ = AR + RQ
−a + 3p = −4ha + 4hp + 3ka + 3k p
−a + 3p = (−4h + 3k )a + (4h + 3k )p
Equating the coefficients of a and of b respectively, we have:
+
−4h + 3k = −14h + 3k = 3
6k = 2
k = 13
KK
From : −4h + 3 1
3⎛
⎝ ⎜
⎞
⎠ ⎟ = −1
−4h = −2
h =12
Given that
OP =
13
OB ,
∴ OB = 3OP .
Given AQ =
14
AB
From (a) (ii),
AQ = −a + 3p
59
3 (a)
(i)
AC = AO + OC= −6a + 8b
(ii)
BD = BC + CD= 5b + (8a − 5b)= 8a
(b) If the points A, E and C are collinear, then
AC = k AE
−6a + 8b = k( AO + OB + BE)
−6a + 8b = k(−6a + 3b + hBD)−6a + 8b = k[−6a + 3b + h(8a)]−6a + 8b = −6ka + 3kb + 8hka−6a + 8b = (−6k + 8hk )a + 3kb
Equating the coefficients of b :
3k = 8
k =83
Equating the coefficients of a :
−6 = −6k + 8hk
−6 = −6 83
⎛
⎝ ⎜
⎞
⎠ ⎟ + 8h 8
3⎛
⎝ ⎜
⎞
⎠ ⎟
−18 = −48 + 64h30 = 64h
h =1532
(c)
OC = 8b = 8b = 8(2) = 16
OA = 6a = 6a = 6(2) = 12
AC = OC2
+ OA2
= 162 + 122
= 20
Given that OB =
38
OC , thus OC =
83
OB .
Therefore, OC =
83
OB =83
(3b) = 8b .
This gives BC = 8b − 3b = 5b.
60
4
(a) PR = PQ + QR
= 4u +32
PS
= 4u +32
(12v)
= 4u + 18v
(b) (i) TX = mPQ
= m(4u)= 4mu
(ii) P, X and R are collinear.
PX = k PR
PT + TX = k(4u +18v)34
PS + 4mu = 4ku +18kv
34
(12v) + 4mu = 4ku +18kv
9v + 4mu = 4ku +18kv
Equating the coefficients of v ,
9 = 18k
k =12
Equating the coefficients of u ,
4m = 4k
4m = 4 12
⎛
⎝ ⎜
⎞
⎠ ⎟
m =12
Given that
PS =
23
QR ,
thus
QR =
32
PS .
Given From (b) (i)
61
Paper 1 1
(a) cot (−θ )
=1
tan (−θ )
=1
−tan θ
= −1t
(b) cos (90° − θ )
= sin θ
=t
1+ t 2
2 6 sec2 θ − 20 tan θ = 0
6(1+ tan2 θ ) − 20 tan θ = 06 + 6 tan2 θ − 20 tan θ = 06 tan2 θ − 20 tan θ + 6 = 03 tan2 θ −10 tan θ + 3 = 0(3 tan θ −1)(tan θ − 3) = 0
tan θ =13
or 3
When tan θ =13
,
θ = 18.43°, 198.43°
When tan θ = 3,θ = 71.57°, 251.57°
∴θ = 18.43°, 71.57°, 198.43°, 251.57°
3 2 sin2 x + cos x = 1
2(1− cos2 x) + cos x −1 = 02 − 2 cos2 x + cos x −1 = 0−2 cos2 x + cos x +1 = 0
2 cos2 x − cos x −1 = 0(2 cos x +1)(cos x −1) = 0
cos x = −
12
or cos x = 1
When cos x = −12
,
x = 120°, 240°
When cos x = 1,x = 0°, 360°
∴ x = 0°, 120°, 240°, 360°
4 cos 2θ − 3 sin θ = 2
1− 2 sin2 θ − 3 sin θ − 2 = 0−2 sin2 θ − 3 sin θ −1 = 0
2 sin2 θ + 3 sin θ +1 = 0(2 sin θ +1)(sin θ +1) = 0
sin θ = −
12
or sin θ = −1
When sin θ = −12
,
Basic ∠ = 30°θ = 210°, 330°
When sin θ = −1,θ = 270°
∴θ = 210°, 270°, 330°
Form 5: Chapter 16 (Trigonometric Functions) SPM Flashback
Fully-Worked Solutions
tan θ = t
62
5 15 cos2 x − 7 cos x = 4 cos 60°
15 cos2 x − 7 cos x = 4(0.5)15 cos2 x − 7 cos x = 2
15 cos2 x − 7 cos x − 2 = 0(3 cos x − 2)(5 cos x +1) = 0
