2018_specialist_unit_1_solutions_ (1).docx
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MATHEMATICS
SPECIALIST
UNIT 1
Semester One
2018
SOLUTIONS
Calculatorfree Solutions
1. (a) b-a=13-21=-12
∴a∙b-a=21∙-12=-2+2=0
∴a⊥b-a
(b) (i) OC=AB=b-a=-i+2j
(ii) AC=OC-OA=-3i+j
∴AC=10
OR
AC=OB
∴AC=13=10 [6]
2. (a) (i) 13!+12!13!-12!=13×12!+12!13×12!-12!=12!13+112!13-1
=1412=76
(ii) 10C4 8C4 =10!4!×6!÷8!4!×4!
=10×9×8!×4!×4!8!×6×5×4!=105×96 =2×32=3
(b) LHS =knk=k×n!k!n-k!
=kk×n!k-1!n-k!
=n×n-1!k-1!n-1-(k-1)!
=nn-1k-1= RHS (Or from RHS to LHS) [8]
3. (a) -3>-5 but -32=9>-52=25 is false
(b) “If the triangle is not equilateral, then the triangle does not
have three equal sides.”
Yes is it always true since the original implication is always
true by definition of equilateral triangles.
(c) “If n is divisible by 3, then n is divisible by 6”.
The converse is not always true, because for n to be
divisible by 6 is must be divisible by both 2 and 3.
(d) FOR ALL natural numbers p, EXISTS a real number q,
such that q is the square root of p. [7]
4. (a) a+b=2-1+-14=13
∴a+b=±11013
∴u=210×±11013=±26
(b) (i) c=ka
-4α=k2-1 → k=-2 → α=2
(ii) -4α--14=32-1
∴9+α-42=35
9+α-42=9×5
α-42=36
α=4±6 → α=10 or -2 [10]
5. (a) AC∙OB=c-a∙a+c=0
c∙a+c∙c-a∙a-a∙c=0
c2-a2=0 → a=c
∴ OABC is a rhombus
(b) LHS =AC2+OB2=c-a2+a+c2
=c-a∙c-a+a+c∙a+c
=c∙c-2a∙c+a∙a+a∙a+2a∙c+c∙c
=2a2+2c2
=OA2+AB2+BC2+OC2 as required [7]
6. (a) ∠AED = 90°
Triangle in a semi-circle is always right angled.
(b) ∠ABE = ∠ADE = 60°
Angles within the same segment are congruent.
(c) ∠CAE = 80°
Opposite angles in a cyclic quadrilateral are supplementary
(d) ∠TCE = ∠CBE = 100°
The alternate segment theorem. [8]
7. AB=23AC
∴b-a=23c-a
13-x-1=234-xy+1
∴x=-5 and y=5 [4]
Calculatorassumed Solutions
8. (a) Assume all non-repeated numbers are selected from
both sets: 3, 8, 4, 6 = 4 digits
Plus all remaining digits from one set: 1, 2, 5, 7 = 4 digits
Plus one more digit to make the first repetition
∴ 4 + 4 + 1 = 9 digits minimum
(b) Assume the largest numbers are chosen first:
8 + 7 + 7 + 6 = 28
one more digit could include the number 5, making
the sum over 30.
∴ 4 digits max [54
9. (a) (i) 22C8=319 770
(ii) 12C4× 10C4=103 950
(iii) 22C8- 20C6=281 010
(iv) 2C2× 20C6+ 2C0× 20C8=38 760+125 970=164 730
(b) (i) 8!=40 320
(ii) 3!×6!=4 320
(iii) 8!-2!×7!=30 240 [12]
10. (a) II and III
(b) 10C1× 60P6× 42C1 OR 10C1× 60C6× 42C1×6!
(c) LHS =n!r!×n-r!+n!r+1!×n-r-1!
=n!r!×n-r-1!×1n-r+1r+1
=n!r!×n-r-1!×r+1+n-rn-rr+1
=n!×n+1r!×r+1×n-r-1!×n-r
=n+1!r+1!×n-r!
=n+1!r+1!×n+1-r+1!
= n+1Cr+1= RHS [10]
11. (a) (i) p+q
(ii) 2p-q
(iii) -3p+2q
(iv) 2q-p
(b)
[6]
12. (a) AC=2k--20=4k=4i+kj
BC=2k-1-3=1k+3=i+k+3j
(b) 4k=1k+3
∴42+k2=12+k+32
∴k=1
(c) D=-2+12,0-32=-0.5,-1.5
(d) DC=21--0.5-1.5=2.52.5
DB=1-3--0.5-1.5=1.5-1.5
∴DC∙DB=2.52.5∙1.5-1.5=2.5×1.5-2.5×1.5=0
∴DC⊥DB → ∠CDB is right angled [7]
13. (a) (i) True.
If a number is divisible by 6, then it is also divisible
by both 2 and 3.
(ii) False.
It must be divisible by both 2 and 3.
(iii) True.
The conjunction AND means that it is divisible by both
2 and 3, and therefore it is also divisible by 6.
(b) Assume that n is odd AND that 3n+5 is also odd
∴∃ k∈N such that n=2k+1
∴3n+5 =32k+1+5
=6k+8=23k+4= even
Since 3n+5 is both odd and even simultaneously, this is
a contradiction, implying that n must be even.
(c) OE+AD+BF
=b+12a-b+-a+12b+-b+12a
=12a-a+12a+b-12b-12b=0 [15]
14. (a) F1cos30°=F2cos45°
∴32F1=12F2
F1sin30°+F2sin45°=250
∴12F1+12F2=250
(b) F2=62F1
∴12F1+32F1=250
∴F1=5001+3=183.01N
∴F2=62×5001+3=25061+3=224.14N
(c) If F1=200N then F2=62×200=244.95N>200N
∴ Cable 2 exceeds its maximum load, hence Cable 1
must not reach its 200N maximum rating
If F2=200N then F1=26×200=163.30N<200N
Max Force =F1sin30°+F2sin45°
=4006×12+200×12=223.07N [12]
15. (a) nD∪C=nD+nC-n(D∩C)
810=400+500-n(D∩C)
∴nD∩C=90
(b) nD∪C∪B=nD+nC+nB
-nD∩C-nD∩B-n(C∩B)
+n(D∩C∩B)
900=400+500+210-90-60-110+nD∩C∩B
∴nD∩C∩B=50 [6]
16. (a) 3C12= 3C2+ 4C2
5C12= 5C2+ 6C2
(b) a=2
b=n+1
(c) 6C12= 8C3- 6C3
(d) nC12= n+2C3- nC3 [8]
17. (a) OE=a+12b-a=12a+b
OF=c+12b-c=12b+c
(b) DE=OE-OD=12a+b-12a=12b
GF=OF-OG=12b+c-12c=12b=DE
DG=OG-OD=12c-12a=12c-a
EF=OF-OE=12b+c-12a+b=12c-a=DG
∴DE=GF and DG=EF ⇒ DEFG is a parallelogram [7]
18. BF×FD=CF×FA
∴x2y=2x6 → y=6 cm
MD×MB=MT2
∴4×4+12+x=762 → x=3 cm
NT2=NC×NC+FC+FA
∴z2=6×6+6+6 → z=63 cm [6]
19. Let n be a unit vector perpendicular to b
then, a=u+utan60°n
u=31.5=325
n∙42=0 → let n=±-12
∴n=±15-12
∴a=31.5+325×3×±15-12=31.5±323-12
∴x=3±323 and y=32∓33 [6]