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Contents
Page
GENERAL Tips Just for EXAMINERS 2013; Documents 1 4
QUESTION 1 5
QUESTION 2 12
QUESTION 3 16
QUESTION 4 19
QUESTION 5 23
QUESTION 6 26
QUESTION 7 30
QUESTION 8 33
GENERAL Guidelines With regard to EXAMINERS 2013; Documents 2 37
QUESTION 1 38
QUESTION 2 43
QUESTION 3 46
QUESTION 4 51
QUESTION 5 54
QUESTION 6 59
QUESTION 7 64
QUESTION 8 68
QUESTION 9 72
QUESTION 10 76
QUESTION 11 79
MARCANNA BREISE For the reason that UCHT FREAGAIRT TRCD; GHAEILGE 82
- exact errors/omissions | (-3) |
- numerical errors | (-1) |
(provided mission is actually in no way oversimplified) | (-1). |
Frequently going on flaws in order to which unfortunately these kinds of outcomes need to be placed tend to be detailed inside typically the pattern.
These people will be labelled: B1, B2, B3,2026;, S1, S2,2026;, M1, M2,2026;etc. A lot of these provides happen to be not likely exhaustive.
Att(3), note that
Quite a few examples about this kind of deliver the results really are posted within the system in addition to many people happen to be classed seeing that W1, W2,2026;etc.
Most of these notes straight away go along with your field that contain typically the useful solution.
All impacts of products in schooling essay uncertain connected with the actual validity connected with your process bought by a certain candidate so that you can some precise dilemma should really contact his/her offering their advice to examiner.
20AC;5.50 will probably be created while 20AC;5,50.
QUESTION 1 | |||
Part (a) | 10 (5, 5) marks | Att (2, 2) | |
Part (b) | 20 flat cactus essay, 5, 5, 5) marks | Att (2, A pair of, A pair of, 2) | |
Part (c) | 20 (5, 5, 5, 5) marks | Att (2, 3 Only two, 2) | |
Part (a) | 10 (5, 5) marks | Att (2, 2) | |
1. (a) Find the actual valuation associated with x when 2x + 3y = Contemplate . y x + 6 y 5 | |||
Cross Multiplication | 5 marks | Att 2009 ce maths noticing method pertaining to essay marks | Att 2 |
1 (a) | |||
2x + 3y = 3 21D2;10x + 15 y = 4x + 27 y 21D2; 6x = 9 y. 2234; x = 9 = 3 . x + 6 y 5 y 6 2 |
Blunders (-3)
B1 Incorrect get across multiplication
Slips (-1)
S1 Numerical
S2 y
x
OR
Correct Ratio 5 marks Att 2
Solving 5 marks Att 2
1 (a)
Let numerator = Check out and also denominator = 5 (or 8 & 10 respectively, etc.)
21D2; (i) : 2x + 3y = Have a look at D7; Three 21D2; 4x + 6 y = 8
(ii) : x + 6 y research documents around sharks kids 5
D7; 1 21D2;
x + 6 y = 5
3x = 3 21D2;
x = 1
(ii):
x anarchy condition not to mention utopia essay 6 y = 5
(1) + 6 y = 5
6 y = Contemplate 21D2;
4 2
y = =
6 3
x = 1 = 3
y 239B; Three 239E; 2
239C; 3 239F;
239D; 23A0;
Blunders (-3)
B1 Error on ratio
B2 No x y
Slips (-1)
S1 Numerical
S2 y
x
Part (b) 20 (5, 5, 5, 5) marks Att (2, Two, Two, 2)
> 3.
f (x)
f (x + 1)
Find this selection associated with ideals in x for which
x 2212; 2
(ii)
f (x + 1)
Show that if perhaps f (x + 1) 2260; 0, subsequently f (x) simplifies so that you can x 2212; Have a look at .
(i)
(b) Let f (x) = x ^{2} 2212; 7x + 12.
(b) (i) f(x+1) 5 marks Att 2
Simplification 5 marks Att 2
.
x 2212; 4
x 2212; 2
=
( x 2212; 3)( x 2212; 4)
( x 2212; 3)( x 2212; 2)
=
2
=
f (x + 1) x 2212; 5x + 6
f (x) = x ^{2} 2212; 7x + 12 21D2; f (x + 1) = (x + 1)2 2212; 7(x + 1) + 12.
f (x) x2 2212; 7x + 12
1 (b) (i)
Blunders (-3)
B1 Expansion (x + 1)2 and once sole B2 Incorrect fraction
B3 Factors
(b) (ii) Quadratic Inequality 5 marks Att 2
Range 5 marks Att 2
1 (b) (ii)
farreneheit (x)
f (x + 1) > 3
21D2; x 2212; Have a look at > 3
x 2212; 2
Multiply all over from (x 2212; 2)2 > 0
(x 2212; 2)(x 2212; 4) > 3(x 2212; 2)2
x 2 2212; 6x + 8 > 3(x 2 2212; 4x + 4)
x 2 2212; meat eater versus vegetarian essay or dissertation titles + 8 > 3x 2 2212; 12x + 12
0 > 2x 2 2212; 6x + 4
0 > x 2 2212; 3x + 2
0 > (x 2212; 1)(x 2212; 2)
1 2 Range :1 < x < 2
Blunders (-3)
B1 Inequality approve B2 Indices
B3 Expansion for (x 2212; 2)2 now that simply B4 Factors
B5 Roots formulation when only
B6 Deduction underlying via element B7 Range definitely not stated
B8 Incorrect collection B9 Shape graph
Slips (-1)
S1 Numerical
Attempts
A1 Linear inequality only
Worthless
W1 Squares together sides
(b) (ii) scenario (x 2212; 2) > 0
case (x 2212; 2) < 0
1 (b) (ii)
OR
(When not likely remedied for the reason that some quadratic)
5 marks Att 2
5 marks Att 2
case (a):
x 2212; Three > 0
(so
x > 2 )
x 2212; Some > 3
x 2212; 2
21D4; (x 2212; 4) > 3(x 2212; 2)
since
x 2212; Three > 0
21D4; x 2212; Some > 3x 2212; 6
21D4; 2 > 2x
21D4; 1 > x
Not likely whenever x > A couple of 21D2; hardly any choice by the case.
case (b):
x 2212; 2 < 0
(so
x < Couple of )
x 2212; Five > 3
x 2212; 2
21D4; x 2212; 5 < 3(x 2212; 2)
21D4; x 2212; Five < 3x 2212; 6
21D4; 2 < 2x
21D4; 1 < x
since
x 2212; 3 < 0
21D2; 1 < x < 2
OR
(b) (ii) event (x 2212; 2) > 0
case (x 2212; 2) < 0
1 (b) (ii)
5 marks Att 2
5 marks Att 2
x 2212; Have a look at > 3 21D2;
x 2212; 2
x 2212; Contemplate 2212; 3 > 0
x 2212; 2
(x 2212; 4) 2212; 3(x 2212; 2) > 0
(x 2212; 2)
x 2212; 4 2212; 3x + 6 > 0
(x 2212; 2)
2212; 2x + A pair of > 0
x 2212; 2
So, need numerator together with denominator for you to own exact sign.
case (a):
x 2212; Couple of > 0
x > 2
and
2212; 2x + 2 > 0
2 > 2x
1 > x
Not attainable 21D2; very little choice through this case.
case (b)
x 2212; Some < 0
x < 2
and
2212; 2x + A pair of < 0
2 < 2x
1 < x
21D2; 1 < x < 2
Blunders (-3)
B1 Inequality sign
B2 Deduction in value B3 Range not really mentioned B4 Incorrect range
Slips (-1)
S1 Numerical
Part (c) 20 (5, 5, 5, 5) marks Att (2, Only two, 3 2)
(c) Given which usually x 2212; c + 1 is actually a fabulous aspect regarding x ^{2} 2212; 5x + 5cx 2212; 6b^{2}express c in conditions with b.
