Mathematics Higher Level

of 70

L.17/20

Pre-Leaving Certificate Examination, 2011

Mathematics Higher Level

Marking Scheme

Paper 1 Pg. 3 Paper 2 Pg. 33

2011 L.17/20_MS 2/72 of 70 examsDEB


Mathematics

Higher Level Marking Scheme

General Instructions

1. Penalties of three types are applied to student’s work, as follows:

• Blunders - mathematical errors / omissions B (−3) • Slips - numerical errors S (–1) • Misreadings (provided task is not oversimplified) M (–1)

Frequently occurring errors to which these penalties must be applied are listed in the scheme. They are labelled B1, B2, B3, ..., S1, S2, S3, ..., M1, M2, M3, ..., etc. Note that these lists are not exhaustive.

2. When awarding attempt marks, e.g. Att. (3), it is essential to note that:

• any correct relevant step in a part of a question merits at least the attempt mark for that part • if deductions result in a mark which is lower than the attempt mark, then the attempt mark

must be awarded • a mark between zero and the attempt mark is never awarded.

3. Worthless work must be awarded zero marks.

4. The phrase “hit or miss” means that partial marks are not awarded - the student receives all of the relevant marks or none.

5. The phrase “and stops” means that no more work of merit is shown by the student.

6. Special notes relating to the marking of a particular part of a question are indicated by a double asterisk. These notes immediately follow the relevant solution.

7. The sample solutions for each question are not intended to be exhaustive - there may be other correct solutions.

8. Unless otherwise indicated, accept the best of two or more attempts - even when attempts have been cancelled.

9. The same error in the same section of a question is penalised once only.

10. Particular cases, verification and answers derived from diagrams (unless requested) qualify for attempt marks at most.

11. A serious blunder, omission or misreading merits the ATTEMPT mark at most.

12. Allow comma for decimal point, e.g. €3.50 may be written as €3,50.

examsDEB

2011 L.17/20_MS 3/72 of 70 examsDEB


Mathematics

Higher Level – Paper 1 Marking Scheme (300 marks)

Attempt SIX QUESTIONS (50 marks each) General Instructions: See Page 2.

1. (a) Simplify 7132

2142

2

−+−+xx

xx. (10, Att. 3)

7132

2142

2

−+−+xx

xx =

)7)(12(

)3)(7(

+−−+

xx

xx

= 12

3

−−

x

x

Blunders (–3) B1: Factors. B2: Roots formula - apply once only. B3: Not like to like. B4: Answer not simplified.

Slips (–1) S1: Numerical slip.

Attempts (3) A1: One equation factorised and stops.

1(b) and are the roots of the equation x2 + 5x + 9 = 0.

(i) Find the value of α1

+ β1

. (10, Att. 3)

+ = –5 = 9

α1

+ β1

= αβ

αβ +

= –9

5

Blunders (–3) B1: + incorrect. B2: incorrect. B3: Incorrect solution given.


Attempts (3) A1: + correct and stops. A2: correct and stops.

examsDEB

2011 L.17/20_MS 4/72 of 70 examsDEB

1(b) (ii) Hence, find the quadratic equation whose roots are α1

+ 3

1 and

β1

+ 3

1. (10, Att. 3)

Sum of roots = α1

+ 3

1 +

β1

+ 3

1

= α1

+ β1

+ 3

2

= –9

5 +

3

2

= 9

)2(35 +−

= 9

65 +−

= 9

1

Product of roots =

+

+

3

11

3

11

βα

= αβ1

+ α3

1 +

β3

1 +

9

1

= αβ1

+

+

βα11

3

1 +

9

1

= 9

1 +

−

9

5

3

1 +

9

1

= 9

2 –

27

5

= 27

5)2(3 −

= 27

56 −

= 27

1

x2 – (sum of roots)x + (product of roots) = 0

x2 – 9

1x +

27

1 = 0

27x2 – 3x + 1 = 0

Blunders (–3)

B1: α1

+ 3

1 +

β1

+ 3

1 incorrect.

B2:

+

+

3

11

3

11

βα incorrect.

B3: Incorrect equation.


Attempts (3)

A1: α1

+ 3

1 +

β1

+ 3

1 correct and stops.

A2:

+

+

3

11

3

11

βα correct and stops.

2011 L.17/20_MS 5/72 of 70 examsDEB

1(c) x – 3 is a factor of both x3 + x2 – 3ax – 3a = 0 and x3 – ax2 + bx – 3b = 0.

