Maths Exam PreMSc 29-10-2009
Answers
a4 · (−3)2 1
1. (3) 4 −3
= − a7 , a 6= 0
−3 · a 9
2. (3) K 4 L − K 2 L3 = K 2 L(K − L)(K + L)
x−y x+y 4xy x2 − 2xy + y 2 − x2 − 2xy − y 2 + 4xy
3. (4) − + 2 = = 0, x 6= ±y
x + y x −√ y x√− y 2 (x − y)(x + y)
1 5− 3 1√ 1√
4. (3) √ √ = = 5− 3
5+ 3 5−3 2 2
3x + 4 3x + 4 − 4x − 8 −x − 4 x+4
5. (4) ≥2⇔ ≥0⇔ ≥0⇔ ≤0
2x + 4 2x + 4 2x + 4 x+2
Signdiagram gives: −4 ≤ x < −2
6. (3) (1.12)5 = 1.7623. Combined percentage is: 76%
2(x+1) 1 2x2 +1
7. (4) 5 − 3 = 5x . LCD is 15x, x 6= 0
(2x+2)·3x 5x 2x2 +1)·3
15x − = 15x ⇔ x = 3
15x
√
8. (4) p + q − q = 5 ⇒ p + q = (q + 5)2 ⇔ p = −q + (q + 5)2 = q 2 + 9q + 25
√
Domain: Because the outcome of . ≥ 0, q + 5 ≥ 0, q ≥ −5.
√
The argument of . must be positive or equal to zero, p + q ≥ 0.
So: p ≥ q ≥ −5
9. (3) 92p−1 = 36 ⇔ (32 )2p−1 = 36 ⇔ 4p − 2 = 6 ⇔ p = 2
10 (6) a = 0 : x2 − 16 = 0 ⇔ (x − 4)(x + 4) = 0 ⇔ x = −4 or x = 4.
a = 6 : x2 + 6x − 16 = 0 ⇔ (x + 8)(x − 2) = 0 ⇔ x = −8 or x = 2
No solution when D = b2 − 4ac < 0 : a2 + 64 > 0 for all real numbers a.
So for every real a the equation has a solution.
(
x + 3y = 5
11 (4) Answer x = 2 and y = 1
2x + y = 5
2 2
12 (6) y = f (x) =
x
Domain: x 6
= ±1. f (0) = 0, f 0 (x) = (1 − x ) · 1 − x(−2x) = 1 + x .
1 − x2 (1 − x2 )2 (1 − x2 )2
f 0 (0) = 1. Equation of the tangent line is: y − 0 = 1(x − 0) or y = x.
90 ln 90−ln 75
13 (3) 750(1.03)x = 900 ⇔ 1.03x = 75 ⇔x= ln 1.05 = 3.74
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Answers
a4 · (−3)2 1
1. (3) 4 −3
= − a7 , a 6= 0
−3 · a 9
2. (3) K 4 L − K 2 L3 = K 2 L(K − L)(K + L)
x−y x+y 4xy x2 − 2xy + y 2 − x2 − 2xy − y 2 + 4xy
3. (4) − + 2 = = 0, x 6= ±y
x + y x −√ y x√− y 2 (x − y)(x + y)
1 5− 3 1√ 1√
4. (3) √ √ = = 5− 3
5+ 3 5−3 2 2
3x + 4 3x + 4 − 4x − 8 −x − 4 x+4
5. (4) ≥2⇔ ≥0⇔ ≥0⇔ ≤0
2x + 4 2x + 4 2x + 4 x+2
Signdiagram gives: −4 ≤ x < −2
6. (3) (1.12)5 = 1.7623. Combined percentage is: 76%
2(x+1) 1 2x2 +1
7. (4) 5 − 3 = 5x . LCD is 15x, x 6= 0
(2x+2)·3x 5x 2x2 +1)·3
15x − = 15x ⇔ x = 3
15x
√
8. (4) p + q − q = 5 ⇒ p + q = (q + 5)2 ⇔ p = −q + (q + 5)2 = q 2 + 9q + 25
√
Domain: Because the outcome of . ≥ 0, q + 5 ≥ 0, q ≥ −5.
√
The argument of . must be positive or equal to zero, p + q ≥ 0.
So: p ≥ q ≥ −5
9. (3) 92p−1 = 36 ⇔ (32 )2p−1 = 36 ⇔ 4p − 2 = 6 ⇔ p = 2
10 (6) a = 0 : x2 − 16 = 0 ⇔ (x − 4)(x + 4) = 0 ⇔ x = −4 or x = 4.
a = 6 : x2 + 6x − 16 = 0 ⇔ (x + 8)(x − 2) = 0 ⇔ x = −8 or x = 2
No solution when D = b2 − 4ac < 0 : a2 + 64 > 0 for all real numbers a.
So for every real a the equation has a solution.
(
x + 3y = 5
11 (4) Answer x = 2 and y = 1
2x + y = 5
2 2
12 (6) y = f (x) =
x
Domain: x 6
= ±1. f (0) = 0, f 0 (x) = (1 − x ) · 1 − x(−2x) = 1 + x .
1 − x2 (1 − x2 )2 (1 − x2 )2
f 0 (0) = 1. Equation of the tangent line is: y − 0 = 1(x − 0) or y = x.
90 ln 90−ln 75
13 (3) 750(1.03)x = 900 ⇔ 1.03x = 75 ⇔x= ln 1.05 = 3.74
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