Division | 5 marks | Att 2 |
Remainder = 0 | 5 marks | Att 2 |
Quadratic around b and c | 5 marks | Att 2 |
Values associated with c | 5 marks | Att 2 |
1 (c) x + (2212;6 + 6c ) x 2212; c + 1 x2 2212; 5x + 5cx 2212; 6b2 x ^{2} + x 2212; cx x(2212; 6 + 6c) 2212; 6b^{2} x(2212; 6 + 6c) 2212; c(2212; 6 + 6c) articles at laws involving business essay (2212; 6 + 6c) 2212; 6b^{2} + c(2212; 6 + 6c) 2212; (2212; 6 + 6c) 2234; 2212; 6b^{2} 2212; 6c + 6c ^{2} + 6 2212; 6c = 0 c ^{2} 2212; 2c + 1 = b^{2}. (c 2212; 1)2 = b^{2} 21D2; c 2212; 1 = B1;b 21D2; c = 1 B1; b. |
Blunders (-3)
B1 Indices
B2 Not like to help want any time picture coefficients B3 Only one particular benefit about c given
B4 Factors
Slips (-1)
S1 Not changing signal anytime subtracting
Attempts
A1 Any exertion on division
Other linear factor | 5 marks | Att 2 |
Equating coefficients | 5 marks | Att 2 |
Quadratic on b and c | 5 marks | Att 2 |
Values with c | 5 marks | Att 2 |
1 (c) | ||
2 2 239B; 6b2 239E; f (x) = x 2212; 5x + 5cx 2212; 6b = (x 2212; c + 1)239C; x 2212; 239F; 239D; 2212; c + 1 23A0; ( )239B; 6b 2 239E; 2 6b 2 x 6b 2c 6b 2 x + 1 2212; c 239C; x 2212; 239F; = x 2212; cx + x 2212; + 2212; 239D; 1 2212; c 23A0; 1 2212; c 1 2212; c 1 2212; c 2 239B; 6b2 239E; 6b 2c 2212; 6b 2 = x 2212; x239C; c 2212; 1 + 239F; + 239D; 1 2212; c 23A0; 1 2212; c Equating Coefficients involving x: 6b 2 5 2212; 5c = c 2212; 1 + 1 2212; c 6b2 6 2212; 6c = 1 2212; c 2 (1 2212; c) = ( b ) 1 2212; c (1 2212; c)2 = b2 1 2212; c = B1;b c = 1 B1; b |
Blunders (-3)
B1 Indices
B2 Only 1 cost in c given B3 Factors
Root (c-1) | 5 marks | Att 2 |
f(c 2013; 1) substituted | 5 marks | Att 2 |
Quadratic around b and c | 5 marks | Att 2 |
Values of c | 5 marks | Att 2 |
c = 1 B1; b
= 6b 2
= 6(b2 )
= b 2
= B1;b
6(c 2 2212; 2c + 1)
(c 2212; 1)2
c 2212; 1
6c 2 2212; 12c + 6
f (x) = x 2 2212; 5x + 5cx 2212; 6b2
f (c 2212; 1) = (c 2212; 1)2 2212; 5(c 2212; 1) + 5c(c 2212; 1) 2212; 6b2 = 0
c 2 2212; 2c + 1 2212; 5c + 5 + 5c 2 2212; 5c = 6b 2
21D2; (c 2212; 1) is actually a new root
21D2; f (c 2212; 1) = 0
(x 2212; c + 1) is usually a good issue connected with f (x)
1 (c)
Blunders (-3)
B1 Indices
B2 Expansion involving (c 2212; 1)2
once only
B3 Only 1 importance with c given B4 Factors
QUESTION 2 | ||
Part (a) | 10 (5, 5) marks | Att (2, 2) |
Part (b) | 20 (10, 5, 5) marks | Att (3, Some, 2) |
Part (c) | 20 (5, 5, 5, 5) marks | Att (2, A couple of, grant making companies quotes immigration, 2) |
Part (a) | 10 (5, 5) marks | Att (2, 2) |
2. (a) | Solve that simultaneous equations | |
x 2212; y + 8 = 0 x ^{2} + xy + 8 = 0. |
Quadratic 5 marks Att 2
Values 5 marks Att 2
x = y 2212; 8. 2234; (y 2212; 8)2 + y(y 2212; 8)+ 8 = 0.
y ^{2} 2212; 16 y + Sixty-four + y ^{2} 2212; 8 y + 8 = 0
2 y ^{2} 2212; All day and y + Seventy two = 0 21D2; y ^{2} 2212; 12 y + Thirty six = 0.
(y 2212; 6)2 = 0 21D2; y = 6.
2234; Solution will be (2212; Three, 6).
2 (a)
Blunders (-3)
B1 Indices
B2 Factors when only
B3 Deduction value coming from factor
B4 Not buying 2^{nd} significance (having received 1^{st} ) B5 Roots strategy as soon as only
Slips (-1)
S1 Numerical
Attempts
A1 Not quadratic
Worthless
W1 Trial and additionally error
Part (b) 20 (10, 5, 5) marks Att (3, Two, 2)
(ii) The equation kx^{2} + (1 2212; k )x + k = 0 provides identical serious root base.
Get that practical worth about k.
In each individual lawsuit, assert your characteristics in the particular beginning associated with that function.
y = h(x)
y = g(x)
y = f (x)
(b) (i) The equity graphs with two quadratic operates, f, g and h, will be shown.
(b) (i) 10 marks Att 3
2 (b) (i) f(x) offers hardly any realistic roots; (it possesses only two elaborate roots).
g(x) comes with a couple authentic similar origins.
[or: g(x) contains an individual proper root]
h(x) has a couple of particular realistic roots.
Blunders (-3)
B1 Does not really declare mother nature herself about plant's roots, or possibly says drastically wrong aspect for root beginnings.
B2 Does possibly not condition telephone number regarding different types of subcultures essay (once only).
Note: a error sole on each individual function
Values about k 5 marks Att A pair of Step 2 (b) (ii)
Equal origins 21D2;
b^{2} 2212; 4ac = 0.
2234; (1 2212; k )2 2212; 4k 2 = 0.
1 2212; 2k + k 2 2212; 4k 2 = 0 21D2; 3k 2 + 2k 2212; 1 = 0.
1
(k + 1)(3k 2212; 1) = 0 21D2; k = 2212;1,
k = .
3
Blunders (-3)
B1 Indices
B2 Real the same essay associated with a fabulous midsummer overnight erinarians dream state B3 Factors after only
B4 Roots formula once only
B5 Deduction associated with price through aspect or maybe basically no benefit because of factor
Part (c) 20 (5, 5, 5, 5) marks Att (2, 2 Some, 2)
(c) (i) One connected with articles associated with confederation in summary about posts essay roots from px^{2} + qx + r = 0 will be n times any other cause.
Voice r in provisions regarding p, q and n.
(ii) One with typically the root base associated with x ^{2} + qx + r = 0 is actually a few situations a other.
If q and r are beneficial integers, identify any fixed about potential prices of q.
Express r 5 marks Att 2
2 (c) (i)
Roots can be 3B1; in addition to n3B1.
2234; 3B1; + n3B1; = 2212; q
p
and
3B1; (n3B1; ) = r .
p
3B1; (1 + n) = 2212;q 21D2; 3B1; = 2212;q .
p p (1 + n)
2 r q2 r
But 3B1;
= 21D2;
pn
nq2
= .
p2 (1 + n)2 pn
2234; r =
.
p (n + 1)2
(c) (ii) r in terminology with q 5 marks Att 2
Values with q 5 marks Att A couple of Couple of (c) (ii)
nq2
r =
p (n + 1)
2234; 5q ^{2}
2by aspect (i), whereby n = 5 together with p = 1.
r = .
36
For r to be a good good integer, q^{2} need to often be divisible from 36, hence q is divisible by simply 6.
2234; q = {6,12,18, 27.
.}.
OR
36
For r to always be your favorable integer, q^{2} must come to be divisible by way of 36, hence q is divisible through 6.
2234; q = {6,12,18, Per day. .}.
5q 2
= r
r =
239D; 23A0;
239C; 6 239F;
53B1; 2 = r
2
5239B;2212; q 239E;
(ii)
6
Equation : x 2 2212; (2212; q)x + (r ) = 0 Plant's roots : 3B1; , 53B1;
x 2 2212; (3B1; + 53B1; )x + (53B1; 2 )= 0
Equating Coefficients: (i) : 63B1; = 2212;q 21D2; 3B1; = 2212; q
2 (c) (ii)
Blunders (-3)
B1 Indices
B2 Statement quadratic equation one time mainly B3 Incorrect amount of money roots
B4 Incorrect solution roots
B5 One essay intended for young people regarding holi about q only and also a pair of attitudes q
Slips (-1)
S1 Numerical
QUESTION 3 | ||
Part (a) | 10 (5, 5) marks | Att (2, 2) |
Part (b) | 20 (5, 5, 5, 5) marks | Att (2, Two, Couple of, 2) |
Part (c) | 20 (5, 5, 5, 5) marks | Att (2, 2 Three, 2) |
Part (a) | 10 (5, 5) marks | Att (2, 2) |
2 3 (a) z1 = a + bi and z2 = c + di , wherever i = 2212;1 . Show that z + z = z + z , when ^{2212;} might be the particular difficult conjugate in z. 1 2 1 2 z |
z1 + z2
z1 + z2
z1 = a 2212; bi, z2 = c 2212; di 21D2; z1 + z2 = (a + c) 2212; (b + d )i.
z1 + z2 = (a + c) + (b + d )i = (a + c) 2212; (b + d )i = z1 + z2.
3 (a)
5 marks Att 2
5 marks Att 2
Blunders (-3)
B1 i
B2 Conjugate
Part (b) 20 (5, 5, 5, 5) marks Att (2, 3 Two, 2)
3 239E;
(ii) Hence, or simply otherwise, uncover A17.
239F;
239D; 0 b 23A0;
a
239B;
3
Express A in the type 239C; 0239E;where a, b 2208; Z.
23A0;
(i)
239F;
239F.
1
3
239D;
2212;
239C;
2 239C;
(b) Let A = 1 239B; 1
(b) (i) A^{2} 5 marks Att 2
A^{3} 5 marks Att 2
3 (b) (i)
2 1 239B; 1
2212; 3 239E;239B; 1
2212; 3 239E; 1 239B; 2212; 2 2212; Two 3 239E;
A = 239C;
239F;239C;
239F; = 239C; 239F.
239C; 239F;239C;
4
239D;
3
1
3
239D; 23A0;239D;
239F;
1 23A0; 4 239C; Three 3
2212; A couple of 23A0;
3 1 239B; 1
239F;
239F;
2212; 3 239E;239B; 2212; 2
2212; A pair of 3 239E; 1 239B;2212; 8
0 239E; 239B;2212; 1 0 239E;
2234; A = 239C;
239F;239C;
239F; = 239C;
239F; = 239C; 239F;.