(i) Find the value of a. (5, Att. 2)

x – 3 is a factor of x3 + x2 – 3ax – 3a = 0 f (x) = x3 + x2 – 3ax – 3a = 0

f (3) = 0 33 + 32 – 3a(3) – 3a = 0 27 + 9 – 9a – 3a = 0 36 – 12a = 0 –12a = –36 a = 3

Blunders (–3) B1: –3 used (incorrectly). B2: Error in indices.


Attempts (2) A1: 3 filled into incorrect equation.

(ii) Find the roots of the equation x3 + x2 – 3ax – 3a = 0. (10, Att. 3)

x2 + 4x + 3 x – 3 ) x3 + x2 – 9x – 9 −x3 + 3x2 . 4x2 – 9x – 9 –4x2 + 12x . 3x – 9 –3x + 9. 0

(x – 3)(x2 + 4x + 3) = 0 (x – 3)(x + 3)(x + 1) = 0 x – 3 = 0 x = 3

x + 3 = 0 x = –3

x + 1 = 0 x = –1

Blunders (–3) B1: Indices. B2: Not like with like. B3: Factors. B4: Roots formula - apply once only. B5: Missing root.

Slips (–1) S1: Numerical slip. S2: Not changing sign when subtracting in division.

Attempts (3) A1: Attempt at division.

2011 L.17/20_MS 6/72 of 70 examsDEB

1(c) (iii) Find the range of values of b ∈ ℝ for which all the roots of the equation x3 – ax2 + bx – 3b = 0 are real. (5, Att. 2)

x3 – ax2 + bx – 3b = 0 x3 – 3x2 + bx – 3b = 0

x2 + b . x – 3 ) x3 – 3x2 + bx – 3b −x3 + 3x2 . bx – 3b –bx + 3b. 0

(x – 3)(x2 + b) = 0 x – 3 = 0 x = –3

x2 + b = 0 x2 = –b

For all roots to be real –b > 0 b < 0

Blunders (–3) B1: Not like with like. B2: Indices. B3: No value of b given.

Slips (–1) S1: Numerical slip. S2: Not changing sign when subtracting in division.

Attempts (2) A1: Attempt at division.

2011 L.17/20_MS 7/72 of 70 examsDEB

2. (a) Solve for x and y:

2x + 5y = 11 ax + 2y = 3a + 2,

where a ∈ ℝ. (10, Att. 3)

2x + 5y = 11 (× 2) ax + 2y = 3a + 2 (× –5)

4x + 10y = 22 –5ax – 10y = –15a – 10 (4 – 5a)x = –15a + 12 (4 – 5a)x = 3(4 – 5a)

x = a

a

54

)54(3

−−

= 3

2x + 5y = 11 2(3) + 5y = 11 6 + 5y = 11 5y = 11 – 6 = 5 y = 1

Blunders (–3) B1: Multiplying one side of equation only. B2: Second variable not found.

Slips (–1) S1: Numerical slip. S2: Not changing sign when subtracting.

Worthless (0) W1: Trial and error only.

2(b) (i) Express )3(2

131

+ 31

3 + )3(2

331

in the form qp

3 , where p, q ∈ ℤ. (10, Att. 3)

)3(2

131

+ 31

3 + )3(2

331

= 31

3 (2

1 + 1 +

2

3)

= 31

3 (2

4 + 1)

= 31

3 (2 + 1)

= 31

3 (3)

= 311

3+

= 34

3

Blunders (–3) B1: Indices. B2: Like to like. B3: Not in required form.

B4: 1 + 2

3 ≠

3

4.


Attempts (3)

A1: 2

1 + 1 +

2

3 = 3 and stops.

2011 L.17/20_MS 8/72 of 70 examsDEB

2(b) (ii) Solve the inequality

2

3

−+

x

x < 3, where x ∈ ℝ and x ≠ 2. (10, Att. 3)

2

3

−+

x

x < 3

(x – 2)(x + 3) < 3(x – 2)2 x2 + x – 6 < 3(x2 – 4x + 4) x2 + x – 6 > 3x2 – 12x + 12 2x2 – 13x + 18 > 0 (2x – 9)(x – 2) > 0

Consider: (2x – 9)(x – 2) = 0 2x – 9 = 0 2x = 9

x = 2

9

x – 2 = 0 x = 2

x < 2 and x > 2

9

Blunders (–3) B1: Inequality sign. B2: Indices. B3: Expansion of (x – 2)2. B4: Factors - apply once only. B5: Roots formula - apply once only. B6: Range not stated. B7: Incorrect range.


