2017_specialist_unit_1_solutions.docx
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MATHEMATICS
SPECIALIST
UNIT 1
Semester One
2017
SOLUTIONS
Calculator−free Solutions
1. (a) ✔
✔
✔
(b) must form a closed loop, e.g.: ✔
(c) III and IV ✔✔ [6]
2. (a) ✔
✔✔
(b) ✔
✔
✔
(c) 10 houses have married couples with children and pets,
therefore, 41 houses must be selected to obtain at least one
with both children and pets. ✔
The Pigeon Hole Principle. ✔ [8]
3. (a)
vector location of
3. (b) ✔
✔
✔ [5]
4. (a) (i) 20 ✔
(ii) ✔
(b) (i) ; since 70 is in the 4th column of the 8th row ✔
(ii) ; since 15 is in the 4th column of the 6th row ✔
(iii) ; since in row 7, elements on columns 2 and 5 are equal ✔
(iv) ; since in row 8, elements on columns 2 and 6 are equal ✔
(c) (i) ✔
(ii) ✔✔
✔
(d) (i) ✔
(ii) ✔
✔ [13]
5. (a)
, hence ✔✔
✔
Hence, ✔
OR
✔✔
(b) , hence ✔
Unit vector ✔ [6]
6. (a) ✔
✔
✔
(b) (angle at circumference is half angle at the centre) ✔
✔
(c) from the alternate-segment theorem ✔
✔ [7]
7. (a) ✔
✔
(b) Show that opposite sides are congruent and parallel:
✔
as required. ✔
, and EFGH is a parallelogram ✔ [5]
Calculator−Assumed Solutions
8.
diagram(s)
✔
✔
✔
✔ [5]
9. (a) ✔✔
(b) (i) ✔✔
(ii) only choices are AB, DE, DF and EF ✔
✔✔ [7]
10. (a) LHS ✔
✔
RHS ✔
(b) (i) ✔
(ii) ✔✔
(iii) ✔✔ [8]
11. (a) ✔
✔
(b) ✔
✔✔
(4, 2) is the midpoint of both OB and AC ✔
Therefore, OB and AC bisect each other.
(c) since OABC is a parallelogram
✔
✔
(d) From (c):
✔
(e) Area OABC = 2 × Area ∆OAC
✔
units2 ✔ [11]
12. (a) (i) For all natural/counting numbers represented by n ✔
there exists another natural number m ✔
such that n is the square of m
OR such that m is a whole number root of n. ✔
12. (a) (ii) 5 is natural and there is no natural number that
when squared gives 5.
(any acceptable answer) ✔
(b) A rhombus has two pairs of parallel sides,
therefore B ⇒ A is a valid statement. ✔
Not all parallelograms are rhombi, e.g. rectangles,
therefore A ⇒ B is not a valid statement. ✔
Hence, A is not equivalent to B, i.e. A ⇔ B is invalid. ✔
(c) (i) If a triangle inscribed in a circle is right angled, then
the triangle has the diameter as one of its sides. ✔
(ii) Yes, because ALL right-angled triangles inscribed
in a circle will have the diameter as the hypotenuse. ✔
(iii) If a triangle inscribed in a circle does not have the
diameter as one of its sides, then the triangle is not
right angled. ✔
(iv) Yes, because if a statement is true then so is the
contrapositive of that statement ✔ [11]
13. (a)
Combined force vector:
✔✔✔
✔
and ✔
bearing = ✔
13. (b) ✔
✔
maximum force = ✔
bearing =
∴ 79.30°T ✔ [10]
14. (a) Assume that n is even and is odd. ✔
Then such that ✔
Thus,
which implies that must be even. ✔
Since cannot be both even and odd, this is a contradiction ✔
and therefore n must be even.
(b) ✔
✔
and hence as required ✔
if O,B,D are collinear OR DB=diameter ✔ [8]
15. (a) ✔
✔
(b) In a rhombus the diagonals are perpendicular.
✔
hence ✔
(c) ✔
✔
✔
as required. [7]
16. (a)
wind vector plane vector flight direction and resultant vector using parallelogram method
(b)
68° angle between wind and plane speeds
(c) ✔
km/h ✔
✔
✔
Hence, bearing = 132°– 8.89° = 123.11°T ✔
and flight duration = ✔ [11]
17. ✔
✔
since A,B,C collinear then ✔
Given AB:BC = 1:2 then ✔
✔✔ [6]
18. (a) ✔
✔
✔
(b) LHS ✔
✔
✔
✔
= RHS [7]
19. (a) Let and
✔
✔
and hence ✔
(b)
✔
✔
and hence ✔
(c) ∆PBD is right angled since ∆ABD is right angled (triangle in semicircle)
✔
(d) ∆ABD is right angled in semicircle
✔
hence, radius is ✔ [9]