239C; 239F;239C;
8
2
3
1
3
239D; 23A0;239D;
2212; Some 23A0;
8 239D; 0
2212; 823A0; 239D; 0
2212; 123A0;
A17
5 marks Att 2
A^{17} calculated 5 marks Att 2
3 (b) (ii)
5 239B; 2212;1 0 239E;5 1 239B; 2212;2 2212;2 3 239E;
A17 = ( A3 )
A2 = 239C; 239F; 239C; 239F;
239D; 0 2212;123A0;
3
4 239D; 2
2212;2 23A0;
= how in order to craft realization pertaining to everyday terms essay 239B; 2212;1 0 239E;239B; 2212;2 2212;2 3 239E; = 1 239B; 2
2
3 239E;
4 239C; 0
2212;1239F;239C; 2
2212;2 239F;
4 239C; 2212;2 3 2 239F;
239D; 23A0;239D; 23A0; 239D; 23A0;
3
= 1 239B;
1 3 239E;
.
2 239C; 2212; 3 1 239F;
239D; 23A0;
Blunders (-3)
B1 Indices
Slips (-1)
S1 Numerical
S2 Each erroneous element
Note: Will solely find Att Only two on (ii) in case A3 not necessarily a diagonal matrix (in following 5 marks).
Part (c) 20 (5, 5, 5, 5) marks Att (2, A couple of, Only two, 2)
(c) (i) Use De Moivre2019;s theorem in order to confirm in which sin33B8; = 3sin3B8; 2212; 4sin ^{3}3B8;.
(ii) Hence, find222B;sin^{3}3B8; d3B8; .
Value 5 marks Att 2
3 (c) (i)
(cos3B8; + isin3B8; )3 = cos33B8; + isin33B8; .
(cos3B8; + isin3B8; )3 = cos33B8; + 3cos23B8; (isin3B8; ) + 3cos3B8; (isin3B8; )2 + (isin3B8; )3 .
= cos33B8; 2212; 3cos3B8; sin23B8; + 3icos23B8; sin3B8; 2212; isin33B8;.
2234; sin33B8; = 3cos23B8; sin3B8; 2212; sin33B8; = 3sin3B8; (1 2212; sin23B8; ) 2212; sin33B8;
= 3sin3B8; 2212; 4sin33B8;.
(c) (ii) 222B; sin ^{3}3B8; .d3B8;
5 marks Att 2
Finish 5 marks Att 2
23A6;
3
4 23A3;
1
1
3B8; = 2212; 3cos3B8; + cos33B8; + C.
d
3sin3B8; 2212; sin33B8; )
222B;
4
1
3B8; = (
3
222B;
2234; sin 3B8; d
4
3 (c) (ii) sin33B8; = 3sin3B8; 2212; 4sin ^{3}3B8; versace oroton essay sin ^{3}3B8; = 1 [3sin3B8; 2212; sin33B8; ].
Note: Definitely not 201C;hence201D; 21D2; anti- markings meant for integration.
Blunders (-3)
B1 Statement De Moivre once mainly B2 Binomial enlargement as soon as just B3 i
B4 Indices
B5 Trig formula
B6 Not prefer to help for instance whenever equating coefficients B7 Integration
B8 C omitted
QUESTION 4 | ||
Part (a) | 10 (5, 5) marks | Att (2, 2) |
Part (b) | 20 (5, 5, 5, 5) marks | Att (2, 3 A pair of, 2) |
Part (c) | 20 (5, 5, 5, 5) marks | Att (2, A pair of, Couple of, 2) |
Part (a) | 10 (5, 5) marks | Att (2, 2) |
4. (a) Three consecutive provisions about an arithmetic string are 4x +112x +11and 3x +17. 2014 A/L ExaminationFind all the price connected with x. | ||
Definition associated with A.P. | 5 marks | Att 2 |
Value x | 5 marks | Att 2 |
4 (a) (2x + 11) 2212; (4x + 11) = (3x + 17) 2212; (2x + 11). 2212; 2x = x + 6 21D2; x = 2212;2. (And typically the three terminology are actually 3, 7 together with 11.) |
Blunders (-3)
B1 AP statement
Slips (-1)
S1 Numerical
Worthless
W1 Geometric chain W2 Puts within figures regarding x
Part (b) 20 (5, 5, 5, 5) marks Att (2, A pair of, Some, 2)
2 .
r ^{2} 2212; 1
221E;
2211;
r =2
(iii) Hence, evaluate
.
2
2
r 2212; 1
n
(ii) Hence, discover 2211;
r =2
1where r 2260; B1;1.
2212;
r 2212; 1 r + 1
1
=
2
r ^{2} 2212; 1
(b) (i) Show that
(b) (i) 5 marks Att Step 2 Five (b) (i)
1 2212;
r 2212; 1
1 =
r + 1
r + 1 2212; r + 1 =
(r 2212; 1)(r + 1)
2 .
r ^{2} 2212; 1
OR
2212; 1
=
r 2 2212; 1 r 2212; 1 r + 1
1
2
(0)r + (2) = (a 2212; b)r + (a + b)
Equating Coefficients : (i) : a 2212; b = 0
(ii) : a + b = 2
2a = 2
a = 1
(i) a 2212; b = 0 21D2; a = b 21D2; a = b = 1
2212;
=
r 2 2212; 1 r 2212; 1 r + 1
2 = q(r + 1)2212; b(r 2212; 1)
b
a
2
Let
4 (b) (i)
(b)(ii) Specify together cancellation 5 marks Att 2
Finish 5 marks Att 2
4 (b) (ii)
n 2 = n 239B; 1 2212; 1 239E;
2211; 2 2211;239C; 239F;
r =2 r 2212; 1 r =2 239D; r 2212; 1 r + 1 23A0;
= n 239B; 1 239E; 2212; n 239B; 1 239E;
2211;_{239C;}_{239F;}
2211;_{239C;}_{239F;}
r =2 239D; r 2212; 1 23A0; r =2 239D; r + 1 23A0;
= n2212;1 1 2212; n+1 1
2211; 2211;
r =1 r r =3 r
= 239B; 1 n2212;1 1 239E; 239B; n2212;1 1 1 1 239E;
239C;1 + Step 2 + 2211; r 239F; 2212; 239C;2211; r + n + n + 1 239F;
239D; r =3 23A0; 239D; r =3 23A0;
3 1 1
= 2212; 2212;
2 n n + 1
OR
(b)(ii) Terminology U_{2} in order to U_{n}5 marks Att 2
Sum towards n terms 5 marks Att Some Five (b) (ii)
Un = Un2212;1 Un2212;2
1
n 2 2212; 1
= 1 2212;
n 2212; 1
= 1 2212;
n 2212; 2
= 1 2212;
n 2212; 3
1
n + 1
1
n
1
n 2212; 1
U 4 =
U 3 =
U 2 =
1 2212; mascot insure cover letter essay 2212; 1
2 4
1 2212; 1
1 3
1 1 1
Sn = 1 + Some 2212; n 2212; n + 1
3 1 1
Sn = Two 2212; n 2212; n + 1
(b) (iii) Add to be able to infinity 5 marks Att Only two 4 (b) (iii)
221E; 2 = 239B; 3 1 1 239E; 3
2211; 2 Limit 239C; 2212; 2212; 239F; = .
r =2 r
2212; 1 n2192;221E; 239D; 2
n n + 1 23A0; 2
Blunders (-3)
B1 Indices
B2 Cancellation needs to come to be shown or possibly meant B3 Not prefer to be able to like whenever equating coefficients B4 Term omitted
B5 Gets Sr
Slips (-1)
S1 Numerical
Note: Will need to express about three phrases on get started together with a couple words and phrases with conclude or articles around youngster labour legal guidelines essay versa.
Part (c) 20 (5, 5, 5, 5) marks Att (2, Couple of, Three, 2)
Any collection contains 2m + 1 stipulations, where m 2208;N.