Attempts (3) A1: Linear equation A2: (x – 2)2 and stops.

x

y

2011 L.17/20_MS 9/72 of 70 examsDEB

2(c) Let f (x) = 2x and g (x) = 2−x.

(i) Given that f (x + 3) + f (x + 1) = kf (x), find the value of k. (5, Att. 2)

f (x + 3) + f (x + 1) = kf (x) 2x + 3 + 2x + 1 = k2x (2x)(23) + (2x)(2) = k2x 8(2x) + 2(2x) = k2x 10(2x) = k2x 10 = k

Blunders (–3) B1: Indices. B2: Like to like. B3: Value of k not stated.


Attempts (2) A1: 2x + 3 = (2x)(23) and stops. A2: 2x + 1 = (2x)(2) and stops.

(ii) Given that g (x – 5) + g (x – 3) = lg (x), find the value of l. (5, Att. 2)

g (x – 5) + g (x – 3) = lg (x) 2–(x – 5) + 2–(x – 3) = l(2–x) 2–x +5 + 2–x + 3 = l(2–x) (2–x)(25) + (2–x)(23) = l(2–x) 32(2–x) + 8(2–x) = l(2–x) 40(2–x) = l(2–x) 40 = l

Blunders (–3) B1: Indices. B2: Like to like. B3: Value of l not stated.


Attempts (2) A1: 2–x +5 or 2–x + 3 and stops.

(iii) Find the value of x for which

f (x + 3) + f (x + 1) = g (x – 5) + g (x – 3). (10, Att. 3)

kf (x) = lg (x) 10(2x) = 40(2–x)

2x = 4

x2

1

(2x)2 = 4 22x = 22 2x = 2 x = 1

Blunders (–3) B1: Indices. B2: Like to like. B3: Value of x not stated.


Attempts (3) A1: 10(2x) = 40(2–x) and stops.

2011 L.17/20_MS 10/72 of 70 examsDEB

3. (a) Given that (z + 1)(2 – i) = 3 – 4i, where i2 = –1, express z in the form a + bi, where a, b ∈ ℝ. (10, Att. 3)

(z + 1)(2 – i) = 3 – 4i

z + 1 = i

i

−−

2

43

= i

i

−−

2

43.

i

i

++

2

2

= 2

2

4

4836

i

iii

−−−+

= )1(4

)1(456

−−−−− i

= 5

510 i−

= 2 – i z = 2 – i – 1 z = 1 – i

Blunders (–3) B1: Error in conjugate. B2: i2 ≠ –1. B3: 2 – i and stops.


Attempts (3)

A1: i

i

−−

2

43 and stops.

3(b) (i) Given that A =

− 31

12, find A–1. (5, Att. 2)

A =

− 31

12

A–1 = )1)(1()3(2

1

−−

−21

13

= 16

1

+

−21

13

= 7

1

−21

13

Blunders (–3) B1: Incorrect or no determinant of matrix. B2: Like to like.


Attempts (2) A1: Determinant or matrix correct.

2011 L.17/20_MS 11/72 of 70 examsDEB

3(b) (ii) Express the simultaneous equations

2a + b = 5 –a + 3b = 1

in matrix form and use matrix methods to solve them. (10, Att. 3)

2a + b = 5 –a + 3b = 1

− 31

12

b

a =

1

5

7

1

−21

13

− 31

12

b

a =

7

1

−21

13

1

5

10

01

b

a =

7

1

−25

115

b

a =

7

1

7

14

b

a =

1

2

a = 2 b = 1

Blunders (–3) B1: Incorrect matrix. B2: Incorrect or no determinant of matrix. B3: A–1A ≠ I. B4: Pre-multiply.


Attempts (3) A1: Attempt at multiplying A–1A.

(iii) Find the values of c and d which satisfy the matrix equation

−−dc

cd

− 31

12 =

− 70

53 (5, Att. 2)

−−dc

cd

− 31

12 =

− 70

53

−−dc

cd

− 31

12.7

1

−21

13 =

− 70

53.7

1

−21

13

−−dc

cd

10

01 =

7

1

−−+−+

14070

10359

−−dc

cd =

7

1

−− 147

714

−−dc

cd =

−− 21

12

c = –1 d = –2

Blunders (–3) B1: Incorrect matrix. B2: Incorrect or no determinant of matrix. B3: AA–1 ≠ I. B4: Post-multiply.


Attempts (2) A1: Attempt at multiplying AA–1.

