Upload
toby-greer
View
221
Download
0
Embed Size (px)
Citation preview
THE BINOMIAL THEOREMTHE BINOMIAL THEOREM
Robert YenRobert YenHurlstone Agricultural High SchoolHurlstone Agricultural High School
This presentation is accompanied by workshop notesThis presentation is accompanied by workshop notes
INTRODUCTIONINTRODUCTION
Expanding (a + x)n
Difficult topic: high-level algebra
Targeted at better Extension 1 students
Master this topic to get ahead in the HSC exam
No shortcuts for this topic
BINOMIAL EXPANSIONS AND PASCAL’S TRIANGLEBINOMIAL EXPANSIONS AND PASCAL’S TRIANGLE
Coefficients
(a + x)1 = a + x 1 1
(a + x)2 = a2 + 2ax + x2 1 2 1
(a + x)3 = a3 + 3a2x + 3ax2 + x3 1 3 3 1
(a + x)4 = a4 + 4a3x + 6a2x2 + 4ax3 + x4 1 4 6 4 1
(a + x)5 = a5 + 5a4x + 10a3x2 + 10a2x3 + 5ax4 + x5 1 5 10 10 5 1
nnCCkk, A FORMULA FOR PASCAL’S TRIANGLE, A FORMULA FOR PASCAL’S TRIANGLE
1 0C0
1 1 1C0 1C1
1 2 1 2C0 2C1
2C2
1 3 3 1 3C0 3C1
3C2 3C3
1 4 6 4 1 4C0 4C1
4C2 4C3
4C4
1 5 10 10 5 1 5C0 5C1
5C2 5C3
5C4 5C5
1 6 15 20 15 6 1 6C0 6C1
6C2 6C3
6C4 6C5
6C6
1 7 21 35 35 21 7 1 7C0 7C1
7C2 7C3
7C4 7C5
7C6
7C7
1 8 28 56 70 56 28 8 1 8C0 8C1
8C2 8C3
8C4 8C5
8C6 8C7
8C8
nCk gives the value of row n, term k,
if we start numbering the rows and terms from 0
nnCCkk, A FORMULA FOR PASCAL’S TRIANGLE, A FORMULA FOR PASCAL’S TRIANGLE
n
kCk
n
CALCULATINGCALCULATING
MentallyMentally
102
20
123
3453
5
C
35C
1026
120
!2!3
!53
5
C
CALCULATINGCALCULATING
MentallyMentally
FormulaFormula
35C
102
20
123
3453
5
C
!!
!
knk
nCk
n
1026
120
!2!3
!53
5
C
CALCULATINGCALCULATING
MentallyMentally
FormulaFormula
because ...because ...
102
20
123
3453
5
C
!2!3
!5
12
12
123
345
123
3453
5
C
!!
!
knk
nCk
n
35C
THE BINOMIAL THEOREMTHE BINOMIAL THEOREM
(a + x)n = nC0 an + nC1 an-1 x + nC2 an-2 x2 + nC3 an-3 x3
+ nC4 an-4 x4 + ... + nCn xn
=
kknn
kk
n xaC
0
Don’t worry too much about writing in notation: just have a good idea of the general term
The sum of terms from k = 0 to n
PROPERTIES OF nCk
1. nC0 = nCn = 1 1st and last
2. nC1 = nCn-1 = n 2nd and 2nd-last
3. nCk = nCn-k Symmetry
4. n+1Ck = nCk-1 + nCk Pascal’s triangle result: each
coefficient is the sum of the two
coefficients in the row above it
n+1Ck = nCk-1 + nCk Pascal’s triangle result
1 0C0
1 1 1C0 1C1
1 2 1 2C0 2C1
2C2
1 3 3 1 3C0 3C1
3C2 3C3
1 4 6 4 1 4C0 4C1
4C2 4C3
4C4
1 5 10 10 5 1 5C0 5C1
5C2 5C3
5C4 5C5
1 6 15 20 15 6 1 6C0 6C1
6C2 6C3
6C4 6C5
6C6
1 7 21 35 35 21 7 1 7C0 7C1
7C2 7C3
7C4 7C5
7C6 7C7
1 8 28 56 70 56 28 8 1 8C0 8C1
8C2 8C3
8C4 8C5
8C6 8C7
8C8
15 = 10 + 56C4 = 5C3 + 5C4
Example 1Example 1
(a) (a + 3)5 =
Example 1Example 1