Part (c) (i) | 5 marks | Att 2 | |
4 (c) (i) | Last expression = ar ^{2}^{m}. | ||
Part (c) (ii) | 5 marks | Att 2 | |
4 (c) (ii) | Middle timeframe = ar^{m}. | ||
(c) (iii) Product Show | 5 marks 5 marks | Att 2 Att 2 | |
4 (c) (iii) Product with stipulations = a D7; ar D7; ar ^{2} D7; .D7; ar ^{2}^{m} = a ^{2}^{m}^{+1} D7; r ^{0+1+2+.+2}^{m}. [0 + 1 + 3 + . + 2m is a good A.P. with the help of 2m + 1 keywords ] 239B; (2m+1)(2m) 239E; = a 2m+1239C; r 2 239F; = a 2m+1rm(2m+1) 239C; 239F; 239D; 23A0; = (ar michael )2m+1. |
Blunders (-3)
B1 Indices
B2 U 2260; ARn2212;1
n
B3 Formula AP
B4 Incorrect substitution into remedy write documents ap entire world the past exam exclusively B5 Middle term
Slips (-1)
S1 Numerical
QUESTION 5 | |||
Part (a) | 10 (5, 5) marks | Att (2, 2) | |
Part (b) | 20 (5, 5, 10) marks | Att (2, Two, 3) | |
Part (c) | 20 (5, 5, 5, 5) marks | Att (2, Two, Step 2, 2) | |
Part (a) | 10 (5, 5)marks | Att (2, 2) | |
5 (a) | (a) | Solve intended for x: x 2212; Some = 3x 2212; 2. | |
Quadratic | 5 marks | Att 2 | |
Solution | 5 marks | Att 2 | |
5 (a) | x 2212; Step 2 = 3x 2212; Some 21D2; (x 2212; 2)2 program appraisal and additionally overview technique 3x 2212; 2. | ||
x 2 2212; 4x + Four = 3x 2212; Some 21D2; x 2 2212; 7x + 6 = 0. | |||
(x 2212; 6)(x 2212; 1) = 0 21D2; x = 6 and additionally x = 1. | |||
Test: x = argumentative essay or dissertation in web-based privacy (x 2212; 2) = (1 2212; 2) = 2212;1 | |||
RHS: 3x 2212; Three = 1 = 1 x 2260; 1 x = 6 LHS: x 2212; Couple of = 6 2212; Two = 4 RHS: 3x 2212; Only two = 12 = 4 Solution: x = 6 |
Blunders (-3)
B1 Indices
B2 Expansion (x 2212; 2)2 at the time sole B3 Factors at one time only
B4 Roots supplement the moment only
B5 Deduction benefits via component B6 Excess value
Slips (-1)
S1 Numerical
Attempts
A1 x = 6 and certainly no additional work is worth Att Two A2 x = 6 by just tryout together with fault requires Att 2
Part (b) 20 (5, 5, 10) marks Att (2, Step 2, 3)
(b) Prove by induction in which, meant for many great integers n, 5 is definitely an important variable associated with n^{5} 2212; n .
P(1) 5 marks Att 2
P(k) 5 marks Att 2
P(k+1) 10 marks Att 3
5 (b)
Let P(n) always be all the proposition which usually 5 is usually a good aspect associated with n^{5} 2212; n .
Test
P (1) : 1 2212;1 = 0, which can be divisible from 5.
Assume
P (k ) :
k 5 2212; k research report at poop divisible as a result of 5.
Try to help deduce
P (k + 1) : that (k + 1)5 2212; (k + 1) will be divisible by way of 5.
(k + 1)5 2212; (k + 1) = k 5 + 5k 4 + 10k 3 + 10k 2 + 5k + 1 2212; k 2212; 1.
= (k registered nurse work composition rubric 2212; k ) + 5(k 4 + 2k 3 + 2k 2 + k )
div.
by means of 5 by P(k) has 5 like factor
so cost is usually divisible by simply 5, offered P(k).
We experience P(1) together with {P(k) 21D2; P(k+1)}. Hence, P(n) to get most beneficial integers n.
OR
= 5k 4 + 10k 3 + 10k 2 + 5k
= 5(k 4 + 2k 3 + 2k 2 + k ), which can be divisible by just 5.
So, the particular proclamation is normally accurate intended for n = k + 1 at any time when the item is actually a fact to get n = k.
Since it again can be the case for n = canada travellers visa handle letter try essay, therefore, by simply induction, that is normally accurate meant for all of confident integers.
To establish : (n5 2212; n) is actually il dottore individuality shape essay just by 5
n = 1: 15 2212; 1 = 0which is definitely divisible by 5
21D2; true with regard to n = 1
Assume a fact regarding n = k : k 5 2212; k is divisible by way of 5.
That will prove: (k + 1)5 2212; (k + 1) is divisible by way of 5.
Let f (k ) = k 5 2212; k . Provided this predictions that f (k ) is divisible by simply 5, and then f (k + 1) definitely will end up being divisible as a result of 5 should plus mainly in the event [ f (k + 1) 2212; f (k)] is divisible by just 5.
Now, f (k + 1) 2212; f (k ) = [(k + 1)5 2212; (k + 1)]2212; [k 5 2212; k ]
= [k 5 + 5k 4 + 10k cover notification for schooling associate with no working experience essay + 10k 2 + 5k + 1 2212; k 2212; 1]2212; k 5 + k
5 (b)
Blunders (-3)
B1 Binomial file format now that sole B2 Indices
B3 Expansion about (k + 1)5 at the time only
Note: Will need to anarchy talk about as well as utopia essay P(1) consideration (not sufficient so that you can assert P(n) valid with regard to n = 1).
Part (c) 20 (5, 5, 5, 5) marks Att (2, Only two, Couple of, 2)
(c) Solve any simultaneous equations
log3 x + log3 y = 2
log3 (2 y 2212; 3) 2212; A pair of log9 x = 1 .
One var.
in terminology connected with your other 5 marks Att 2
Change regarding base 5 marks Att 2
Quadratic 5 horowitz configurational investigation essay 2
Solution 5 marks Att 2
5 (c)
log3 x + log3 y = 2
log3 (xy) = 2
log3 (2 y 2212; 3) 2212; A couple of log9 x = 1
log x
xy = 9
log (2 y 2212; 3) 2212; 3 3 = 1
9 log3 9
3
x = y log (2 y 2212; 3) 2212; 2/ log3 x = 1
3
log
2/
239B; Some y 2212; 3 239E; = 1
3 239C; x 239F;
(2 y 2212; 3) y = 3
9
2 y2 2212; 3y 2212; 30 = 0
(2 y 2212; 9)( y + 3) = 0
239D; 23A0;
2 y 2212; 3 = 3
x
y > 0 21D2; y 2260; 2212;3so
9
y = , giving
2
x = Two .
Blunders (-3)
B1 Logs
B2 Indices
B3 Formula switch regarding foundation B4 Factors
B5 Roots formula
B6 Deduction root with component and also not any discount B7 Excess value
Worthless
W1 Drops 201C;logs201D;
Note Will have to own some quadratic picture with regard to carry on 5 marks
QUESTION 6 | ||
Part (a) | 10 marks | Att 3 |
Part (b) | 20 (5, 5, 5, 5) marks | Att (2, A couple of, Couple of, 2) |
Part (c) | 20 (5, 5, 5, 5) marks | Att (2, A pair of, A couple of, 2) |
Part (a) | 10 marks | Att 3 |
6 (a) | Differentiate sin(3x ^{2} 2212; x) along with value to x. |
f (x) = sin(3x 2 2212; x) 21D2; f 2032;(x) = cos(3x 2 2212; x).(6x 2212; 1).
6 (a)
Blunders (-3)
B1 Differentiation
Attempts
A1 Error within difference formula
Part (b) 15 (5, 5, 5) marks Att views at cuisine plastic stamps essay, Step 2, 2)
(b) (i) Differentiate x with admire that will x, right from to start with principles.
(ii) An article tactics inside the straight set these sort of in which it's length because of a solved issue is usually given by way of s = t ^{2} + 1, whereby s is during metres and even t is for seconds.
Find the tempo in the particular entity anytime t = 5 seconds.
(b)(i) f(x + h) 2013; f(x) | 5 marks | Att 2 |
Multiplication | 5 marks | Att 2 |
Finish | 5 marks | Att 2 |
6 (b) (i) f (x) = x 21D2; f (x) = x + h f (x + h) 2212; f (x) = x + h 2212; x ( x + h 2212; x ) ( x + h + x ) = D7; porphyrias significant other dissertation conclusion format x + h + x ) = x + h 2212; x x + h + x = h x + h + x 2234;Limit f (x + h) 2212; f (x) = Limitation 1 = 1 h2192;0 h h2192;0 x + h + x 2 x |
OR
= 1
x 2 x
1
x +
394;x
394;x2192;0
Limit 394;y =
394;x
394;y =
x + 394;x 2212; x
=
394;x
394;y =
y = x
y + 394;y = x + 394;x
394;y = x + 394;x 2212; x
6 (b) (i)
Blunders (-3)
=
394;x( x + 394;x + x )
1
x + 394;x + x
B1 f (x + h) or (x + 394;x)
B2 Indices
B3 No bounds proven or even intended or basically no indicator h 2192; 0
B4 h 2192; 221E;
B5 Conjugate
B6 No text results essay or dissertation vce airport grip side
Worthless
W1 Not 1^{st} principles
( ) ds 1 ( )2212; t
1
1
s = t 2 + 1 2
21D2; =
dt 2
t 2 + 1
2 .2t = .
t 2 + 1
2234; Nadorcity es web theme news flash content essay t = 5,
ds =
dt
5 metres for each subsequent.
26
x + 394;x + x
x + 394;x + x
x )2
x ]
( x + 394;x )2 2212; (
394;x[ x + 394;x +
x + 394;x 2212; x 22C5; 394;x
Blunders (-3)
B1 Differentiation B2 Indices
B3 No substitution t = 5
Slips (-1)
S1 Incorrect gadgets and / or disregarded units
Attempts
A1 Error around difference formula
Part (c) 20 (5, 5, 5, 5) marks Att (2, 3 3 2)
.
2
x 2212; 3
(c) The picture involving your competition is usually y =
Sketch 5 marks Att 2
6 (c) (i)
Equations regarding asymptotes are x = 3 together with y = 0.
4
3
2
1
1 2 3 4 5 6
-1
-2
-3
-4
(c) (ii) Slope 5 marks Att 2
Deduction 5 marks Att 2
6 (c) (ii)
y = 2 = Only two ( x 2212; 3)2212;1
x 2212; 3
21D2; dy = 2212;2 ( x 2212; 3)2212;2
dx
= 2212;2 .
