2017_specialist_units_12_solutions_.docx
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SEMESTER TWO
MATHEMATICS
SPECIALIST
UNITS 1 & 2
2017
SOLUTIONS
Calculatorfree Solutions
1. (a) 1000 = 310 + 650 + 440 – 170 – 150 – 180 + x
x = 100
(b) 150 – 100 = 50
(c) (i) 44
(ii)
n(B) = 34 [6]
2. (a) (i) Substitute z = 2i to get (2i)4 – 2(2i)3 + 7(2i)2 –8(2i) + 12
which reduces to 0
(ii) z = –2i (the conjugate) is the other root.
(b)
[5]
3. (a) (i)
(ii) This statement is false
(b) (i) = 10
(ii) 2 x 4! = 48 [8]
4. (a) p = 4, q = 0.2
(b) y = 1 – x becomes y = 4[ 1 – 0.2 x]
i.e. y = 4 – 0.8x
or, if (c) is done before (b), gradient is – 0.8 and intercept is 4
y = 4 – 0.8x
(c) A = ( 5, 0 ) and B = ( 0, 4 )
(d)
(e)
(f) Reflection across y axis
i.e. g(x) becomes – g(x)
(g) A = 0.5 ( ms – rn ) [12]
5. (a)
(b) line
of intersection and accuracy
(c) sin x
(d) 2cos x . sin x = sin 2x [8]
6. (a) Let the numbers be 2k – 1, 2k + 1, 2k + 3, 2k + 5, 2k + 7
2k 1 + 2k + 1 + 2k + 3 + 2k + 5 + 2k + 7
= 10k + 15
Since 10k + 15 = 5(2k + 3) then divisible by 5.
(b) Assume that is rational, hence =
But – a and b are integers, so is rational.
This contradicts the supposition, and
therefore by contradiction must be irrational. [7]
7. For :
True for n = 1
Assume true for :
ie
Prove true for :
Proof:
as required
Therefore, True for n = k + 1, and since true for n =1,
true for all whole numbers. [5]
Calculatorassumed Solutions
8. wz = (2 + ai)(3b + i) = 4
6b + 2i + 3abi – a = 4
6b – a = 4 and 2 + 3ab = 0
2 +3(6b – 4)(b) = 0
9b2 – 6b +1 = 0
b = and a = –2 [5]
9. (a) RHS =
=
= = LHS
(b) (i)
and
=
= |p| |q| [ cos A cos B + sin A sin B ]
(ii) cos(A + B) = cos(A – (–B)) = cos A cos (–B) + sin A sin (–B)
= cos A cos B – sin A sin B
(iii) cos2 A + [cos(120 + A)]2 +[cos(120– A)]2
= cos2 A + [cos 120cos A – sin120sin A]2 +[cos120cos A + sin120sin A]2
= cos2 A + [ + [
= 1.5 cos2 A + 1.5 sin2 A
= 1.5 [13]
10. (a) (3a – b) (a + 3b) = 3a a – b a + 9a b – 3b b
= 3|a|2 + 8ab – 3|b|2
= 3 + 8|a||b|cos – 3
= 8cos as required
(b) = (4,1)
Unit vector on x axis = (1,0)
Length of projection = | PQ. | = |4| = 4 [7]
11. (a) Let the cost of a bottle of orange concentrate cost x
Let the cost of a bottle of banana concentrate cost y
5x + 1y = 19
2x + 3y = 18
(b)
= [4]
12. (a)
(b)
= 30 Bearing is 060 T
(c) Speed in Easterly direction is
Time taken is minutes [8]
13. (a) (i) Rotation of 180
(ii) Rotation of 270 clockwise
(b) P = 6(B – 2A) x B –1
(c) BA =
BAX =
Coordinates are (14, 5)
(d) Det B = 22 Area = 25 x 22 = 550
(e) Singular matrix has det = 0
Area = 0 i.e. A line [13]
14. (a) 3(2i + 3j) – (mi – 5j) = 8i + 14j
(6 – m)i + 14j = 8i + 14j
6 – m = 8
m = – 2
(b) 2i + 3j = k(mi – 5j)
2 = km and 3 = –5k
k = – 0.6 and by substitution, m =
(c)
2m – 15 = 0
m = 7.5 [6]
15. (a)
hence,
and
therefore,
(b) (i)
(ii) for
hence [7]
16. (a) (equilateral triangle)
(central angle theorem)
(b) (i)
AOB = 60 (proved in (a))
(theorem)
Similarly, DBC = 30
(ii) Assume E is the centre.
All angles of ABE = 60
and all angles of BEC = 60
But AEB = 2ACB which is impossible if they are both 60
E is not the centre. [8]
17.
(i)
(ii)
(iii)
(iv) 13
[8]
18. (a) 2m
(b)
t = 30 secs
(c) 138m
(d)
t = 9.68, 20.32
10.64 minutes [7]
19. (a) It is given that
Since AB BA,
(b)
as required [6]
20. (a)
(b)
(c) Let k be the midpoint of .
Then K = (2, 2)
So
K is the midpoint of [7]
Diagonals bisect each other.