(a) (a + 3)5 = 5C0 a5 + 5C1 a4 31 + 5C2 a3 32 + 5C3 a2 33 + 5C4 a1 34
+ 5C5 35
= a5 + 5a4.3 + 10a3.9 + 10a2.27 + 5a.81 + 243
= a5 + 15a4 + 90a3 + 270a2 + 405a + 243
(b) (2x – y)4 =
Example 1Example 1
(a) (a + 3)5 = 5C0 a5 + 5C1 a4 31 + 5C2 a3 32 + 5C3 a2 33 + 5C4 a1 34
+ 5C5 35
= a5 + 5a4.3 + 10a3.9 + 10a2.27 + 5a.81 + 243
= a5 + 15a4 + 90a3 + 270a2 + 405a + 243
(b) (2x – y)4 = 4C0 (2x)4 + 4C1 (2x)3(-y)1 + 4C2 (2x)2(-y)2
+ 4C3 (2x)1(-y)3 + 4C4 (-y)4
= 16x4 + 4(8x3)(-y) + 6(4x2)(y2) + 4(2x)(-y3) + y4
= 16x4 – 32x3y + 24x2y2 – 8xy3 + y4
Example 2 (2008 HSC, Question 1(d), 2 marks)Example 2 (2008 HSC, Question 1(d), 2 marks)
(2x + 3y)12 = 12C0 (2x)12 + 12C1 (2x)11(3y)1 + 12C2 (2x)10(3y)2 + ...
General term Tk = 12Ck (2x)12-k(3y)k = 1 mark
For coefficient of x8y4, substitute k = ?
Example 2 (2008 HSC, Question 1(d), 2 marks)Example 2 (2008 HSC, Question 1(d), 2 marks)
(2x + 3y)12 = 12C0 (2x)12 + 12C1 (2x)11(3y)1 + 12C2 (2x)10(3y)2 + ...
General term Tk = 12Ck (2x)12-k(3y)k = 1 mark
For coefficient of x8y4, substitute k = 4:
T4 = 12C4 (2x)8(3y)4
= 12C4 (28)(34) x8y4
Coefficient is 12C4 (28)(34) or 10 264 320.
It’s OK to leave the coefficient unevaluated, especially if the
question asks for ‘an expression’.
TTkk is not the is not the kkthth term term
Tk is the term that contains xk
Simpler to write out the first few terms rather than memorise the notation
Better to avoid referring to ‘the kth term’: too confusing
HSC questions ask for ‘the term that contains x8’ rather than ‘the 9th term’
Example 3 (2005 HSC, Question 2(b), 3 marks)Example 3 (2005 HSC, Question 2(b), 3 marks)
General term Tk = 12Ck (2x)12-k
= 12Ck 212-k x12-k (-x-2)k
= 12Ck 212-k x12-k (-1)k x-2k
= 12Ck 212-k x12-3k (-1)k
For term independent of x:
...1
21
221
22
210
212
1
211
11212
012
12
2
xxC
xxCxC
xx
k
x
2
1
Example 3 (2005 HSC, Question 2(b), 3 marks)Example 3 (2005 HSC, Question 2(b), 3 marks)
General term Tk = 12Ck (2x)12-k
= 12Ck 212-k x12-k (-x-2)k
= 12Ck 212-k x12-k (-1)k x-2k
= 12Ck 212-k x12-3k (-1)k
For term independent of x:12 – 3k = 0 k = 4
T4 = 12C4 28 x0 (-1)4
= 126 720
...1
21
221
22
210
212
1
211
11212
012
12
2
xxC
xxCxC
xx
k
x
2
1
FINDING THE GREATEST COEFFICIENT
(1 + 2x)8 = 1 + 16x + 112x2 + 448x3 + 1120x4
+ 1792x5 + 1792x6 + 1024x7 + 256x8
Example 4Example 4
(a) (1 + 2x)8 = 8C0 18 + 8C1 17 (2x)1 + 8C2 16 (2x)2 + 8C3 15 (2x)3 +...
General term Tk = 8Ck 18-k (2x)k
= 8Ck 1 (2k) xk
= 8Ck 2k xk
tk = 8Ck 2k Leave out xk as we are only interested in the coefficient
Example 4Example 4
(a) (1 + 2x)8 = 8C0 18 + 8C1 17 (2x)1 + 8C2 16 (2x)2 + 8C3 15 (2x)3 +...