cos x =
23
or cos x = −15
When cos x =23
Basic ∠ = 48.19°∴ x = 48.19°, 311.81°
When cos x = −15
Basic ∠ = 78.46°∴ x = 101.54°, 258.46°
∴ x = 48.19°, 101.54°, 258.46°, 311.81°
63
Paper 2 1 (a) LHS
= tan x2+ cot x
2
=sin x
2
cos x2
+cos x
2
sin x2
=sin2 x
2+ cos2 x
2
sin x2
cos x2
=1
sin x2
cos x2
=2
2 sin x2
cos x2
=2
sin x= 2 cosec x= RHS
(b) (i)
(ii) sin 3
2x =
34π
x −1
2 sin 3
2x =
32π
x − 2
The solutions to the equation
2 sin 3
2x =
32π
x − 2 are given by the
x-coordinates of the intersection
points of the graphs of y = 2 sin 3
2x
and y =
32π
x − 2.
Hence, the equation of the straight line for solving the equation
sin 3
2x =
34π
x −1 is y =
32π
x − 2 .
Number of solutions = Number of intersection points = 3
This is a y = sin θ graph
with 1 1
2 cycles
because θ = 1 1
2x.
x 0 2π y −2 1
64
2 (a) The sketch of the graph of y = cos 2x for 0 ≤ x ≤ π is as shown below:
(b) 2 sin2 x = 2 − x
π
1− cos 2x = 2 − xπ
−cos 2x = 1− xπ
cos 2x =xπ−1
The straight line that has to be drawn is
y = x
π− 1.
x 0 π y −1 0
Hence, the number of solutions to the
equation 2 sin2 x = 2 − x
π for 0 ≤ x ≤ π
= Number of intersection points = 2
3 (a) LHS = −2 cos2 x + cosec2 x − cot2 x
= −2 cos2 x +1
= −(2 cos2 x −1)= −cos 2x= RHS
(b) (i) The sketch of the graph of
y = −cos 2x is as shown below.
(ii) 2(−2 cos2 x + cosec2 x − cot2 x) = x
π−1
2(−cos 2x) = xπ−1
−cos 2x =x
2π−
12
The straight line that has to be
sketched is y =
x2π
−12
.
x 0 2π
y −
12
12
Number of solutions = Number of intersection points = 4
If cot2 x +1 = cosec2 x , then cosec2 x − cot2 x = 1.
cos 2x = 1− 2 sin2 x∴ 2 sin2 x = 1− cos 2x
65
4 (a), (b)
sin 2x + 2 cos x = 0
sin 2x = −2 cos x
Number of solutions = Number of intersection points = 2
66
Paper 1 1 Number of codes that can be formed
= 6P3 × 4P2 = 120×12 = 1440. 2 (a) Number of teams that can be formed if
each team consists of 3 boys (and 5 girls)
= 7C3 × 6C5
= 35× 6= 210
(b) Available: 7 boys and 6 girls
Needed: 7 boys and 1 girl or 8 boys and 0 girls Hence, the number of teams that can be formed if each team consists of not more than 1 girl = 7C7 × 6C1 = 6
3 (a) Number of arrangements = 5! = 120 (b) Step 1
If the letter 'O ' and 'E ' have to be side by side, they will be counted as 1 item. Together with the letters 'P ', 'W ' and 'R ', there are altogether 4 items, as illustrated below:
Number of arrangements = 4! Step 2 But the letters 'O ' and 'E ' can also be arranged among themselves in their group.
Number of arrangements = 2!
