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WATP 2020_specialist_unit_1_solutions_.docx

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SEMESTER ONE

MATHEMATICS

SPECIALIST

UNIT 1

2020

SOLUTIONS

Calculatorfree Solutions

1. (a) (i) b

(ii)

(b)

=

= [5]

2. (a) (i) 5

(ii) 21

(b) (i) x = 3

(ii) x = 4

(iii) x = 7

(iv) x = 2

(c) (i)

(ii)

= 20

(d) (i)

(ii) [12]


3. (a) (i)

(ii) OA x OB = OC2

But OC = OX

OA x OB = OX2

This is the converse of the secant/tangent theorem

OX is a tangent

OA x OB = OC2

But OC = OY

OA x OB = OY2

This is the converse of the secant/tangent theorem

OY is a tangent

(b) or

(c) (i)

(ii) 3! = 6 ways [10]

4. (a) (i) i 13j

(ii) 5

(iii)

(iv)

(b)

(c)

[12]

5. (a)

Thale’s Theorem

(b)

Revolution equal to 360 degrees

(c)

Equal angles at base of isosceles triangle [6]

6.

This is a contradiction as (x – 3y) cannot both belong and not belong

to the integer set.

:. False, there are no integers for which this statement is true. [5]

CalculatorAssumed Solutions

7.

v = 0.46 km/h [6]


8. (a) (i) LM parallel to x–axis so b = 5

M = (5, 5)

(ii)

(b) (i) x = 5

y = 2

(ii) If y = 2 & x ≥ 6, then BC // AD

If AB // DC

(iii)

[13]


9. (a)

(b)

Bearing = 110T

(c)

Time taken = [7]

10. (a)

50 like tea only

(b) None like neither

(c)

100 = 70 + 50 + 60 20 40 30 + x

= 90 + x

x = 10

(d) (i)

(ii)

(e) 41 [8]


11. (a)

(b)

[8]

12. (a) 3 > 2 but 3 < 2 is false

(b) “If the triangle is not isosceles, then the triangle does not

have two equal sides.”

Yes is it always true since the original implication is always

true by definition of isosceles triangles.

(c) “If n is divisible by 4, then n is divisible by 8”.

The converse is not always true, as shown by the

counter-example where n = 12

(d) FOR ALL natural numbers p, EXISTS a real number q,

such that q is one less than p. [7]


13. (a)

Let BAD = x

reflex BOD = 2x Angle at the centre theorem

Similarly let BCD = y

BOD = 2y Angle at the centre teorem

Since 2x + 2y = 360

then x + y = 180

Hence BAD + BCD = 180

QED

(b) EAG + BCD = 180 Cyclic quadrilateral

EAG ECF

In EAG and ECF

AEG CEF Bisector

EAG ECF Proven

CFE BFG Vertically Opposite

AGF BFG QED

(c) (i) Converse

(ii) Yes [10]

14. (a) (i)

x = 1

(ii)

0.13

(iii) a b = 13i 4j

| a b | =


(b) (i)

(ii) cos(0) = 1

So, the dot product of a vector with itself is its magnitude

squared multiplied by 1.

(iii)

[14]

15. (a)

F1 + F2 + F3 =

276T

(b) a + b = 5i + j

c = 5i j

| c | =

Direction = 101T [10]

16.

[4]


17.

[5]

18. (a) Independent term is the middle term.

8064

(b)

Hence,

[8]