General term Tk = 8Ck 18-k (2x)k
= 8Ck 1 (2k) xk
= 8Ck 2k xk
tk = 8Ck 2k
(b)
Leave out xk as we are only interested in the coefficient
kk
kk
k
k
C
C
t
t
2
28
11
81
Example 4Example 4
(b)
Ratio of consecutive factorials
kk
kk
k
k
C
C
t
t
2
28
11
81
!
!1
n
n
Example 4Example 4
(b)
Ratio of consecutive factorials
kk
kk
k
k
C
C
t
t
2
28
11
81
1
82
2.1
8.
1
1
2.!7
!8.
!1
!
2!8!
!8
!18!1
!8 1
k
k
k
k
k
k
k
k
kkkk
81...567
1...5678
7!
8! eg
1!
!1
nn
n
Example 4Example 4
(c) For the greatest coefficient tk+1, we want:
tk+1 > tk
16 – 2k > k + 1
-3k > -15
k < 5
k = 4 k must be a whole number
1
1
82
11
k
k
t
t
k
k
for the largest possible integer
value of k
Example 4Example 4
(c) Greatest coefficient tk+1 = t5
= 8C5 25
= 56 32
= 1792
tk = 8Ck 2k
THE BINOMIAL THEOREM FOR (1 + THE BINOMIAL THEOREM FOR (1 + xx))nn
(1 + x)n = nC0 + nC1 x + nC2 x2 + nC3 x3 + nC4 x4 +
... + nCn xn
=
kn
kk
n xC0
Example 5Example 5
(1 + x)n = nC0 + nC1 x + nC2 x2 + ... + nCn xn
Sub x = ?
[Aiming to prove: [Aiming to prove: ] ]n
k
n
k
nC 20
Example 5Example 5
(1 + x)n = nC0 + nC1 x + nC2 x2 + ... + nCn xn
Sub x = 1:
(1 + 1)n = nC0 + nC1 (1) + nC2 (12) + ... + nCn (1n)
2n = nC0 + nC1 + nC2 + ... + nCn
This will make the x’s disappear and make the LHS become 2n
k
n
k
nn C
0
2
Example 6Example 6
(1 + x)2n = 2nC0 + 2nC1 x + 2nC2 x2 + ... + 2nC2n x2n
Term with xn = 2nCn xn
Coefficient of xn = 2nCn
[Aiming to prove ][Aiming to prove ] n
nn
kk
n CC 2
0
2
Example 6Example 6
(1 + x)2n = 2nC0 + 2nC1 x + 2nC2 x2 + ... + 2nC2n x2n
Term with xn = 2nCn xn
Coefficient of xn = 2nCn
(1 + x)n = nC0 + nC1 x + nC2 x2 + ... + nCn xn
(1 + x)n.(1 + x)n = (nC0 + nC1 x + nC2 x2 + ... + nCn xn)
(nC0 + nC1 x + nC2 x2 + ... + nCn xn)
If we expanded the RHS, there would be many terms
[Aiming to prove ][Aiming to prove ]
nn
n
kk
n CC 2
0
2
Example 6Example 6
(1 + x)n.(1 + x)n = (nC0 + nC1 x + nC2 x2 + ... + nCn xn)
(nC0 + nC1 x + nC2 x2 + ... + nCn xn)
Terms with xn
= nC0(nCn xn) + nC1 x (nCn-1 xn-1) + nC2 x2 (nCn-2 xn-2) + ...