2011 L.17/20_MS 12/72 of 70 examsDEB

3(c) (i) Express –2 + 2 3 i in the form r(cos + isin ), where i2 = –1,

and hence, express (–2 + 2 3 i)4 in the form 4n(–2 + 2 3 i), where n ∈ ℕ. (10, Att. 3)

–2 + 2 3 i = r(cos + i sin )

r = 22 )32()2( +−

= 124 +

= 16 = 4

tan = 2

32

= 3

= tan–1 3

= 3

π

= π –

= π – 3

π

= 3

2π or 120°

–2 + 2 3 i = r(cos + i sin )

= 4(cos

3

2π + i sin

3

2π ) or 4(cos 120° + i sin 120°)

(–2 + 2 3 i)4 = 44(cos

3

2π + i sin

3

2π )4

= 44(cos

3

8π + i sin

3

8π )

= 44(cos

3

2π + i sin

3

2π )

= 44(–cos

3

π + i sin

3

π)

= 44(–

2

1 + i

2

3)

= 43(–2

4 + i

2

34)

= 43(–2 + 2 3 i)

Blunders (–3) B1: Error in argument. B2: Error in modulus. B3: Error in trig. definition. B4: Error in application of De Moivre Theorem. B5: Error in indices. B6: Answer not in required form.


Attempts (3) A1: Correct r or .

2011 L.17/20_MS 13/72 of 70 examsDEB

3(c) (ii) Find the four complex numbers z such that z4 = –2 + 2 3 i. Give your answer in the form a + bi, with a and b fully evaluated. (10, Att. 3)

z4 = –2 + 2 3 i

= (–2 + 2 3 i) 4

1

= (4[cos (2nπ + 3

2π) + i sin(2nπ +

3

2π)] ) 4

1

= 4 4

1

[cos

π+π

3

26n + i sin

π+π

3

26n] 4

1

= (22) 4

1

[cos

12

26 π+πn + i sin

12

26 π+πn]

= 2 2

1

[cos

6

3 π+πn + i sin

6

3 π+πn]

Let n = 0

z = 2 [cos

6

π + i sin

6

π]

= 2 [2

3 + i

2

1] or

2

1[ 3 + i ] or

2

3 +

2

1i

Let n = 1

z = 2 [cos

6

4π + i sin

6

4π]

= 2 [cos

3

2π + i sin

3

2π]

= 2 [–cos

3

π + i sin

3

π]

= 2 [–2

1 +

2

3i ] or

2

1[–1 + 3 i ] or

2

1− +

2

3i

Let n = 2

z = 2 [cos

6

7π + i sin

6

7π]

= 2 [–cos

6

π – i sin

6

π]

= 2 [–2

3 –

2

1i] or

2

1[– 3 – i] or –

2

3 –

2

1i

Let n = 3

z = 2 [cos

6

10π + i sin

6

10π]

= 2 [cos

3

5π + i sin

3

5π]

= 2 [cos

3

π – i sin

3

π]

= 2 [2

1 –

2

3i] or

2

1[1 – 3 i] or

2

1 –

2

3i

Blunders (–3) B1: Modulus. B2: Argument. B3: Trig definition. B4: Formula De Moivre - apply only once. B5: Application of De Moivre. B6: Trig not evaluated. B7: No general solution.

Slips (–1) S1: Numerical slip. S2: Trig. value.

Attempts (3) A1: One solution only.

2011 L.17/20_MS 14/72 of 70 examsDEB

4. (a) The sum of the first n terms of a series is given by Sn = 3n – 1. Find the first three terms of the series. (10, Att. 3)

Sn = 3n – 1 S1 = 31 – 1 = 3 – 1 = 2 T1 = 2

S2 = 32 – 1 = 9 – 1 = 8 T2 = S2 – S1 = 8 – 2 = 6

S3 = 33 – 1 = 27 – 1 = 26 T3 = S3 – S2 = 26 – 8 = 18

Blunders (–3) B1: Indices. B2: S2 but no T2 . B3: S3 but no T3 .


4(b) The sequence u1, u2, u3,..., is defined by un + 1 = aun + b, where u1 = 3, u2 = 2 and u3 = 4.

(i) Find the value of a and the value b. (10, Att. 3)

un + 1 = aun + b u2 = au1 + b 2 = 3a + b

u3 = au2 + b 4 = 2a + b

2a + b = 4(× –1) 3a + b = 2

–2a – b = –4 3a + b = 2 a = –2

2a + b = 4 2(–2) + b = 4 –4 + b = 4 b = 4 + 4 = 8

Blunders (–3) B1: u2 incorrect. B2: u3 incorrect. B3: Finds a or b only.


Attempts (3) A1: 3a + b = 2 or 2a + b = 4 and stops.