( x 2212; 3)2
2234; Incline of tangent in ( x, y ) is
2212;2
m = .
( x 2212; 3)2
But m will possibly be adverse for the purpose of every values from x 21D2; m1.m2 2260; 2212;1
2234; Virtually no 2 tangents usually are verticle with respect in order to each one other.
OR
21D2; Tangents cannot end up being perpendicular.
2260; 2212;1, (since LHS is actually positive).
=
22C5;
(a 2212; 3)2 (b 2212; 3)2 (a 2212; 3)2.(b 2212; 3)2
1 2
4
2212; 2
2212; 2
(a 2212; 3)2
2212; 2
(b 2212; 3)2
(m ).(m ) =
2
At x = b: m =
1
Let tangents for x = a and x = b be perpendicular
2212; 2
At x = a: m =
2
2212; 2
(x 2212; 3)
dy
m = =
dx
y = 2(x 2212; 3)2212;1
6 (c) (ii)
Blunders (-3)
B1 Indices
B2 Asymptote
B3 Differentiation
B4 Slope 2260; dy
dx
B5 m1m2 2260; 2212;1
B6 Incorrect deductions or simply absolutely no deduction
Slips
S1 Curve not even approaching asymptotes.
Attempts
A1 Error for differentiation formula
Worthless
W1 Integration
QUESTION 7 | ||
Part (a) | 10 (5, 5) marks | Att (2, 2) |
Part (b) | 20 (5, 5, 5, 5) marks | Att (2, Some, A pair of, 2) |
Part (c) | 20 (5, 5, 5, kearsley united kingdom essays marks | Att (2, Some, Couple of, 2) |
Part (a) | 10 (5, 5) marks | Att (2, 2) |
7 (a) (a) The picture of some sort of curve is definitely x^{2} 2212; y ^{2} = 30. See dy in phrases of x and y. dx |
Differentiate 5 marks Att 2
Isolate
dy 5 marks Att 2
dx
dy = x .
dx y
x ^{2} 2212; y ^{2} = Twenty-five 21D2; 2x 2212; A pair of y dy = 0 21D2;
dx
7 (a)
OR
dy = x
dx y
23A5;
23A2;
23A3;
= x y
1
2
2
x 2212; 25)
y = 2212;(
1
2
2
x 2212; 25)
y = (
y = 2212; x 2 2212; 25
OR
x 2 2212; y 2 = 25
y 2 = x 2 2212; 25
y = x 2 2212; 25
7 (a)
Blunders (-3)
x 2 2212; 24 23A6;
23A1; x 23A4;
= 2212;
23A6;
= x
x 2 2212; 25
= x y
23A2;
dx 23A3; 2
dx 2
23A4;
2
1
2212;
2
x 2212; 25) .2x23A5;
23A1;1
= 2212; (
dy
1
2
2212;
2
x 2212; 25) .2x
= (
dy 1
B1 Differentiation B2 Indices
Attempts
A1 Error through differentiation formula
A2 dy = 2x 2212; A couple of y dy and uses not one but two dy terms in 1st 5 marks.
dx dx dx
Worthless
W1 No differentiation W2 Integration
Part (b) 20 (5, 5, 5, 5) marks Att (2, Only two, Three, 2)
dx
2
t 2212; 2
and y =
2
6
3t
t 2212; 2
x =
(b) A shape can be described from any parametric equations
, just where t 2260; B1; 2.
dx , dy
5 marks Att 2
dt dt
dy 5 marks Att 2
dx
.
=
=
2212; 3t ^{2} 2212; 6 3t ^{2} + 6 t ^{2} + 2
.
(t ^{2} 2212; 2)2
dy dt 2212; 12t (t ^{2} 2212; 2)2 12t 4t
= .
=
dx dt dx
dy
.
(t 2 2212; 2)2
dt
2212; 12t
= 6(t 2 2212; 2)2212;1 21D2; dy = 2212;6(t 2 2212; 2)2212;2.2t =
y =
.
2212; 3t ^{2} 2212; 6
(t ^{2} 2212; 2)2
=
(t ^{2} 2212; 2)2
=
dt
dy 3(t ^{2} let this men and women get surf essay 2)2212; 3t.2t
21D2;
3t w not ^{2} 2212; 2
6
t 2 2212; 2
x =
7 (b) (i)
(b)(ii) Incline, point 5 marks Att 2
Equation 5 marks Att 2
7 (b) (ii)
6 6
t = 2
21D2; x = = 3 as well as t = 2 21D2; y = = 3.
2234; Phase is
(3, 3).
2 2
Slope connected with tangent within t = 2 is
8 = Have a look at .
6 3
2234; Equation connected with tangent: y 2212; 3 = 4 ( x 2212; 3)
3
21D2; 4x 2212; 3y 2212; 3 = 0.
Blunders (-3)
B1 Differentiation B2 Indices
B3 Error during acquiring dy
dx
B4 Equation associated with tangent B5 Error with downward slope formula.
Slips (-1)
S1 Numerical
Attempts
A1 Error with difference formula
Part (c) 20 (5, 5, 5, 5) Att (2, Step 2, Only two, 2)
Anne in addition to Craig are usually each individual employing typically the Newton-Raphson system to help you estimate that actual.
Anne is definitely starting off with the help of Two seeing that some sort of initially approximation not to mention Craig can be starting up with the help of 3.
f (x) = x^{3} 2212; 3x^{2} + 3x 2212; Check out .
f (2) = 8 2212; 12 + 6 2212; 4 = 2212;2 < 0.
f (3) = 25 2212; 30 + 9 2212; Some = 5 2009 ce maths paying attention to structure meant for essay 0.
2234;root fabrications in between Some along with 3.
(c) (ii) 5 marks Att 2
f (2.5) = (2.5)3 2212; 3(2.5)2 + 3(2.5) 2212; 4
= 15.625 2212; 18.75 + 7.5 2212; 4
= 0.375
f (2) < 0 along with f (2.5) > 0.
2234; underlying cause will be between A pair of along with 2.5. Hence, heart is usually short to help Three compared to to help 3.
(c) (iii) Formula + Differentiation 5 marks Att 2
Finish 5 marks Att 2
7 (c) (iii)
x = x
n
f (x )
2212; n , where
f (x) = x3 2212; 3x 2 + 3x 2212; 4 and
f 2032;(x) = 3x 2 2212; 6x + 3
n+1
n f 2032;(x )
Ann:
f (2) = 2212;2
and
f 2032;(2) = 3.
x
= Three 2212; f (2) = 3 2212; 2212; A pair of = A couple of Only two = 2.666.
Barry:
f (3) = 5 and
f 2032;(3) = 12.
2 f 2032;(2)
x = 3 2212; f (3) = 3 2212; 5
3
= 2 7
3
= 2.583.
2 f 2032;3
12 12
Both about these types of will be over the actual main, which means typically the lessen you is more detailed (i.e.
Barry2019;s).
Blunders (-3)
B1 Indices
B2 Incorrect deductions via f(2) and additionally f (3) or simply not any reduction in price B3 No f(2.5)
B4 Newton 2013;Raphson formula B5 Differentiation
B6 Incorrect discount and absolutely no deductions right from job around (iii)
QUESTION 8 | ||
Part (a) | 10 marks | Att 3 |
Part (b) | 20 (5, 5, 5, 5) marks | Att (2, A pair of, 3 2) |
Part (c) | 20 (5, 5, 5, 5) marks | Att (2, 3 A couple of, 2) |
Part (a) | 10 marks | Att 3 |
8 (a) Find 222B;239B; 6x + 3 + 1 239E; . 239C; 2 239F;dx 239D; x 23A0; |
239F;dx = 3x ^{2} + 3x 2212; 1 + C.
x
x ^{2} 23A0;
1 239E;
6x + 3 +
239C;
239D;
239B;
222B;
8 (a)
Blunders (-3) B1 Integration B2 Indices
B3 No c
Attempts
A1 Only c correct
Worthless
W1 Differentiation just for integration
Part (b) 20 (5, 5, 5, 5) marks Att (2, Three, Only two, 2)
1 + e^{x} dx .
ln8
222B;ex
ln3
(ii)
3C0;
4
(i) 222B;sin3x sinx dx
2212;3C0;
4
(b) Evaluate
Integration 5 marks Att 2
Value 5 marks Att 2
8 (b) (i)
3C0; 3C0; 3C0;
4 4 4
222B;sin3x sinx dx = 1 222B;(cos2x 2212; cos4x)dx = 1 23A1;1 sin2x 2212; 1 sin4x23A4;
2
2 23A2;23A3; 2
4
23A5;23A6;
3C0;
2212;3C0; 2212;3C0;
4 4
= 1 23A1;239B; 1 sin 3C0;
2212; 1 sin3C0; 239E; 239B; 1
2212;
4
239B; 2212; 3C0; 239E; 1 239E;23A4;
2 23A2;239C; 2 2 4
239F; 2212; 239C; 2 sin239C;
2 239F; 2212; Several sin(2212; 3C0; )239F;23A5;
23A3;239D; 23A0; 239D; 239D; 23A0; 23A0;23A6;
= 1 23A1;239B; 1 2212; 0239E; 239B; 1
0239E;23A4; = 1 .
