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MATHEMATICS

SPECIALIST

UNIT 1

Semester One

2019

SOLUTIONS

Calculatorfree Solutions

1.

parallelogram with 2v and u as sides parallelogram with 2v and u as sides

parallelogram with v and –u as sides parallelogram with v and –u as sides

vector u in the direction of –v with magnitude |u| vector u in the direction of –v with magnitude |u|

[6]

2. (a) (i) 3-5=k4α

∴k=34 → α=-5k=-203

(ii) 4α∙11=0

∴4+α=0 → α=-4

(iii) PQ=1α+5

since the x–coordinate is already 1 unit in length, then

the y–coordinate must be zero.

∴α=-5

(iv) PQ as base ⇒ |OP| = |OQ|

3-5=4α → 34=16+α2

∴α2=18 → α=±32

2. (b) (i) Solving simultaneously (any method, elimination shown below):

uv == -6i2i -2j+3j ×3×2 → 3u2v == -18i4i -6j+6j ↓+

∴ 3u+2v=-14i → i=-314u-17v

similarly (or by substitution):

uv == -6i2i -2j+3j ×1×3 → u3v == -6i6i -2j+9j ↓+

∴ u+3v=7j → j=17u+37v

(ii) r=14-314u-17v+717u+37v

∴r=-2u+v [14]

3. (a) (i) 20!-18!=20×19×18!-18!

=20×19-1×18!

=380-1k=379k

(ii) 20P3 21C3=20!3!÷21!3!×18!

=20!3!×3!×18!21×20!=k21

(b) RHS =nn-r=n!n-r!×n-n-r!

=n!n-r!r!

=nr= LHS [6]

4. (a) If m<1, then m>m2.

It is NOT always true because it does not work for negatives.

e.g. m=-2<1 → m2=4>m ∴ false

The converse is always true for 0<m<1

(b) If the parallelogram is not a rectangle, then it does not have congruent diagonals.

Yes it is always true as only squares and rectangles have congruent diagonals.

(c) For all rational numbers , there exists two integer numbers a and b

such that p is the quotient of a and b. [8]

5. (a) (i) 26=1+6+15+20+15+6+1=64

(ii) 115=10+15

=105+5×104+10×103+10×102+5×10+15

=100 000+50 000+10 000+1 000+50+1

=161051

(b) (i) x=3 since 6C3=20

(ii) x=7 since 7C5=21

(iii) x=8 since 8C2= 8C6

(c) 2x-y5

=2x5+52x4-y1+102x3-y2+102x2-y3+52x1-y4+-y5

=32x5-80x4y+80x3y2-40x2y3+10xy4-y5

(d) (i) 8C5=56

(ii) 2C2× 6C3=1×20=20

(iii) 3C2× 5C3+ 3C3× 5C2

=3×10+1×10=30+10=40 [13]

6. (a) AD=25AB=25b-a

CD=CO+OA+AD=-12b+a+25b-a

∴CD=35a-110b

(b) OC+CE=OE → OC+βCD=αOA given

∴12b+β35a-110b=αa

×10 → 5b+6βa-βb=10αa

→ 6β-10αa=β-5b

since a and b are non–parallel, then:

β-5=0 → β=5

6β-10α=0 → α=35β=3 [6]

Calculatorassumed Solutions

7. (a) ABC collinear ⇒ AB // BC

AB=-24-4-5=-69

BC=-610--24=-46

∴-69=k-46 → k=-6-4=32k=96=32

Since k is unique, then AB // BC and hence ABC collinear.

(b) AB=-69=3-23 and BC=-46=2-23

∴AB:BC=3:2 [5]

8. (a) ∠PFO = 35°

Because ΔOFP is isosceles since |OP| = |OF| = radii

(b) ∠FEP = 55°

Since ∠FOP = 110° from ΔOFP, and the angle at the centre

is double the size of the angle at the edge.

(c) ∠PQF = ∠FEP = 55°

Angles at the circumference within the same segment

are congruent.

(d) ∠CFP = ∠FEP = 55°

The alternate segment theorem

(e) |GC| = |CF| = 11 – |FB| = 11 – 8 = 3 cm

Tangents to a circle from the same external point

are congruent.

(f) |AM|×(|AM| + 2×radius) = |AH|2

|AM|×(|AM|+8) = 52

|AM|2+8|AM| – 25 = 0

CAS ⇒ |AM| = -4±41

∴ |AM| = 41-4 ≈2.40 cm only solution [13]

9. (a) (i) Divisible by 3 and 5 = divisible by 15

100 ÷ 15 = 6.6 ⇒ only 6 elements are divisible by 15

Therefore, assuming every other element is chosen

instead of those 6, we need 100 – 6 +1 = 95 elements

(ii) Divisible by 3 = 100 ÷ 3 = 33.3 ⇒ 33 elements

Divisible by 5 = 100 ÷ 5 = 20 elements

Divisible by 3 or 5 = 33 + 20 – 6 = 47 elements

Assuming the other 53 elements are chosen first,

then 53 + 1 = 54 elements must be chosen

(b) Assuming the highest numbers are chosen first:

100 + 99 + 98 + … + 91 + 90 = 955

If 89 is chosen next then the sum exceeds 1000.

