MENJAWAB SOALAN SPM
ADDITIONAL MATHEMATICS
Kertas 1 dan 2
Skema Jawapan & Pemarkahan
Siri Ceramah Di:
(i) SMK Dato’ Klana Putra, Lenggeng, Negeri Sembilan,
(ii) SMK Engku Husain, Semenyih, Selangor,
(iii) SMK Dengkil, Kuala Langat, Selangor,
(iv) SMK Taman Jasmin 2, Kajang, Selangor,
(v) SMK Jalan 4, Bandar Baru Bangi, Selangor,
(vi) SMK Dato’ Ahmad Razali, Amapang, Selangor.
Q.1:
(a) -1, 1 / -1 and 1 / {-1, 1}
….√(1)
reject: 1 or -1 / (-1, 1) / [-1, 1]
(b) many-to-one / ….√(1)
many with one /
many → one
Q.2(a):
(a)
f(x) = 4x + 2
f -1(x) = y
f(y) = x
= 4y + 2 √ -----------(1)
x = 4y + 2
x - 2
y = ——— √------------(2)
4
Q.2(b):
(b) gf(2) = g[f(2)]
= g[4(2) + 2] √………..………..(1)
= g(10)
= 102 – 3(10) – 1
= 69 √…………………..(2)
f(2) = 4(2) + 2
= 10 √ ………………….(1)
Q.3:
gh(x) = g[h(x)]
= mx2 + n – 3 ……..(P1)
= 3x2 + 7
Bandingkan:
m = 3 ……...(1)
n – 3 = 7 ………(P1)
n = 10 ………(2) (Kedua-duanya)
Q.4:
2x2 + p + 2 = 2px + x2
x2 – 2px + p + 2 = 0 √ ………..(P1)
a = 1, b = - 2p, c = p +2
b2 - 4ac > 0
(-2p)2 – 4(1)(p + 2) > 0
4p2 – 4p – 8 > 0 √ ………..(P2)
p2 – p – 2 > 0
Let: p2 – p – 2 = 0
(p – 2)(p+ 1) = 0
p = 2 atau p = - 1
Therefore: p > 2 atau p < -1 √ ..…….(3)
-1
2
x
Q.5:
(x – ⅔)(x + 4) = 0 ….…….(P1)
x2 – ⅔ x – 4x – 8/3 = 0
3x2 + 10x – 8 = 0 …………(2)
OR:
x2 – (⅔ - 4)x + (⅔)(-4) = 0 …………(P1)
3x2 + 10x – 8 = 0 …………(2)
Q.6:
(a) p = 3 (axis of symmetry) √...(1)
(b) x = 3 √...(1)
(c) (3, -1)
√...(1)
Note: Minimum value = -1
Q.7:
32x + 1 = 4x
(2x + 1)log10 3 = xlog10 4 √ …..…….(P1)
2xlog10 3 + log10 3 = xlog10 4
2x(0.4771) – x(0.6021) = - 0.4771 √.………..(P2)
0.9542x – 0.6021x = - 0.4771
0.3521x = - 0.4771
- 0.4771
x = ————
0.3521
= - 1.355 √ ………...(3)
Q.