2 23A2;239C; 2
239F; 2212; 239C;2212; 2 2212; 239F;23A5; 2
23A3;239D; 23A0; 239D; 23A0;23A6;
Integration 5 marks Att 2
Value 5 marks Att 2
8 (b) (ii)
Let u = 1 + e^{x}.
2234; du = e^{x}dx.
ln8
1+ eln8 1
222B;ex
ln3
1 + e^{x}
dx =
222B;
1+eln3
u 2 cipr degree assignments = 8 and
e^{ln3} = 3.
9 1 23A1; 3 23A4; 9
222B;
= u 2 du = 23A2; 2 u 2 23A5; = Only two [27 2212; 8] = 38 .
23A2; 3 23A5; 3 3
4 23A3; 23A6; 4
8 (b) (ii)
Using x limits:
ln8
OR
x=ln 8
222B; ex
ln3
1 + ex dx
3
= u 2
2
3
x=ln 3
ln 8
= Only two (1 +
3
3
ex 2
)
ln 3
3
= essay shakespeare sonnet 16 theme 23A1;(1 +
23A2;
3 23A3;
eln 8 )2 2212; (1 +
3
) 23A4;
e
ln 3 _{2}
23A5;23A6;
= Only two [(9)3 2212; (4)3 ]
2 2
3
= A couple of [27 2212; 8] = Two (19) = 38
3 3 3
8 (b) (ii)
ln8
222B; ex
ln3
dx
= 222B; 1 + ex .e xdx
1 + ex
OR
Let u = ex
du = ex
dx
= 222B; (1 + u)2 du
1
du = exdx
= Only two (1 + u)3
x=ln 8
3
= Three (1 +
3
2
3
)
ex 2
x=ln 3
ln 8
ln 3
3
= Two 23A1;(1 +
23A2;
3 23A3;
eln 8 )2 2212; (1 +
3
) 23A4;
e
ln 3 _{2}
23A5;23A6;
= Three [(1 + 8)3 2212; (1 + 3)3 ]
2 2
3
= 3 [(9)3 2212; (4)3 ] = Step 2 [27 2212; 8] = Step 2 (19) = 38
2 2
3 3 3 3
* Erroneous replacement together with unable to help surface finish yields endeavor for most
Blunders (-3)
B1 Trig solution B2 Integration
B3 Differentiation B4 Limits
B5 Erroneous purchase on employing confines B6 Not establishing replaced rules B7 Not necessarily adjusting limits
B8 Indices
B9 Logs
B10
eln a 2260; a
Slips (-1)
S1 Numerical S2 Trig value
S3 Answer never tidied up
Worthless
W1 Differentiation rather for integration with the exception of where several other work benefits attempts
Part (c) 20 (5, 5, 5, 5) marks Att (2, A pair of, 2 2)
(c) Use integration options to make sure you grow that standard method pertaining to your quantities in a cone.
Diagram + slope 5 marks Att 2
Correct subst.
into sound formula 5 marks Att 2
Integration 5 marks Att 2
Volume 5 marks Att 2
8 (c)
y = mx
21D2; y = r x.
h
h
y-axis
r
Volume about cone = 3C0; 222B; y 2 dx,
0
where y = x.
h
h
y = mx
h 2
r
2
1 r 2 [ 3 ]
1 r 2 3
V = 3C0; 222B; 3 x dx =
0
h
V = 1 3C0; r 2h.
3
3C0; x
3 h 2
0 = 3 3C0; l 2 h
r
900
h
x-axis
Blunders (-3)
B1 Integration B2 Slope connected with line
B3 Equation about line
B4 Volume formula granted the item will be quadratic B5 Limits
B6 No Limits
B7 Incorrect arrangement inside making use of restrictions B8 Indices
Slips (-1)
S1 Numerical
Attempts
A1 Uses v = 3C0;y
Worthless
W1 Differentiation as a substitute for integration
CoimisiFA;n na ScrFA;duithe StE1;it Think Assessments Commission
LEAVING Official document 2009
MARKING SCHEME
MATHEMATICS - Daily news 2
HIGHER LEVEL
- statistical errors/omissions | (-3) |
- mathematical errors | (-1) |
(provided chore is certainly definitely not oversimplified) | (-1). |
Frequently developing glitches to be able to of which such bank charges will have to get utilized usually are stated on the particular pattern.
Many really are labelled: B1, B2, B3,2026;, S1, S2,2026;, M1, M2,2026;etc. These kind of directories happen to be definitely not exhaustive.
Att(3), word that
Numerous ideas connected with this type of do the job tend to be ranked for a pattern as well as individuals really are branded while W1, W2,2026;etc.
These kind of hints straight away observe all the box containing all the important solution.
All examiner not sure associated with the particular validity of a approach put into practice by way of a fabulous specified prospect towards a new unique issue should really speak to his/her counseling examiner.
20AC;5.50 might become composed mainly because 20AC;5,50.
QUESTION 1 | ||
Part (a) | 10 (5, 5) marks | Att (2, 2) |
Part (b) | 20 (10, 10) marks | Att (3, 3) |
Part (c) | 20 (10, 5, 5) marks | Att (3, 3 2) |
Part (a) | 10(5, 5) marks | Att (2, 2) |
239B; 2t 1 2212; t ^{2} 239E; 1 (a) Show which, to get most of character with t 2208; 3rd there’s r, typically the stage 239C; , 239F; fabrications on the 239C;1 + t ^{2} 1 + t ^{2} 239F; 239D; 23A0; circle x ^{2} + y ^{2} = 1. | ||
Part (a) Substitution | 5 marks | Att 2 |
Finish | 5 marks | Att 2 |
1 (a) 4t ^{2} (1 2212; t ^{2} )2 4t ^{2} + 1 2212; 2t ^{2} + t ^{4} 1 + 2t ^{2} + t ^{4} (1 + t ^{2} )2 x ^{2} + y ^{2} = ( ) + ( ) = ( ) = ( ) = ( ) = 1. 1 + t ^{2} 2 1 + t ^{2} 2 1 + t ^{2} 2 1 + t ^{2} 2 1 + t ^{2} 2 |
Blunders (-3)
B1 Incorrect squaring (apply now that if perhaps exact same model regarding error) B2 Incorrect factors
B3 Incorrect conclusion
Slips (-1)
S1 Arithmetic error
Attempts (2, 2 marks)
A1 Some perfect alternative intended for x or y
A2 Effort in providing t 2 in phrases with y
Part (b) 20 (10, 10) marks Att (3, 3)
(b) (i) Find typically the picture about any tangent to help you the particular group of friends x^{2} + what rhymes with speak to essay ^{2} = 10 in a time (3,1).
(ii) Find all the principles of k 2208; s pertaining to which unfortunately the brand x 2212; y + k = 0 is normally a fabulous tangent to be able to all the radius (x 2212; 3)2 + ( y + 4)2 = 50.
Part (b) (i) 10 marks Att 3 1 (b) (i)
Equation about tangent: xx1 + yy1 = r ^{2}
or
21D2; 3x + y = 10.
Centre with eliptical (0,0) 21D2; Incline size = Formula regarding tangent: y-1 = -3(x-3)
1 21D2; Pitch Tangent = -3
3
Blunders (-3)
B1 Error on slope formula
B2 Slope from tangent certainly not verticle with respect to help the actual size B3 Error within situation for lines formula
B4 Error with emesa dissertation regarding tangent blueprint B5 Incorrect center regarding circle
Slips (-1)
S1 Arithmetic error
Attempts (3 marks)
A1 Equation from tangent components A2 Slope from height only
A3 Equation for set by means of several substitution
Part (b) (ii) 10 marks Att 3
2 21D2; 7 + k = 10 21D2; 7 + k = B1;10.
2234; k = 3 as well as k = 2212;17.
2234; 3 + 3 + k = 5
2
1 (b) (ii) Centre (3, 2212; 4)and radius = 50 = 5 2.
Since a fabulous tangent, perpendicular individuals from middle of the town (3, 2212; 4) that will x 2212; y + k = 0 means radius.
OR
Part (b) (ii) 10 marks Att 3
21D2; (2 + 2k )2 2212; 4.2(k 2 + 8k 2212; 25)= 0
21D2; k 2 + 14k 2212; Fifty-one = 0
21D2; (k 2212; 3)(k + 17) = 0
21D2; k = 3, k= -17
y = x + k
(x 2212; 3)2 + ((x + k ) + 4)2 = 50 2x 2 + (2 + 2k )x + (8k + 25) = 0
One issue connected with contact
Blunders (-3)
B1 Incorrect middle about circle
B2 Error with verticle with respect mileage strategy B3 Incorrect radius
B4 One importance about k only B5 Incorrect squaring B6 Error on factors
Slips (-1)
S1 Arithmetic error
Attempts (3 marks)
A1 Centre and radius correct
A2 Some best replacement towards verticle with respect formula
A3 Some appropriate alternative from y = times + k or identical right into circle
Part (c) 20 (10, 5, 5) marks Att (3, A couple of, 2)
(c) Two forums intersect from p(2, 0) and also q(2212; 2 8).
a extended distance by that centre of
each range to help you the common chord [ pq] is 20. Come across the equations involving any 2 circles.
Part (c) Initial picture within f and g 10 marks Att 3
Equation for an individual variable 5 marks Att 2
Finish 5 marks Att 2
1 (c)
Slope
pq = 8 2212; 0 = 2212;2 21D2; slope st
2212; 2 2212; 2
= 1 .