Step 3 Hence, using the rs multiplication principle, the number of arrangements of all the letters of the word 'POWER ' in which the letters O and E have to be side by side = 4!× 2!= 24 × 2 = 48
4 (a) If there is no restriction, the number of
ways the dancing groups can be formed
= 16C6
= 8008
(b) If the group of dancers must consist of 1
Form Five student and exactly 2 Form Six students, the number of Form Four students required is 3. Hence, the number of ways the dancing groups can be formed
= 7C3 × 5C1 × 4C2
= 1050
5 Number of arrangements
N 7P3
T 7P3
G 7P3
R 7P3
L 7P3
Hence, the total number of arrangements = 7P3 × 5 = 1050
Form 5: Chapter 17 (Permutations and Combinations) SPM Flashback
Fully-Worked Solutions
7C7 × 6C1 Impossible
7 Form Four + 5 Form Five + 4 Form Six students
67
Paper 1
1 P(Blue) =
25
k6 + k
=25
5k = 12 + 2k3k = 12k = 4
2 P(both students participate in the same game)
= P(TT or BB or HH )= P(TT ) + P(BB) + P( HH )
=5
14×
413
⎛
⎝ ⎜
⎞
⎠ ⎟ +
614
×5
13⎛
⎝ ⎜
⎞
⎠ ⎟ +
314
×2
13⎛
⎝ ⎜
⎞
⎠ ⎟
=4
13
3 Let A − Azean D − Dalilah N − Nurur P(only one of them will qualify)
= P( A D N ) + P( A D N ) + P( A D N )
=13
×25
×47
⎛
⎝ ⎜
⎞
⎠ ⎟ +
23
×35
×47
⎛
⎝ ⎜
⎞
⎠ ⎟ +
23
×25
×37
⎛
⎝ ⎜
⎞
⎠ ⎟
=8
105+
835
+435
=44
105
Form 5: Chapter 18 (Probability) SPM Flashback
Fully-Worked Solutions
Initially, there are 5 students playing table tennis out of 14 students.
After 1 student is chosen, it is left with 4 students playing table tennis out of 13 students.
68
Paper 1 1
P(Z > k ) = 0.5 − 0.2580 = 0.2420
2 X − Number of students who passed
X ~ B(7, 0.6)
P( X = 5)= 7C5(0.6)5 (0.4)2
= 0.2613
3 (a) X ~ N (4.8, 1.22 )
Z =X − µ
σ
Z =4.2 − 4.8
1.2Z = −0.5
(b) P(4.8 < X < 5.2)
= P 4.8 − 4.81.2
< Z <5.2 − 4.8
1.2⎛
⎝ ⎜
⎞
⎠ ⎟
= P(0 < Z < 0.333)= 0.5 − 0.3696= 0.1304
4 X − Volume, in ml X ~ N(650, 252)
(a) Z =
25
X − 65025
=25
X − 650 = 10X = 660
Hence, the volume which is equivalent to
the standard score of 25
is 660 ml .
(b) P( X > 620)
= P Z >620 − 650
25⎛
⎝ ⎜
⎞
⎠ ⎟
= P(Z > −1.2)= 1− 0.1151= 0.8849
Hence, the percentage of bottles of soy sauce that have volumes of more than 620 ml
= (0.8849 ×100)%= 88.49%
5 P(Z > m) = 0.5 − 0.1985
P(Z > m) = 0.3015
∴ m = 0.52
Form 5: Chapter 19 (Probability Distributions) SPM Flashback
Fully-Worked Solutions
69
Paper 2 1 (a) X − Number of children
X ~ B(8, 0.4)
(i) P( X ≥ 2)
= 1− P( X = 0) − P( X = 1)= 1− 8C0 (0.4)0 (0.6)8 − 8C1(0.4)1 (0.6)7
= 1− 0.01680 − 0.08958= 0.8936
(ii) Variance = 192
npq = 192n(0.4)(0.6) = 192
n = 800
Hence, the population of the housing estate is 800.