+ nCn xn (nC0)
Coefficient of xn
= nC0 nCn + nC1
nCn-1 + nC2 nCn-2 + ... + nCn
nC0
= (nC0)2 + (nC1)2 + (nC2)2 + ... + (nCn)2
by symmetry of Pascal’s triangle
[Aiming to prove ][Aiming to prove ]
nCk = nCn-k
nn
n
kk
n CC 2
0
2
Example 6Example 6
By equating coefficients of xn on both sides of
(1 + x)2n = (1 + x)n.(1 + x)n
2nCn = (nC0)2 + (nC1)2 + (nC2)2 + ... + (nCn)2
n
kk
nn
n CC0
22
Example 7 (2006 HSC, Question 2(b), 2 marks)Example 7 (2006 HSC, Question 2(b), 2 marks)
(i) (1 + x)n = nC0 + nC1 x + nC2 x2 + ... + nCn xn
Differentiating both sides:
n(1 + x)n-1 = 0 + nC1 + 2 nC2 x + ... + r nCr xr-1 + ... + n nCn xn-1
= nC1 + 2 nC2 x + ... + r nCr xr-1 + ... + n nCn xn-1
(ii) Substitute x = ? to prove result:
[Aiming to prove: n 3n-1 = nC1 + ... + r nCr 2r-1 + ... + n nCn 2n-1 ]
The general term
Example 7 (2006 HSC, Question 2(b), 2 marks)Example 7 (2006 HSC, Question 2(b), 2 marks)
(i) (1 + x)n = nC0 + nC1 x + nC2 x2 + ... + nCn xn
Differentiating both sides:
n(1 + x)n-1 = 0 + nC1 + 2 nC2 x + ... + r nCr xr-1 + ... + n nCn xn-1
= nC1 + 2 nC2 x + ... + r nCr xr-1 + ... + n nCn xn-1
(ii) Substitute x = 2 to prove result:
n(1 + 2)n-1 = nC1 + 2 nC2 2 + ... + r nCr 2r-1 + ... + n nCn 2n-1
n 3n-1 = nC1 + 4 nC2 + ... + r nCr 2r-1 + ... + n nCn 2n-1
Example 8 (2008 HSC, Question 6(c), 5 marks)Example 8 (2008 HSC, Question 6(c), 5 marks)
(i) (1 + x)p+q = p+qC0 + p+qC1 x + p+qC2 x2 + ... + p+qCp+q xp+q
The term of independent of x will be
q
qp
x
x 1
Example 8 (2008 HSC, Question 6(c), 5 marks)Example 8 (2008 HSC, Question 6(c), 5 marks)
(i) (1 + x)p+q = p+qC0 + p+qC1 x + p+qC2 x2 + ... + p+qCp+q xp+q
The term of independent of x will be
qqp
q
qp
Cx
xC
is,that ,
q
qp
x
x 1
Example 8 (2008 HSC, Question 6(c), 5 marks)Example 8 (2008 HSC, Question 6(c), 5 marks)
(i) (1 + x)p+q = p+qC0 + p+qC1 x + p+qC2 x2 + ... + p+qCp+q xp+q
The term of independent of x will be
(ii) (1 + x)p = pC0 + pC1 x + pC2 x2 + ... + pCp xp
qqp
q
qp
Cx
xC
is,that ,
q
qp
x
x 1
Example 8 (2008 HSC, Question 6(c), 5 marks)Example 8 (2008 HSC, Question 6(c), 5 marks)
(ii)
If we expanded the RHS, there would be many terms
Terms independent of x
=
Example 8 (2008 HSC, Question 6(c), 5 marks)Example 8 (2008 HSC, Question 6(c), 5 marks)
(ii)
If we expanded the RHS, there would be many terms
Terms independent of x
= pC0 qC0 + pC1 x qC1 + pC2 x2 qC2 + ... + pCp xp qCp
= 1 + pC1 qC1 + pC2 qC2 + ... + pCp
qCp
px
1
x
1
2
1
x
p ≤ q
Example 8 (2008 HSC, Question 6(c), 5 marks)Example 8 (2008 HSC, Question 6(c), 5 marks)
(ii)
Terms independent of x
= pC0 qC0 + pC1 x qC1 + pC2 x2 qC2 + ... + pCp xp qCp
= 1 + pC1 qC1 + pC2 qC2 + ... + pCp
qCp
By equating the terms independent of x on both sides of
p+qCq = 1 + pC1 qC1 + pC2 qC2 + ... + pCp
qCp
1 + pC1 qC1 + pC2 qC2 + ... + pCp
qCp = p+qCq
px
1
x
1
2
1
x
q
pq
qp
xx
x
x
111
1
From (i)
And now ...And now ...
A fairly hard identity to prove:A fairly hard identity to prove:
Example 9 from 2002 HSCExample 9 from 2002 HSC
Question 7(b), 6 marksQuestion 7(b), 6 marks
Example 9 (2002 HSC, Question 7(b), 6 marks)Example 9 (2002 HSC, Question 7(b), 6 marks)
(i) (1 + x)n = nC0 + nC1 x + nC2 x2 + ... + nCn xn
= c0 + c1 x + c2 x2 + ... + cn xn [1]
Identity to be proved involves (n + 2) 2n-1 so try ...