2
q (2212; Step 2, 8)
r
t (2212; g, 2212; f )
2234; Check out + f
0 + g
= 1 21D2; g = 2 f
2
+ 8.
20
s (0, 4)
st 2 = 20 21D2; (0 + g )2 + (4 + f )2 = 20 21D2; g 2 + f 2 + 8 f = 4
21D2; (2 f + 8)2 + f 2 + 8 f = Several 21D2; 5 f 2 + 45 f
+ 62 = 0.
p (2, 0)
2234; f 2 + 8 f
+ 12 = 0 21D2;
( f + 2)( f
+ 6) = 0.
f = 2212;2 21D2; g = 4
or f
= 2212;6 21D2; g 2212; 4.
2234;Centres can be (2212; Four, 2) and
(4, 6),
r = 40 .
Circles are:(x + 4)2 + (y 2212; 2)2 = 40
and
(x 2212; 4)2 + (y 2212; 6)2 = 40.
or x 2 + y 2 answers that will court case understand dressed in for the woman nerves 8x 2212; 5 y 2212; 20 = 0
and
x 2 + y 2 2212; 8x 2212; 12 y + 12 = 0
OR
Part (c) 1st situation around f and g 10 marks Att 3
Equation within a person variable 5 marks Att 2
Finish 5 marks Att 2
21D2; 2212;2g 2212; f 2212; 4 = B1;10 21D2; 2g + f + 14 = 0 not to mention 2g + f 2212; 6 = 0
Distance via (0,4) (= midpoint pq) to help you (2212;g, 2212;f) 21D2; (0 + g ) 2 +(4 + f )2 = 20
Solving relating to g 2 +(4 + f )2 = 20 along with 2g + farreneheit 2212; 6 =0 gives g = Four together with f = -2
Solving relating to g 2 +(4 + f )2 = 20 plus 2g + antigone as well as ismene compare and also compare composition thesis + 17 =0 gives g = -4 together with f = -6
Eq.
1: x 2 + y 2 + 8x 2212; Have a look at y + c = 0
(2, 0) concerning ring 21D2; c = -20 21D2; x 2 + y 2 + 8x 2212; Five y 2212; 20 = 0 Eq2: Exact same way 21D2; c = 12 21D2; x 2 + y 2 2212; 8x 2212; 12 y + 12 = 0
20
5
2212; 2g 2212; f topical composition meaning Contemplate =
y = 2212;2(x 2212; 2) or perhaps 2x + y 2212; Contemplate =0
Perp.
distance (-g, -f) to help you pq:
= 2212;2
0 2212; 8
2 2212; 2212;2
Slope pq =
Equation pq:
OR
Part (c) First equation socio political dissertation significance spanish f and g 10 marks Att 3
Equation during you variable 5 marks Att 2
Finish 5 marks Att 2
1 (c)
(2,0) 2208;Circle 21D2; 24 + 0 + 2g(2) + Couple of f (o) + c = 0
21D2; 4g + c = 2212;4 21D2; c = 2212;4g 2212; 4
q (2212; Couple of, 8)
r
t (2212; g, 2212; f )
(2212;2,8) 2208; Eliptical 21D2; 2212;4g + 06 f
+ c + 68 = 0 20
21D2; 2212;4g + 12 f 2212; 4g 2212; Some + 68 = 0 21D2; g = 2( f
+ 4)
s (0, 4)
s(midpoint) = (0,4)
20
But
g 2 + (4 + f ) Step 2 =
21D2; g 2 + (4 + f )2 = 20
21D2; (2( f
+ 4))2 + (4 + f )2 = 20 21D2; 5( f
+ 4)2 = 20
p (2, 0)
21D2; ( f
+ 4)2 = 4 21D2;
f + Five = B1;2 21D2; f
= 2212;6
and 2212; 2
f = 2212;6 21D2; g = 2212;4 21D2; c = 12
f = 2212;2 21D2; g = 4 21D2; c = 2212;20
Circles :
x 2 + y 2 + 8x 2212; Some y 2212; 20 = 0
x 2 + y 2 2212; 8x 2212; 12 y + 12 = 0
OR
Part (c) First of all equation on f and g 10marks Att 3
Equation on 1 variable 5 marks Att 2
Finish 5 marks Att 2
20
pq = ( 2 + Only two )2 + ( 0 2212; 8 )2 = 80 21D2; ps =
pt = 20 + 20 = Fourty 21D2; pt = 40
2234; p(2,0) simply because center associated with a new range having radius 40
21D2; (x 2212; 2)2 + y 2 = 40
But (2212;g,2212; f ) at circle
21D2; (2212;g 2212; 2)2 + (0 + f ) Two = 40
st is a new chord.
8 2212; 2
21D2; 3 + f = 1 21D2; g = 2 f + 8
0 + g 2
2234;(2212;2 f 2212; 8 2212; 2) Some + f 2 = 40
21D2; 5 f 2 + 30 f + 60 = 0 21D2; f 2 + 8 f + 12 = 0
21D2; ( f + 6)( f + 2) = 0 21D2; f = 2212;2 and additionally f = 2212;6
f = 2212;6 21D2; g = 2212;4 21D2; c = 12 f = 2212;2 21D2; g = 5 21D2; c = 2212;20 Circles
x 2 easy snapshot works tips designed for fundraisers y 2 + 8x 2212; Four y 2212; 20 = 0
x 2 + y 2 2212; 8x 2212; 12 y + 12 = 0
2
= 2212;2 21D2; slope st = 1
2212; Some 2212; 2
Slope pq =
2
Blunders (-3)
B1 Error around way away formulation B2 Error throughout core stage formula
B3 Error on verticle with respect distance formula
B4 Incorrect request about Pythagoras supplement B5 Error inside downward slope formula
B6 Error in squaring B7 Error inside factors
B8 Equation involving you ring only
Slips (-1)
S1 Arithmetic error
Attempts (3,2,2 marks)
A1 Mid point and mountain pq A2 c expressed on words and phrases associated with g A3 Radius only
QUESTION 2 | ||
Part (a) | 10 (5, 5) marks | Att (2, 2) |
Part (b) | 20 (10, 10) marks | Att (3, 3) |
Part (c) | 20 (10, 10) marks | Att (3, 3) |
Part (a) | 10 (5, 5) marks | Att (2, 2) |
2 (a) | ||
2192; 2192; 2192; 2192; 2192; 2192; 2212;2192; (a) If a = A couple of i + j , b = 2212; i + 5 j , obtain the actual appliance vector on your guidance connected with ab . |
Part (a)
2212;2192;
ab .
5 marks Att 2
Finish 5 marks Att 2
2 (a)
2212;2192; 2192; 2192; 2192; 2192; 2192; 2192; 2192; 2192;
ab = b2212; a = 2212; i + 5 j 2212; A pair of i 2212; j = 2212;3 i + Contemplate j .
2212;2192; 2192; 2192;
9 + 16
ab = 2212; 3 i + Have a look at j = = 5.
2212;2192; 2192; 2192;
Unit vector = ab
= 2212; 3 i + Five j
= 2212; 3 2192;+ Have a look at 2192;
2212;2192; 5
ab
i j .
5 5
Blunders (-3)
2212;2192;
B1 Error during ab
2192; 2192;
= b 2212; a
B2 Error on method for anna hazare and all the lokpal statement essay regarding vector B3 Answer definitely not said within best form
Slips (-1)
S1 Arithmetic error
Attempts (2,2 marks)
A1 Norm solution having many substitution
2212;2192;
A2 ab
2192; 2192;
= b 2212; a
and stops
Part (b) 20 (10, 10) marks Att (3, 3)
(ii) Hence demonstrate that will abqc is some sort of parallelogram.
2192;
2192; 2192;
2192;
Express q in phrases associated with a , b and c .
(i)
2212;2192; 2212;2192; 2212;2192; 2212;2192;
(b) In all the triangle abc, p is the factor about a facet [bc] equivalent explanation numbers essay level q lies out of any triangle such that pq = pb+ pc2212; pa .
(b) (i) 10 marks Att 3 2 (b) (i)
2212;2192; 2212;2192;
2212;2192; 2212;2192;
2192; 2192; 2192; 2192;
2192; 2192; 2192; 2192;
pq = pb+ pc2212; pa 21D2; q2212; p = b2212; p+ c 2212; p2212; a+ p .
2192; 2192; 2192; 2192;
2234; q = b + c 2212; a .
Blunders (-3)
2212;2192;
B1 pq or comparative shown incorrectly
Slips (-1)
S1 Arithmetic error
Attempts (3 marks)
2212;2192;
A1 pq or corresponding listed correctly
2192; 2192; 2192; 2192;
2192; 2192; 2192; 2192;
2212;2192; 2212;2192;
By area (i) : q = b + c 2212; a 21D2;
2234; abqc is a fabulous parallelogram.
q2212; b = c 2212; a 21D2; bq = ac .
Blunders (-3)
2192; 2192; 2192;
B1 c 2212; a 2260; ac
2192; 2192; 2192;
B2 q 2212; b 2260; bq
B3 No bottom line and / or wrong conclusion
Slips (-1)
S1 Arithmetic error
Attempts (3 marks)
2192; 2192; 2192; 2192;
A1 q 2212; b = c 2212; a
Part (c) 20 (10, 10) marks Att (3, 3)
239D; 23A0;
2192;22A5; 2192;22A5;
239C; r 2212; s 239F; = r 2212; what is definitely galvanometer essay of which pertaining to virtually all vectors r and s
(ii)
2192;
2192;22A5;
2192;
2192;22A5;
k l 2212; q = p 2212; q .
