Page 72 - PRE-U STPM MATHEMATICS (T) TERM 1

Mathematics Term 1 STPM Answers

Long division: The set values of x: x ∈ (3, ∞).
2
ax – (3a + 1)x + 9
4x – 9
3
x – 3 ax – (1 + 6a)x + 3(3a + 4)x – 27 (ii) 4x – 9 = (x – 3)(2x – 7x + 9)
2
2
p(x)
ax – 3ax 2
3
= A + Bx + C
2
–(3a + 1)x + 3(3a + 4)x x – 3 2x – 7x + 9
2
–(3a + 1)x + 3(3a + 1)x 4x – 9 = A(2x – 7x + 9) + (Bx + C)(x – 3)
2
2
9x – 27 When x = 3, 4(3) – 9 = A[2(3) – 7(3) + 9]
2
9x – 27 3 = 6A
1
Alternative solution: A = 2
ax – (1 + 6a)x + 3(3a + 4)x – 27 2
2
3
= (x – 3)(ax + px + 9) Coefficients of x : 0 = 2A + B
2
1
Comparing coefficients of x , 2 0 = 21 2 + B
2
–(1 + 6a) = p – 3a B = –1
p = 3a – 1 – 6a
= –3a – 1 Constants: –9 = 9A – 3C
1
= –(3a + 1) –9 = 91 2 – 3C
2
The quotient = ax – (3a + 1)x + 9 9 2
C =
2
(c) Remainder Theorem: p(4) = 13 –x + 9
1
2
2
3
p(4) = a(4) – (1 + 6a)(4) + 3(3a + 4)(4) – 27 = 13 Hence, 4x – 9 = 2(x – 3) + 2x – 7x + 9
p(x)
2
64a – 16 – 96a + 36a + 48 – 27 = 13
4a = 8 or 4x – 9 = 1 + 2 9 – 2x .
a = 2 p(x) 2(x – 3) 2(2x – 7x + 9)
2
(d) (i) p(x) = (x – 3)(2x – 7x + 9) = 0
→
(x – 3) = 0 8. Given that ON = 2i, OS = 4k and OR = 3j + 4k
→
→
x = 3 (real root)
(2x – 7x + 9) = 0 (a) Given that SR : NM = 3 : 5
2
→
→
→
2
–(–7) ± (–7) – 4(2)(9) OM = ON + NM
x =
2(2) → → 5 →
OM = ON + SR
7 ± –23 3
= (complex roots) 2 0 0
1 2
3 1 2 1 2
4 OM = 0 + 3 – 0
5
→
Therefore p(x) = 0 has only one real root. (shown) 0 4 4
2
Alternative solution: → 1 2
p(x) = (x – 3)(2x – 7x + 9) = 0 OM = 5 = 2i + 5j
2
(x – 3) = 0 0
x = 3 (real root) Alternative solution:
(2x – 7x + 9) = 0 →
2
2
2
b – 4ac = (–7) – 4(2)(9) OR = 3j + 4k
→
→
= –23 < 0 OS + SR = 4k + 3j
→
(no real root) \ SR = 3j
Therefore p(x) = 0 has only one real root. (shown)
→
→
Since SR parallel to NM.
2x – 7x + 9 SR : NM = 3 : 5,
2
9
7
→
= 21 x – x + 2 therefore NM = 5j
2
2 2
→
→
→
9
7 2
= 231 x – 2 – 49 + 4 OM = ON + NM
4 16 2 = 2i + 5j
7 2
= 21 x – 2 + 23
4 8 2 0 2
1 2 1 2 1 2
→ → → → →
(b) OT = ON + NT = ON + OS = 0 + 0 = 0
2
For (x – 3)(2x – 7x + 9) . 0, 0 4 4
7 2
(x – 3)3 21 x – 2 + 23 4 . 0. 0 2 –2
1 2 1 2 1 2
→
→
→
4 8 TR = OR – OT = 3 – 0 = 3
7 2
7 2
Since 1 x – 2 > 0, 3 21 x – 2 + 23 4 . 0 4 4 0
4
8
4
for x ∈ R.
Therefore (x – 3) . 0.
291
Answers STPM Math T S1.indd 291 3/28/18 4:25 PM