2011 L.17/20_MS 15/72 of 70 examsDEB

4(b) (ii) Find un + 2 and un + 3 in terms of un. (5, Att. 2)

un + 1 = aun + b = –2un + 8

un + 2 = –2un + 1 + 8 = –2(–2un + 8) + 8 = 4un – 16 + 8 = 4un – 8

un + 3 = –2un + 2 + 8 = –2(4un – 8) + 8 = –8un + 16 + 8 = –8un + 24

Blunders (–3) B1: un + 2 only. B2: un + 3 only.


Attempts (2) A1: un + 2 and stops. A2: un + 3 and stops.

(iii) Hence, show that 2un + 1 – un + 2 = un + 3. (5, Att. 2)

2un + 1 – un + 2 = 2(–2un + 8) – (4un – 8) = –4un + 16 – 4un + 8 = –8un + 24 = un + 3

Blunders (–3) B1: Fails to finish.


Attempts (2) A1: Fills in un + 2 or un + 3.

2011 L.17/20_MS 16/72 of 70 examsDEB

4(c) The first three terms of a series are log a, log

2

3a and log

4

9a.

(i) Show that the series is arithmetic. (10, Att. 3)

log

2

3a – log a = log

4

9a – log

2

3a

log

a

a

2

3 = log

4

9a ×

a3

2

log

2

3 = log

2

3

series is arithmetic

or

2

31 TT + =

24

9loglog

aa +

= 2

1log

4

9 2a

= log21

4

9 2

a

= log

2

3a

= T2 series is arithmetic

Blunders (–3)

B1: log a – log b ≠ log

b

a.

B2: nlog a ≠ log an. B3: Fails to show.


Attempt (3)

A1: log

2

3a – log a or log

4

9a – log

2

3a.

A2: 2

4

9loglog

aa +

.

2011 L.17/20_MS 57/72 of 70 examsDEB

4(c) (ii) For what value of n is the sum of the first n terms of the series

equal to 2[log a2 + log

8

27]? (10, Att. 3)

a = log a

d = log

2

3a – log a

= log

a

a

2

3

= log

2

3

Sn = 2

n[2a + (n – 1)d ]

= 2

n[2(log a) + (n – 1)(log

2

3)]

= 2

n[log a2 + log

1

2

3−

n

]

2

n[log a2 + log

1

2

3−

n

] = 2[log a2 + log

8

27]

2

n = 2

n = 4

1

2

3−

n

= 8

27

= 3

2

3

n – 1 = 3 n = 3 + 1 = 4

Blunders (–3) B1: Sn incorrect. B2: a or d incorrect. B3: Log rules. B4: Fails to find n.


Attempt (3) A1: Sn with substitution.

2011 L.17/20_MS 58/72 of 70 examsDEB

5. (a) (i) Find the first three terms in the expansion of (1 + x)9. (5, Att. 2)

(1 + x)9 =

0

9x0 +

1

9x1 +

2

9x2 + …

= 1 + 9x + )1)(2(

)8)(9(x2 + …

= 1 + 9x + 36x2 + …

Blunders (–3) B1: Error in indices. B2: nCr incorrect. B3: Written in descending order.


Attempts (2) A1: 9Cr correct.

(ii) Hence, deduce the first three terms in the expansion of 9

31

− x

. (5, Att. 2)

9

31

− x

= 1 + 9

−

3

x + 36

2

3

− x

+ …

= 1 – 3x + 9

36x2 + …

= 1 – 3x + 4x2 + …

Blunders (–3) B1: Error in indices. B2: Not deduced. B3: Not three terms.

Slips (–1) S1: Numerical slip. Attempts (2)

A1: Fills –3

x into equation and stops.

2011 L.17/20_MS 59/72 of 70 examsDEB

5(b) (i) Write 1

12

4

+−

r

r in the form ar2 + b, where a, b ∈ Z. (5, Att. 2)

1

12

4

+−

r

r =

1

)1)(1(2

22

++−

r

rr

= r2 – 1

Blunders (–3) B1: Factors. B2: Fails to cancel.


Attempts (2) A1: (r2 – 1)( r2 + 1) and stops.

(ii) Find, in terms of n, =

n

r 1

1

12

4

+−

r

r. (10, Att. 3)

=

n

r 1

1

12

4

+−

r

r =

=

n

r 1

r2 – 1

= 6

n(n + 1)(2n + 1) – n

= 6

n(2n2 + 3n + 1 – 6)

= 6

n(2n2 + 3n – 5)

Blunders (–3)

B1: =

n

r 1

r2 incorrect.