(b) X − Mass of a worker
X ~ N(65, 62)
P(58 < X < 70)
= P 58 − 656
< Z <70 − 65
6⎛
⎝ ⎜
⎞
⎠ ⎟
= P(−1.167 < Z < 0.833)
= 1− Q(1.167) − Q(0.833)= 1− 0.1216 − 0.2025= 0.6759
Number of workers whose masses are between 58 kg and 70 kg = 142 P(−1.167 < Z < 0.833) × N = 142
0.6759 × N = 142
N =142
0.6759N = 210
Hence, the total number of workers = 210.
2 (a) X − Number of goals
X − B(8, p)
(i) Mean = 6.4
np = 6.48p = 6.4
p = 0.8
(ii) P( X ≥ 3)
= 1− P( X = 0) − P( X = 1) − P( X = 2)= 1− 8C0 (0.8)0 (0.2)8 − 8C1(0.8)1 (0.2)7
− 8C2(0.8)2 (0.2)6
= 0.9988
(b) X − Body-mass of a student
X ~ N(52, 122)
(i) P( X < 40)
= P Z <40 − 52
12⎛
⎝ ⎜
⎞
⎠ ⎟
= P(Z < −1)= 0.1587
(ii) P( X > m) = 8%
P Z >
m − 5212
⎛
⎝ ⎜
⎞
⎠ ⎟ = 0.08
m − 5212
= 1.406
m = 68.872
70
3 (a) X − Number of students who scored a distinction in Mathematics X ~ B(n, p) X ~ B(8, 0.7)
(i)
P( X = 3) = 8C3(0.7)3 (0.3)5
= 0.04668
(ii) P( X < 3)
= P( X = 0) + P( X = 1) + P( X = 2)= 8C0 (0.7)0 (0.3)8 + 8C1(0.7)1 (0.3)7
+ 8C2 (0.7)2 (0.3)6
= 0.0113
(b) X − Volume of milk, in ml
X ~ N(µ, σ 2) X ~ N(1000, 202)
(i) P( X < 1050)
= P Z <1050 − 1000
20⎛
⎝ ⎜
⎞
⎠ ⎟
= P(Z < 2.5)= 1− 0.00621= 0.9938
(ii) P( X > v) = 0.7
P Z >
v − 100020
⎛
⎝ ⎜
⎞
⎠ ⎟ = 0.7
v − 100020
= −0.524
v = 989.52 ml
v = 0.9895 l
4 X − Mass of a pineapple, in kg X ~ N(1.3, 0.22)
(a) P(grade A)
= P( X > 1.4)
= P Z >1.4 − 1.3
0.2⎛
⎝ ⎜
⎞
⎠ ⎟
= P(Z > 0.5)= 0.3085
(b) P(grade B)
= P(1.2 < x < 1.4)
= P 1.2 − 1.30.2
< Z <1.4 − 1.3
0.2⎛
⎝ ⎜
⎞
⎠ ⎟
= P(−0.5 < Z < 0.5)= 1− 0.3085 − 0.3085= 0.383
Hence, the number of grade B pineapples
= 0.383 ×1000= 383
(c) P( X > m) = 93.32%
P Z >
m − 1.30.2
⎛
⎝ ⎜
⎞
⎠ ⎟ = 0.9332
m − 1.30.2
= −1.5
m = 1.0
71
Paper 2 1 (a) (i)
v = a dt∫
v = (8 − 2t ) dt∫v = 8t −
2t 2
2+ c
v = 8t − t 2 + c
When t = 0, v = 20 m s−1. Thus, c = 20. ∴ v = 8t − t2 + 20 At the maximum velocity,
dvdt
= 0
8 − 2t = 0t = 4
d 2vdt 2
= −2 (negative)
Therefore, v is a maximum. When t = 4, v(max.) = 8(4) − 42 + 20 = 36 m s−1
(ii) When v = 0,
8t − t 2 + 20 = 0,t 2 − 8t − 20 = 0
(t + 2)(t − 10) = 0t = −2 or 10t = −2 is not accepted
∴ t = 10∴ n = 10
(b)
Total distance travelled during the period of 0 ≤ t ≤ 10
= Area below the v − t graph
= v dt0
10
∫= (8t − t2 + 20) dt
0
10
∫=
8t2
2−
t3
3+ 20t
⎡
⎣ ⎢ ⎤
⎦ ⎥ 0
10
= 4t 2 −t3
3+ 20t
⎡
⎣ ⎢ ⎤
⎦ ⎥ 0
10
= 4(10)2 −103
3+ 20(10) − 0
= 266 23
m
Form 5: Chapter 20 (Motion Along a Straight Line) SPM Flashback
Fully-Worked Solutions
72
2 (a) v = 3t(4 − t ) = 12t − 3t2
dvdt
= 12 − 6t
At maximum velocity,
dvdt
= 0
12 − 6t = 0t = 2
d 2vdt 2
= −6 (negative)
Hence, v is a maximum. ∴ vmax = 12(2) − 3(2)2 = 12 m s−1
(b)
s = v dt∫
s = (12t − 3t 2 ) dt∫s =
12t 2
2−
3t 3
3+ c
s = 6t2 − t3 + c
When t = 0, s = 0. Thus, c = 0. ∴ s = 6t2 − t 3 Distance travelled during the 3rd second
= s3 − s2
= [6(3)2 − 33]− [6(2)2 − 23]
= 27 − 16
= 11
= 11 m
(c) When the particle passes through point O again, s = 0.