Therefore, a maximum of 11 elements must be chosen. [8]

10. (a) nM∪C=nM+nC-nM∩C

14 334 = 7 531 + 9 885 – nM∩C

∴nM∩C = 3 082 households

(b) nM∪C∪B=nM+nC+nB

- nM∩C-nM∩B-nC∩B

+nM∩C∩B

∴nM∪C∪B= 7 531 + 9 885 + 4 977 – 3 082 – 2 252 – 4 310 + 1 724

= 14 473 that have all three

Therefore, 16 366 – 14 473 = 1 893 households have neither [6]

11. (a) (i) 36P4=1 413 720 OR 36C4×4!

(ii) 10C2× 26C2×4!=351 000

(iii) 5C1× 34P2× 4C1=22 440

(b) II and III

(c) x+1P3= 4C3× xP2

x+1!x+1-3!=4×x!x-2!

x+1 × x!x-2!=4x! x-2!

x+1=4 → x=3

11. (d) LHS =n!n-2!+2n×n-1!n-1!

=n!n-2!×n-1n-1+2n!n-1!

=n!×n-1+2n!n-1!=n!n-1+2n-1!

=n! ×n+1 n+1-2! =n+1!n+1-2= n+1P2= RHS [14]

12. (a) w=-20cos10°-20sin10°

Hovering speed =-w=20cos10°20sin10°

(b) (i)

wind vector with correct x-axis angle drone vector as a side of parallelogram and angle θ resultant vector pointing towards O, and diagonal of parallelogram wind vector with correct x-axis angle drone vector as a side of parallelogram and angle θ resultant vector pointing towards O, and diagonal of parallelogram

(ii) w=-20cos10°-20sin10° d=-25cosθ25sinθ r=-rcos24.23°rsin24.23

(iii) -rcos24.23°= -20cos10°-25cosθrsin24.23°= -20sin10°25sinθ

∴ 252cos2θ=rcos24.23°+20cos10°2252sin2θ=rsin24.23°+20sin10°2

→252=rcos24.23°+20cos10°2+rsin24.23°+20sin10°2

CAS → r=38.8621 m/s OR r=-5.7897 m/s

→ θ=46.64° OR θ=2.94°

∴time=dv=12002+540238.8621=33.86 seconds

bearing =270°+θ=308.86°T [14]

13. (a) Fy=300sin62°+252sin56°

=473.8N

Since 473.8N<500N the machinery is not moving upwards

(b) No horizontal component needed ⇒ Fx=0

∴400cos62°=xcos56°

→x=400cos62°cos56°=335.82N

(c) Fy=400sin62°+335.82sin56°

=631.59N [7]

14. (a) (i) Let n∈N with n=2k+1= odd

Then n2+1=2k+12+1

=4k2+4k+2

=22k2+2k+1

Since 2 is a factor, then n2+1 is divisible by 2, and

hence the conjecture is true ∀ n∈N

(ii) Contrapositive statement:

“if n2+1 is odd, then n is even.”

Let n2+1=odd=2k+1

∴n2=2k

⇒n2=even ⇒ n=even

Since the contrapositive statement is true ∀ n∈N,

then the original conjecture is true ∀ n∈N

(b) A ⇒ B:

If the quadrilateral has two diagonals that intersect at right angles,

then the quadrilateral is a rhombus, which implies it does have

two pairs of parallel sides.

∴ A ⇒ B is true

B ⇒ A:

If the quadrilateral has two pairs of parallel sides then it is a parallelogram,

which does not necessarily imply it is a rhombus, and therefore it does not necessarily have diagonals that intersect at right angles.

∴ B ⇒ A is false

Therefore, A ⇔ B is a false statement.

(c) Assume that n is odd and n2 is even.

Then ∃ k∈N: n=2k+1

→n2=2k+12=4k2+4k+1

=22k2+2k+1=2m+1=odd

Since n2 is both even and odd simultaneously, this is a contradiction

and therefore the original conjecture must be true ∀ n∈N, n even. [15]

15. (a) (i) AB2=OA2+OB2-2OAOBcosθ

(ii) AB∙AB=OA2+OB2-2OAOBcosθ

LHS =OB-OA∙OB-OA

=OB∙OB-OB∙OA-OA∙OB+OA∙OA

=OB2+OA2-2 OA∙OB

∴OB2+OA2-2 OA∙OB=OA2+OB2-2OAOBcosθ

→-2 OA∙OB=-2OAOBcosθ

→OA∙OB=OAOBcosθ as required

(b) (i) 45∙2-3=45×2-3cosθ

8-15=41×13cosθ

∴cosθ=-7533

Since cosθ<0 ⇒ θ is obtuse

(ii) ⇒ sinθ=22533

∴ area ΔOAB =12OAOBsinθ

=12 41×13×22533=11 units2 [10]

16. P, Q, R and S are the midpoints of their respective sides:

OP=-0.54 OQ=61.5 OR=1.5-4 OS=-5-1.5

Therefore:

PQ=61.5--0.54=6.5-2.5

SR=1.5-4--5-1.5=6.5-2.5

PQ=SR ⇒ ∴ PQ // SR

PS=-5-1.5--0.54=-4.5-5.5

QR=1.5-4-61.5=-4.5-5.5

PS=QR ⇒ ∴ PS // QR

Since PQ // SR and PS // QR ⇒ PQRS is a parallelogram [5]