B2: =

n

r 1

–1 incorrect.

B3: Fails to finish.


Attempts (3) A1: r2 correct and stops.

(iii) Hence, evaluate =

12

1r

1

12

4

+−

r

r. (5, Att. 2)

=

12

1r

1

12

4

+−

r

r =

=

12

1r 6

n(2n2 + 3n – 5)

= 6

12(2(12)2 + 3(12) – 5)

= 2(2(144) + 36 – 5) = 2(288 + 36 – 5) = 2(319) = 638

Blunders (–3) B1: n ≠ 12.


Attempts (2) A1: Fills in 12 and stops.

2011 L.17/20_MS 60/72 of 70 examsDEB

5(c) Use induction to prove that

1 × 3 + 2 × 4 + 3 × 5 + ……… + n(n + 2) = 6

n(n + 1)(2n + 7)

where n is a positive integer. (5, 5, 10, Att. 2, 2, 3)

n = 1

1 × 3 = 6

1(1 + 1)(2(1) + 7)

= 6

1(2)(9)

= 6

1(18)

= 3 ∴ true for n = 1

Assume true for n = k

(1 × 3) + (2 × 4) + (3 × 5) + … + k(k + 2) = 6

k(k + 1)(2k + 7)

To Prove:

Prove true for n = k + 1

(1 × 3) + (2 × 4) + (3 × 5) + … + (k + 1)(k + 3) = 6

)1( +k(k + 2)(2k + 9)

(1 × 3) + (2 × 4) + (3 × 5) + … + k(k + 2) = 6

k(k + 1)(2k + 7)

(1 × 3) + (2 × 4) + (3 × 5) + … + k(k + 2) + (k + 1)(k + 3)

= 6

k(k + 1)(2k + 7) + (k + 1)(k + 3)

= 6

1+k[k(2k + 7) + 6(k + 3)]

= 6

1+k(2k2 + 7k + 6k + 18)

= 6

1+k(2k2 + 13k + 18)

= 6

1+k(k + 2)(2k + 9)

∴ True for n = k + 1

So, P(k + 1) is true whenever P(k) is true. Since P(1) is true, then by induction P(n) is true for all positive integers (n ∈ N, n ≥ 1).

** P(1) (5, Att. 2) ** P(k) (5, Att. 2) ** P(k + 1) (10, Att. 3)

Blunders (–3) B1: Fails to show n = 1 true. B2: Fails to show n = k + 1 true. B3: Indices. B4: Factors. B5: Fails to justify.


Attempts (2, 2, 3) A1: States n = 1 is true (not shown). A2: States n = k + 1 is true (not shown).

2011 L.17/20_MS 61/72 of 70 examsDEB

6. (a) Differentiate (1 – 3x2)6 with respect to x. (10, Att. 3)

dx

d(1 – 3x2)6 = 6(1 – 3x2)5(–6x)

or –36x(1 – 3x2)5

∗∗ Need not simplify derivatives.

Blunders (–3) B1: Error in differentiation. B2: Error in indices.


Attempts (3) A1: 6(1 – 3x2)5 and stops. A1: Attempt at differentiation. A2: Omits ‘–6x’.

6(b) Let f (x) = x4 – 8x2, where x ∈ ℝ.

(i) Find the co-ordinates of the points where the graph of f (x) cuts the x-axis. (5, Att. 2)

f (x) = x4 – 8x2 f (x) ∩ x-axis f (x) = 0 x4 – 8x2 = 0 x2(x2 – 8) = 0 x = 0 x2 – 8 = 0 x2 = 8 x = ± 8 or ± 22 or ±2.828427...

Points = (0, 0), ( 8 , 0), (– 8 , 0)

Blunders (–3) B1: Missing solution. B2: Not in co-ordinate form.


Attempts (2) A1: One correct solution. A2: x = 0 or y = 0.

2011 L.17/20_MS 62/72 of 70 examsDEB

6(b) (ii) Find the co-ordinates of the turning points of the graph of f (x) and determine whether each turning point is a local maximum or a local minimum. (10, Att. 3)

max./min. f ′(x) = 0 f ′(x) = 4x3 – 16x 4x3 – 16x = 0 4x(x2 – 4) = 0 4x = 0 x2 – 4 = 0 x = 0 x2 = 4 x = ±2

f ′′(x) = 12x2 – 16

@ x = 0 f ′′(x) = 12(0)2 – 16 = –16 < 0 Maximum point @ x = 0

f (0) = 04 – 8(0)2 = 0 Maximum point = (0, 0)

@ x = 2 f ′′(x) = 12(2)2 – 16 = 32 > 0 Minimum point @ x = 2

f (2) = 24 – 8(2)2 = –16 Minimum point = (2, –16)

@ x = –2 f ′′(x) = 12(–2)2 – 16 = 32 > 0 Minimum point @ x = –2

f (–2) = (–2)4 – 8(–2)2 = –16 Minimum point = (–2,–16)

Blunders (–3) B1: Differentiation. B2: Factors/roots. B3: Missing turning point. B4: Type of turning point not stated.