When s = 0,6t 2 − t 3 = 0
t 2 (6 − t) = 0t = 0 or 6t = 0 is not accepted
∴ t = 6
(d) When a particle reverses its direction, it is
at instantaneous rest, i.e. v = 0.
When v = 0,3t(4 − t ) = 0
t = 0 or 4t = 0 is not accepted
∴ t = 4
73
3 (a) vP = 4 + 2t − 2t2 When object P travels at a maximum
velocity, dvP
dt= 0 .
When dvP
dt= 0,
2 − 4t = 0
t =12
d 2vP
dt 2= −4 (negative)
Hence, the velocity of object P is a maximum.
∴ vP (max) = 4 + 2 12
⎛
⎝ ⎜
⎞
⎠ ⎟ − 2 1
2⎛
⎝ ⎜
⎞
⎠ ⎟
2
= 4 12
m s−1
(b) At point C, vP = 0.
4 + 2t − 2t 2 = 02 + t − t 2 = 0t 2 − t − 2 = 0
(t − 2)(t +1) = 0
t = 2 or −1 t = −1 is not accepted. ∴ t = 2 For object P,
sP = v P dt∫sP = (4 + 2t − 2t 2 ) dt∫sP = 4t + t2 −
2t 3
3+ c
When t = 0, sP = 0. Thus, c = 0.
∴ sP = 4t + t2 −
2t 3
3
When t = 2,
sP = 4(2) + 22 −2(2)3
3
sP = 6 23
m
For object Q, sQ = (vQ × t) + 24 sQ = −7t + 24
When t = 2,sQ = −7(2) + 24
= 10 m
Hence, the distance between object P and
object Q is 10 − 6 2
3= 3 1
3 m
(c) When object P and object Q meet,
sP = sQ
4t + t2 −2t3
3= −7t + 24
12t + 3t2 − 2t3 = −21t + 722t 3 − 3t 2 − 33t + 72 = 0 (shown)
This displacement is measured from point A.
Displacement = Uniform velocity × Time
We have to "plus 24" here because at the beginning of the motion, object Q is at 24 m to the right of point A.
74
4 (a) (i) v = t 2 − 6t + k
When t = 0, v = 8
8 = 02 − 6(0) + kk = 8
(ii) When v < 0,
t2 − 6t + 8 < 0(t − 2)(t − 4) < 0
Hence, the range of values of t when the particle moves to the left is 2 < t < 4 .
(iii) a =
dvdt
= 2t − 6
When a < 0,2t − 6 < 0
2t < 6t < 3
Hence, the range of values of t when the particle decelerates is 0 ≤ t < 3 .