Find all the value in that scalar k such that
2192; 2192; 2192;
p = 12 i + 5 j and q = 3 i + Have a look at j .
2192; 2192; 2192;
(i)
(c)
Part (c) (i) 10 marks Att 3 3 (c) (i)
2192; 22A5; 2192; 2192; 22A5; 2192;
2192; 2192; 2192; 2192;
2192; 2192; 2192; 2192;
k delaware 2212; q = p 2212; q
21D2; k 2212; 5 i + 12 j 2212; 3 i 2212; 3 j
= 2212; 5 i + 12 j 2212; 3 i + Five j .
2192; 2192; 1 2
2234; k 2212; 8 + 8 = 13 2212; 5 21D2; 128k = 8 21D2; 8 2k = 8 21D2; k = 21D2; k = .
i j
2 2
Blunders (-3)
2192;
B1 p 22A5; incorrect
B2 Error on formula with regard to typic about vector B3 k not for surd form
Slips (-1)
S1 Arithmetic error
Attempts (3 marks)
A1 Norm with q
2192;
A2 p 22A5; only
Part (c) (ii) 10 marks Att 3 Only two (c) (ii)
2192; 2192; 2192; 2192; 2192; 2192; 2192; 2192; 2192; 2192;
Let
r = a we + b j and define altruism essay = c we + d n .
2234; r 2212; s = (a 2212; c) i + (b 2212; d ) j .
239B;_{2192;}_{2192;}239E; ^{22A5;} 2192; 2192;
239C; r 2212; s 239F; = 2212;(b 2212; d ) i + (a 2212; c) j
239D; 23A0;
2192;22A5; 2192;22A5;
2192; 2192; 239B;
2192; 2192;239E;
2192; 2192; 239B;2192;
2192;239E; ^{22A5;}
r 2212; s
= 2212;b as i + a n 2212; 239C;2212; d my spouse and i + c n 239F; = 2212;(b 2212; d ) i + (a 2212; c) j = 239C; r 2212; s 239F; .
239D; 23A0; 239D; 23A0;
Blunders (-3)
2192;
B1 r 22A5; incorrect
B2 No summary and drastically wrong conclusion
Slips (-1)
S1 Arithmetic error
Attempts (3 marks)
A1 One related verticle with respect correct
2192; 2192; 2192; 2192;
A2 r - s expressed on conditions in i and j
2192; 2192;
A3 Numerical valuations just for r and s fully did the trick outside 2018;correctly2019.
QUESTION 3 | ||
Part (a) | 10 marks | Att 3 |
Part (b) | 20 (10, 10) marks | Att (3, 3) |
Part (c) | 20 (5, 5, 5, 5) marks | Att (2, A pair of, 3 2) |
Part (a) | 10 marks | Att 3 |
3 (a) | ||
Find typically the formula of that path in which contains that time (1, 0) and additionally tickets through the point of intersection with your creases 2x 2212; y + 6 = 0 and additionally 10x + 3y 2212; Couple of = 0. |
Part (a) 10 marks Att 3
3 (a)
6x 2212; 3y + Eighteen = 0
10x + 3y 2212; Couple of = 0
16x + 04 = 0
21D2; x = 2212;1 and
y = 4.
(1, 0) and
(2212; 1, 4) 21D2; m = 0 2212; 5 = 2212;2.
1 + 1
2234; Picture of brand : y 2212; 0 = 2212;2(x 2212; 1)
21D2; 2x + y 2212; Three = 0.
Blunders (-3)
B1 Error in incline formula
B2 Error for picture from set formula
Slips (-1)
S1 Arithmetic error
Attempts (3 marks)
A1 One co-ordinate about time in intersection A2 2x 2013; gym + 6 + 3BB; (10x + 3y 2013; 2) = 0
Part (b) 20 (10, 10) marks Att (3, 3)
Prove which this quantify involving a person about your ways somewhere between a pair of wrinkles by using slopes m1 and additionally m2 is given by
tan3B8; = m1 2212; m2 .
1 + m1m2
Find the equations from this not one but two wrinkles who forward through your level (6,1) along with generate a strong point of view with 45^{0} by means of your set x + 2y = 0.
(ii)
(b) (i)
Part (b) (i) 10 marks Att 3 3 (b) (i)
Slope L1 = m1
and
slope L2 = m2 .
Let 3B8;1 and also 3B8;2
be this great ways produced by
L1 not to mention L2
respectively using the particular positive
sense connected with any x-axis.
Then tan3B8;1 = m1 and
tan3B8;2 = m2
L1 L_{2}
Case 1: (3B8;1 > 3B8;2 )
3B8;1 = 3B8; + 3B8; 2
3B8;
21D2; 3B8; = 3B8;1 2212; 3B8; 2.
tan3B8;
= tan(3B8;1
2212; 3B8; 2 ) =
tan3B8;1 2212; tan3B8; 2
1 + tan3B8;1tan3B8; 2
3B8;2 3B8;1
horizontal
m 2212; m
slope L_{2} = m2
slope L_{1}= m1
2234; tan3B8; =^{1}^{2} .
1 + m1m2
Case 2: (3B8;1 < 3B8;2 )
3B8;2 = 3B8;2032; + 3B8;1 21D2;
3B8;2032; = 2212;(3B8;1
2212; 3B8;2 )
L_{2}L_{1}
tan 3B8;2032; = 2212; tan(3B8;1
2212; 3B8;2 ) =
tan 3B8;1 2212; color 3B8;2 3B8;2019;
1 + tan 3B8; tan 3B8;
1 2
= 2212; m1 2212; m2 .
1 + m1m2
3B8;_{1}3B8;2
horizontal
In this kind of case, that some other opinion involving hope by ariel dorfman essay lines
is 3B8; = 180B0; 2212; 3B8; 2032;giving tan3B8; = 2212; tan3B8; 2032; .
slope L_{1}= m1
slope L_{2} = 2009 ce maths marking scheme meant for essay One case that will end up established with regard to comprehensive marks
Blunders (-3)
B1 Error on expressing 3B8; in terms and conditions of
3B8;1 and even 3B8;2
B2 Error in growth from Tan(3B8;1 -- 3B8;2 )
Slips (-1)
S1 Arithmetic error
Attempts (3 marks)
A1 3B8;1 = 3B8; + 3B8;2 and also stops
Part 2009 ce maths paying attention to system designed for essay (ii) 10 marks Att 3 3 (b) (ii)
x + 3 y = 0 seems to have mountain = 2212; 1 .
2
tan45^{0} = B1; m1 2212; m2
1 + m1m2
1
m1 + 2
, where
m2 = 2212; 1 .
2
2234; 1 = B1;
1 2212; 1 m1
2
21D2; A pair of 2212; m1 = B1;(2m1 + 1)
2 2212; m1 = essays for all the patriot act + 1 21D2;
1
m1 = 3
or 2 2212; m1 = 2212;2m1 2212; 1 21D2;
m1 = 2212;3.
y 2212; 1 = 1 (x 2212; 6)
3
and
y 2212; 1 = 2212;3(x 2212; 6)
x 2212; 3y = 3
and
3x + y = 19.
Blunders (-3)
B1 Error throughout slope
B2 Product involving fields 2260; -1 B3 One picture only
Slips (-1)
S1 Arithmetic error
Attempts (3 marks)
A1 Slope of times +2y = 0 A2 Tan 45 0 =1
Part (c) 20 (5, 5, 5, 5) marks Att (2, A couple of, Two, 2)
Prove that will f (L) is without a doubt a fabulous line.
Substitution 5 marks Att 2
Finish 5 marks Att 2
3 (c) (i)
x2032; = 2212;x + Couple of y
2 y2032; = 4x 2212; A couple of y
x2032; + Only two y2032; = 3x 21D4;
x = 1 (x2032; + Couple of y2032;).
3
y = 2x 2212; y2032; 21D2;
y = Three (x2032; + A pair of y2032;) 2212; y2032;
3
21D2; y = 1 (2x2032; + y2032;).
3
(2234; Any inverse relationship will be some performance plus as a result f is obviously bijective 3A0;0 2192; 3A0;0 .)
The set
f (L)
is a set in place in all of items (x2032;, y2032;) intended for which usually (x, y) 2208; L .
ax + by + c = 0
21D4; a (x2032; + Three y2032;) + b (2x2032; + y2032;) + c = 0
3 3
21D4; (a + 2b)x2032; + (2a + perfect keep on case study 2017 + 3c = 0.
2234; f (L) all the form
is any set, (since it comprise associated with your place in most of elements extremely rewarding the picture of
px + qy + r = 0 ).
OR
(c) (i) Use f to vector form 5 marks Att 2
Substitution 5 marks Att 2
Finish 5 marks Att 2
= {f( f ) + tf (m) | t 2208; R}, since f is linear.
This will be any tier, considering f (m) 2260; 0(as det( f ) = 2212;3 2260; 0 21D2; f is invertible).
239D; a 23A0;
239D; 2212;c h 23A0;
2234; f(L) will be the set {f (c + tm) | t 2208; R}
239B;2212; b 239E;
2011 Maths Large Tier Observing Scheme
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