Attempts (3) A1: Differentiation. A2: One correct point.

2011 L.17/20_MS 63/72 of 70 examsDEB

6(b) (iii) Draw a sketch of the graph of f (x). (5, Att. 2)

2�1�2�3 3

x-axis

y-axis

1

�2

�4

�6

�8

�10

�12

�14

�16

f x( )

2

4

6

8

Blunders (–3) B1: Axes not shown. B2: Missing or incorrect point.


Attempts (2) A1: Axes shown. A2: One correct point.

6(c) Given that y = ln( 1+x + 1−x ),

show that dx

dy can be expressed in the form

12 −xb

a, where a, b ∈ ℕ. (10, 10, Att. 3, 3)

y = ln( 1+x + 1−x )

dx

dy =

11

1

−++ xx ×

−+

+ 12

1

12

1

xx

= 2

1 ×

11

1

−++ xx ×

11

11

−+−++

xx

xx

= )1)(1(2

1

−+ xx

= 12

12 −x

** Differentiation (10, Att. 3) ** Completed (10, Att. 3)

Blunders (–3) B1: Differentiation. B2: Indices. B3: Fails to finish.


Attempts (3, 3) A1: Attempt at differentiation. A2: Attempt at common denominator.

2011 L.17/20_MS 64/72 of 70 examsDEB

7. (a) Differentiate sin x with respect to x from first principles. (5, 5 Att. 2, 2)

Let f (x) = sin x f (x + h) = sin (x + h) f (x + h) – f (x) = sin (x + h) – sin x

= 2cos

2

xhx ++ sin

2

xhx −+

= 2cos (2

hx + ) sin

2

h

h

xfhxf )()( −+ =

h

hhx

2sin)

2(cos2 +

f ′(x) = 0

lim→h h

xfhxf )()( −+

= 0

lim→h h

hhx

2sin)

2(cos2 +

= 0

lim→h

2

2sin)

2(cos2

2

1

h

hhx

+

= 0

lim→h

cos (2

hx + ) ×

0lim→h

2

2sin

h

h

= cos (2

0+x ) × (1)

= cos x

∗∗ f (x + h) – f (x) simplified (5, Att. 2) ∗∗ Finish (5, Att. 2)

Blunders (–3) B1: Mishandles f (x + h). B2: Trigonometric simplifying. B3: No limits shown. B4: lim h → ∞.


Attempts (2, 2) A1: f (x + h) or y + Δy.

2011 L.17/20_MS 65/72 of 70 examsDEB

7(b) The equation of a curve is x2 + 2xy + 3y2 = 18.

(i) Find the co-ordinates of the points where the tangents to the curve at these points are horizontal. (10, 5, Att. 3, 2)

x2 + 2xy + 3y2 = 18

2x + 2xdx

dy + 2y + 6y

dx

dy = 0

dx

dy(2x + 6y) = –2x – 2y

dx

dy =

)3(2

)(2

yx

yx

++−

= yx

yx

3

)(

++−

as tangents are horizontal

dx

dy = 0

yx

yx

3

)(

++−

= 0

x + y = 0 y = –x

x2 + 2xy + 3y2 = 18 x2 + 2x(–x) + 3(–x)2 = 18 x2 – 2x2 + 3x2 = 18 2x2 = 18 x2 = 9 x = ±3

x = 3 x = –3 y = –(3) y = –(–3) y = –3 y = 3 point = (3, –3) point = (–3, 3)

(ii) Find the equations of the tangents to the curve at these points. (5, Att. 2)

tangents horizontal slope = 0

equations of tangents y = 3 y = –3

** Differentiation (10, Att. 3) ** Points (5, Att. 2) ** Tangents (5, Att. 2)

Blunders (–3) B1: Differentiation. B2: Factors. B3: Only one value found. B4: Only one tangent found.

Slips (–1) S1: Numerical slip. Attempts (3, 2, 2) A1: Differentiation. A2: One value only. A3: One tangent only.

