02[A Math CD]
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7/23/2019 02[A Math CD]
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1 Penerbitan Pelangi Sdn. Bhd.
Paper 1
1. m= 3k2
m+ 2 = 3k
k=m+ 2
3
Answer: D
2. p2=k+ 5
2
2p2= k+ 5
2p25 = k
k= 2p25
Answer: A
3. 3h1 = k(3h1 )2= k2
9(h1) = k2
h1 =k2
9
h=k2
9+ 1
Answer: D
4. (xy)3= 8
3(xy)3= 38 xy= 2
x= 2 +y
Answer: B
5. hk(4 + h) = kh(4k+ hk) = k
h 4khk= k
hhk= k + 4k
h(1 k) = 5k
h=5k
1 k
Answer: C
6. 2u(w1) + 5w = 4u
2uw2u+ 5w = 4u
2uw+ 5w = 4u+ 2u
w(2u+ 5) = 6u
w =6u
2u+ 5
Answer: C
7. h=m+ 2
3m
3mh= m+ 2
3mhm= 2
m(3h1) = 2
m=2
3h 1
Answer: B
8.p 2
q =
2
3
3(p2) = 2q
3p
6 = 2q 3p = 2q+ 6
p =2q+ 6
3
Answer: A
9. x=4(x y)
y xy= 4(xy)
xy= 4x4y
xy+ 4y= 4x
y(x+ 4) = 4x
y=
4x
x+ 4
Answer: C
10.gh
2
= 2g1
gh= 2(2g 1)
gh= 4g2
2 = 4g gh
2 = g(4 h)
g=2
4 h
Answer: C
CHAPTER
2 Algebraic Formulae
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2
Mathematics SPM Chapter 2
Penerbitan Pelangi Sdn. Bhd.
Paper 1
1. E = 8 1F G
E2 = 8 1F G
2
E 2 = 641
F G
F G =64
E2
F =64
E2+ G
Answer: A
2. p= 5q2 2
p+ 2 = 5q2
p+ 2
5= q2
q= p+ 25
Answer: C
3. 2e= 7f g
2e+ g= 7f
7f= 2e+ g
f=2e+ g
7
Answer: B
4. 2h=4 + h
k 1
2h(k 1) = 4 + h
2hk 2h= 4 + h
2hk 2h h= 4
2hk 3h= 4 h(2k 3) = 4
h=4
2k 3
Answer: B
5. 1 + 3w2= 3(4w2 e)
1 + 3w2= 12w2 3e
1 + 3e= 12w2 3w2
1 + 3e= 9w2
w2=1 + 3e
9
w= 1 + 3e9 w=
1 + 3e
3
Answer: D
6. T=1
4
W
3
T12 =1
4
W
312
12T= 3 4W
4W= 3 12T
W=3 12T
4
Answer: C
7.6h+ k
2= hk3k
6h+ k= 2(hk3k)
6h+ k= 2hk6k
k+ 6k= 2hk6h
7k= h(2k6)
h=7k
2k 6Answer: A
Paper 1
1.p
2q=pq
p2 =pq+ q
p= 2(pq+ q)
p= 2pq+ 2q
p 2pq= 2q
p(1 2q) = 2q
p=2q
1 2q
Answer: B
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3
Mathematics SPM Chapter 2
Penerbitan Pelangi Sdn. Bhd.
2. mk= 5 m
k
mkk= 5 km
kk
mk2= 5km
mk2+ m= 5k
m(k2+ 1) = 5k
m= 5kk2+ 1
Answer: D
3. V= 1 t4
V 2= 1 t4
2
V 2=1 t
4
4V 2= 1 t
4V
2
+ t= 1 t= 1 4V 2
Answer: B
4. 3h4k=2k
h
h(3h4k) = 2k
3h24hk= 2k
3h2= 2k+ 4hk
3h2= k(2 + 4h)
k=3h2
2 + 4h
Answer: B
5. m=n(m 1)
1 n
m(1 n) = n(m1)
mmn= mnn
m= mnn+ mn
m= 2mnn
m= n(2m1)
n=m
2m 1
Answer: A
6. u=1 + 2t
1 2t
u(1 2t) = 1 + 2t
u2ut= 1 + 2t
u1 = 2t+ 2ut
u1 = t(2 + 2u)
t=u 1
2 + 2u
=u 1
2(u+ 1)
Answer: C
7. w=v(2 w)
v 1
w(v1) = v(2 w)
vww= 2vvw
vw2v+ vw= w
2vw2v= w
v(2w2) = w
v= w2w 2
Answer: B
8. p+ q=4p+ 3
q
q(p+ q) = 4p+ 3
pq+ q2= 4p+ 3
pq4p= 3 q2
p(q4) = 3 q2
p=3q2
q 4
Answer: C
9. h=2k+ 1
k 5
h(k5) = 2k+ 1
hk5h= 2k+ 1
hk2k= 1 + 5h
k(h2) = 1 + 5h
k=1 + 5h
h 2
Answer: A
10.nm
1
2m= 3n
nm
2m1
2m2m= 3n2m
2n1 = 6mn
(6n)m= 2n1
m=2n 1
6n
Answer: B
11.3
m2k= k1
3m
= k1 + 2k
3m
= 3k1
3 = m(3k1)
m=3
3k 1
Answer: B
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4
Mathematics SPM Chapter 2
Penerbitan Pelangi Sdn. Bhd.
12.2p
3= 2q 1
2p= 3(2q 1) 2p= 6q 32p+ 3 = 6q
q =2p+ 3
6
q=(2p+ 3)2
36
Answer: B
13.e
2
=4e 7
fef= 2(4e 7)
f =2(4e 7)
e
f=2(4e 7)
e 2
=4(4e 7)2
e2
Answer: C
14.4h
k1
= 3
4h= 3k1 (4h)2= (3k1 )2
16h2= 9(k1)
16h2
9 = k1
k=16h2
9
+1
Answer: B
15.p 2
3q+ 1 = 4r 1
p2 = (4r 1)(3q+ 1)
p2 = 4r(3q ) + 4r 3q 1
p 2 4r+ 1 = 4r(3q ) 3q
p
4r 1 = 3
q(4r 1)
3q =p 4r 1
4r 1
q = p 4r 14r 1 3
Answer: D
16. x= t2 5 y= t+ 4 From , t =y 4 t2 = (y 4)2
Substitute into ,x= t
2 5
x= (y 4)2 5
=y2 8y+ 16 5
=y2 8y+ 11
Answer: D