(b) (i) t (s) 0 1 2 3 4
v (m s−1) 8 3 0 −1 0
(ii) Total distance travelled in the first 4 s
= Area under the v − t graph
= v dt0
2∫ + v dt2
4∫= (t 2 − 6t + 8) dt
0
2∫ + (t 2 − 6t + 8) dt2
4∫
=t 3
3− 3t 2 + 8t
⎡
⎣ ⎢ ⎤
⎦ ⎥ 0
2
+t 3
3− 3t 2 + 8t
⎡
⎣ ⎢ ⎤
⎦ ⎥ 2
4
=23
3− 3(2)2 + 8(2) − 0 +
43
3− 3(4)2 + 8(4) −
23
3− 3(2)2 + 8(2)
⎛
⎝ ⎜
⎞
⎠ ⎟
=203
+163
−203
=203
+ −43
=203
+43
= 8 m
75
Paper 2 1 (a) The total mass of seafood is not less than
20 kg. The inequality is x + y ≥ 20.
x 0 20 y 20 0
The mass of prawns is at most three times that of squids. The inequality is x ≤ 3 y .
x 0 30 y 0 10
The allocation is RM250. The inequality is 10x + 5y ≤ 250 ⇒ 2x + y ≤ 50
x 0 25 y 50 0
(b)
(c) If the restaurant buys 15 kg of squids,
y = 15. From the graph, when y = 15, the minimum value of x = 5. Therefore, if the restaurant buys 15 kg of squids, the minimum mass of prawns it can buy is 5 kg. Total expenditure = 10x + 5y When x = 5 and y = 15, the total expenditure = 10(5) + 5(15) = RM125 Hence, the maximum amount of money that could remain from its allocation is RM250 − RM125 = RM125.
Form 5: Chapter 21 (Linear Programming) SPM Flashback
Fully-Worked Solutions
76
2 (a) The total number of participants is at least 30. The inequality is x + y ≥ 30 .
x 0 30 y 30 0
The number of Mathematics participants is at most twice that of Science. The inequality is y ≤ 2x .
x 0 30 y 0 60
The expenditure for a Science participant and a Mathematics participant are RM80 and RM60 respectively. The maximum allocation is RM3600. The inequality is 80x + 60y ≤ 3600 ⇒
4x + 3y ≤ 180.
x 0 45 y 60 0
(b)
(c) (i) When y = 12, from the graph,
xmin = 18 and xmax = 36 . Hence, when the number of Mathematics participants is 12, the minimum and maximum number of Science participants are 18 and 36 respectively.
(ii) Cost = 80x + 60y
Draw the straight line 80x + 60y = 480
x 0 6 y 8 0
For minimum cost, from the graph, xmin = 10 and ymin = 20. Hence, the minimum cost
= 80x + 60y= 80(10) + 60(20)= RM2000
77
3 (a) For the constraint "the total number of balls bought should not be more than 80", the inequality is x + y ≤ 80 .
x 0 80 y 80 0
For the constraint "the number of footballs bought should not be more than 4 times the number of volleyballs bought", the inequality is y ≤ 4x .
x 0 20 y 0 80
For the constraint "the number of footballs bought should exceed the number of volleyballs bought by at least 15", the inequality is y − x ≥ 15.
x 0 40 y 15 55
(b)
(c) (i) From the graph, if the number of volleyballs bought is 25 (x = 25), the range of the number of footballs bought is 40 ≤ y ≤ 55 .
(ii) Cost = 60x + 80y
Draw the straight line 60x + 80y = 4800 .
x 0 80 y 60 0
From the graph, the optimal point is (16, 64). Hence the maximum cost = 60(16) + 80(64) = RM6080.
78
4 (a) The maximum time for the use of machine P is 12 hours:
60x + 20y ≤ 12 × 603x + y ≤ 36
x 0 12 6 y 36 0 18
The use of machine Q is at least 8 hours:
30x + 40y ≥ 8 × 603x + 4y ≥ 48
x 0 16 y 12 0
The ratio of the number of 'Premier' pewter plates produced to the number of 'Royal' pewter plates produced is at least 1 : 3:
xy
≥13
3x ≥ yy ≤ 3x
x 0 6 y 0 18
(b)
(c) (i) From the graph, if y = 12, xmax = 8
(ii) Profits = 100x + 140y Draw the straight line 100x +140y = 1400
x 0 14 y 10 0
From the graph, the optimal point = (6, 18) Hence, the maximum profit = 100(6) +140(18) = RM3120
(0, 36) is out of the graph paper. So, another point (6, 18) has to be determined.