[Aiming to prove: c0 + 2c1 + 3c2 + ... + (n + 1)cn = (n + 2) 2n-1 ]
To save time, this question asks us to abbreviate nCk to ck
Example 9 (2002 HSC, Question 7(b), 6 marks)Example 9 (2002 HSC, Question 7(b), 6 marks)
(i) (1 + x)n = nC0 + nC1 x + nC2 x2 + ... + nCn xn
= c0 + c1 x + c2 x2 + ... + cn xn [1]
Identity to be proved involves (n + 2) 2n-1 so try ...
Differentiating both sides:
n(1 + x)n-1 = c1 + 2c2 x + ... + ncn xn-1
Substituting x = ? to give 2n-1 on the LHS:
[Aiming to prove: c0 + 2c1 + 3c2 + ... + (n + 1)cn = (n + 2) 2n-1 ]
Example 9 (2002 HSC, Question 7(b), 6 marks)Example 9 (2002 HSC, Question 7(b), 6 marks)
(i) (1 + x)n = nC0 + nC1 x + nC2 x2 + ... + nCn xn
= c0 + c1 x + c2 x2 + ... + cn xn [1]
Identity to be proved involves (n + 2) 2n-1 so try ...
Differentiating both sides:
n(1 + x)n-1 = c1 + 2c2 x + ... + ncn xn-1
Substituting x = 1 to give 2n-1 on the LHS:
n(1 + 1)n-1 = c1 + 2c2 1 + ... + ncn 1n-1
n 2n-1 = c1 + 2c2 + ... + ncn [2]
How do we make the LHS say (n + 2) 2n-1?
[Aiming to prove: c0 + 2c1 + 3c2 + ... + (n + 1)cn = (n + 2) 2n-1 ]
Example 9 (2002 HSC, Question 7(b), 6 marks)Example 9 (2002 HSC, Question 7(b), 6 marks)
(i) (1 + x)n = nC0 + nC1 x + nC2 x2 + ... + nCn xn
= c0 + c1 x + c2 x2 + ... + cn xn [1]Differentiating both sides:
n(1 + x)n-1 = c1 + 2c2 x + ... + ncn xn-1 Substituting x = 1:
n(1 + 1)n-1 = c1 + 2c2 1 + ... + ncn 1n-1
n 2n-1 = c1 + 2c2 + ... + ncn [2]
Add 2 (2n-1) to both sides. But that’s 2n.
[Aiming to prove: c0 + 2c1 + 3c2 + ... + (n + 1)cn = (n + 2) 2n-1 ]
Example 9 (2002 HSC, Question 7(b), 6 marks)Example 9 (2002 HSC, Question 7(b), 6 marks)
(i) (1 + x)n = nC0 + nC1 x + nC2 x2 + ... + nCn xn
= c0 + c1 x + c2 x2 + ... + cn xn [1]
Differentiating both sides:
n(1 + x)n-1 = c1 + 2c2 x + ... + ncn xn-1
Substituting x = 1:
n(1 + 1)n-1 = c1 + 2c2 1 + ... + ncn 1n-1
n 2n-1 = c1 + 2c2 + ... + ncn [2]
To prove the 2n identity, sub x = 1 into [1] above:
(1 + 1)n = c0 + c1 1 + c2 12 + ... + cn 1n
2n = c0 + c1 + c2 + ... + cn [3]
[Aiming to prove: c0 + 2c1 + 3c2 + ... + (n + 1)cn = (n + 2) 2n-1 ]
Example 9 (2002 HSC, Question 7(b), 6 marks)Example 9 (2002 HSC, Question 7(b), 6 marks)
(i) n 2n-1 = c1 + 2c2 + ... + ncn [2]
2n = c0 + c1 + c2 + ... + cn [3]
[2] + [3]:
n 2n-1 + 2n = c1 + 2c2 + ... + ncn + c0 + c1 + c2 + ... + cn
2n-1 (n + 2) = c0 + 2c1 + 3c2 + ... + (n + 1)cn
c0 + 2c1 + 3c2 + ... + (n + 1)cn = (n + 2) 2n-1
[Aiming to prove: c0 + 2c1 + 3c2 + ... + (n + 1)cn = (n + 2) 2n-1 ]
Each coefficient ck increases by 1 as
required
Example 9 (2002 HSC, Question 7(b), 6 marks)Example 9 (2002 HSC, Question 7(b), 6 marks)
(ii) Answer involves dividing by (k + 1)(k + 2) and alternating
–/+ pattern so try integrating (1 + x)n from (i) twice and substituting x = -1.
(1 + x)n = c0 + c1 x + c2 x2 + ... + cn xn [1]
[Aiming to find ] 211
4.33.22.1210
nn
cccc nn
Example 9 (2002 HSC, Question 7(b), 6 marks)Example 9 (2002 HSC, Question 7(b), 6 marks)
(ii) (1 + x)n = c0 + c1 x + c2 x2 + ... + cn xn [1]
Integrate both sides:
To find k, sub x = 0:
[Aiming to find ]
kxn
cx
cx
cxcx
nnnn
13221
01
1321
1
1
Don’t forget the constant of integration
1
1
1321
1
11
1
000011
1
132210
1
1
nx
n
cx
cx
cxcx
n
nk
kn
nnn
n
211
4.33.22.1210
nn
cccc nn
Example 9 (2002 HSC, Question 7(b), 6 marks)Example 9 (2002 HSC, Question 7(b), 6 marks)
(ii)
Integrate again to get 1.2, 2.3, 3.4 denominators:
To find d, sub x = 0:
[Aiming to find ]
dn
x
nn
xcxcxcxcx
nn
nnn
1214.33.221
21
1 142
31
202
211
4.33.22.1210
nn
cccc nn
Example 9 (2002 HSC, Question 7(b), 6 marks)Example 9 (2002 HSC, Question 7(b), 6 marks)
(ii)
Sub x = ? for –/+ pattern:
[Aiming to find ] 211
4.33.22.1210
nn
cccc nn
Example 9 (2002 HSC, Question 7(b), 6 marks)Example 9 (2002 HSC, Question 7(b), 6 marks)
(ii)
Sub x = -1 for –/+ pattern:
[Aiming to find ] 211
4.33.22.1210
nn
cccc nn
Example 9 (2002 HSC, Question 7(b), 6 marks)Example 9 (2002 HSC, Question 7(b), 6 marks)
(ii)
[Aiming to find ] 211
4.33.22.1210
nn
cccc nn
2
1
21
1
21
12
21
1
1
1
21)1(
4.33.22.1
21
1
1
1
21
1)1(
4.33.22.10
210
2210
n
nn
n
nn
n
nnnnn
cccc
nnnnn
cccc
nn
nn
Example 10 (2007 HSC, Question 4(a), 6 marks)Example 10 (2007 HSC, Question 4(a), 6 marks)
(i) P(both green) = 0.1 0.1
= 0.01
(ii) p = 0.1, q = 1 – 0.1 = 0.9, n = 20
P(X = 2) =
Example 10 (2007 HSC, Question 4(a), 6 marks)Example 10 (2007 HSC, Question 4(a), 6 marks)
(i) P(both green) = 0.1 0.1
= 0.01
(ii) p = 0.1, q = 1 – 0.1 = 0.9, n = 20
P(X = 2) = 20C2 0.12 0.918
= 0.28517 ...
0.285
Example 10 (2007 HSC, Question 4(a), 6 marks)Example 10 (2007 HSC, Question 4(a), 6 marks)
(i) P(both green) = 0.1 0.1 = 0.01
(ii) p = 0.1, q = 1 – 0.1 = 0.9, n = 20
P(X = 2) = 20C2 0.12 0.918
= 0.28517 ... 0.285
(iii) P(X > 2) = 1 – P(X = 0) – P(X = 1) – P(X = 2)
= 1 – 20C0 0.10 0.920 – 20C1 0.110.919 – 0.285 from (ii)
= 1 – 0.920 – 20(0.1)0.919 – 0.285 = 0.32325 ... 0.32
HOW TO STUDY FOR MATHS (P-R-A-C)
1. Practise your maths
2. Rewrite your maths
3. Attack your maths
4. Check your maths
WORK HARD AND BEST OF LUCK
FOR YOUR HSC EXAMS!