2011 L.17/20_MS 66/72 of 70 examsDEB

7(c) Let f (x) = 2

xx ee −− – x, where x ∈ ℝ, x > 0.

(i) Show that f ′(x) > 0, where f ′(x) is the first derivative of f (x). (5, 5, Att. 2, 2)

f (x) = 2

xx ee −− – x

f ′(x) = 2

1[ex(1) – e–x(–1)] – 1

= 2

1[ex + e–x] – 1

= 2

1(ex +

xe

1 – 2)

= 2

1

−+x

xx

e

ee 21)( 2

= x

x

e

e

2

)1( 2−

> 0 as ex > 0 and (ex –1)2 > 0

∗∗ Differentiation (5, Att. 2) ∗∗ Show f ′(x) > 0 (5, Att. 2)

Blunders (–3) B1: Differentiation. B2: Factors. B3: Not shown.

Slips (–1) S1: Numerical slip. Attempts (2, 2) A1: Differentiation. A2: Factors.

(ii) Show that f ′′(x) > 0, where f ′′(x) is the second derivative of f (x). (10, Att. 3)

f ′(x) = x

x

e

e

2

)1( 2−

= xe2

1[(ex)2 – 2ex + 1]

= 2

1ex – 1 +

xe2

1

= 2

1ex +

2

1e–x – 1

f ′′(x) = 2

1ex(1) +

2

1e–x(–1) – 0

= 2

1[ex – e–x]

= 2

1(ex –

xe

1)

x > 0 ex > 1

xe

1 < 1

ex > xe

1

ex – xe

1 > 0

2

1(ex –

xe

1) > 0

2011 L.17/20_MS 67/72 of 70 examsDEB

7(c) (continued)

or

f ′(x) = x

x

e

e

2

)1( 2−

f ′′(x) = 2

1

−−−2

2

)(

)1)(()1()1)()(1(2)(x

xxxxx

e

eeeee

= 2

1 ×

x

xxxx

e

eeee2

2 ))1(2)(1( −−−

= 2

1 ×

x

xxxx

e

eeee2

22 )2)(1( +−−

= 2

1 ×

x

xxx

e

eee2

2 ))(1( +−

= 2

1(ex – 1)(1 +

xe

1)

= 2

1(ex + 1 – 1 –

xe

1)

= 2

1(ex –

xe

1)

x > 0 ex > 1

xe

1 < 1

ex > xe

1

ex – xe

1 > 0

2

1(ex –

xe

1) > 0

Blunders (–3) B1: Differentiation. B2: Indices. B3: ex not greater than 1.


Attempts (3) A1: Differentiation.

2011 L.17/20_MS 68/72 of 70 examsDEB

8. (a) Find (i) x dx (5, Att. 2)

x dx =

23

x

2

3 + c

(ii) 4

1

x dx (5, Att. 2)

4

1

x dx = x–3

− 3

1 + c

= –3

3−x + c

** If c shown in either parts (i) or (ii) - do not penalise.

Blunders (–3) B1: Error in integration. B2: No c - penalise first instant only. B3: Error in indices.

Attempts (2, 2) A1: Anything ‘+ c’.

Worthless (0) W1: Differentiation instead of integration.

8(b) (i) Find the real numbers a and b such that

)2)((

252

2

−+−xxx

x =

xx

ax

++

2

1 +

2−x

b. (10, Att. 3)

xx

ax

++

2

1 +

2−x

b =

)2)((

)()2)(1(2

2

−+++−+

xxx

xxbxax

= )2)((

252

2

−+−xxx

x

(ax + 1)(x – 2) + b(x2 + x) = 5x2 – 2 ax2 – 2ax + x – 2 + bx2 + bx = 5x2 – 2 (a + b)x2 + x(–2a + 1 + b) – 2 = 5x2 +0x – 2

a + b = 5 –2a + b + 1 = 0

a + b = 5 (×2) –2a + b = –1 (×1)

2a + 2b = 10 –2a + b = –1 3b = 9 b = 3

a + b = 5 a + 3 = 5 a = 5 – 3 = 2

Blunders (–3) B1: Indices. B2: Like to like. B3: One value only.


Attempts (3) A1: One equation found. A2: Like to like.

2011 L.17/20_MS 69/72 of 70 examsDEB

8(b) (ii) Hence, evaluate 6

3

)2)((

252

2

−+−xxx

x dx. (10, Att. 3)

6

3

)2)((

252

2

−+−xxx

x dx = 6

3

−+

++

2

3122 xxx

x dx

= 6

3

xx

x

++

2

12 dx + 6

3

2

3

−x dx

Let u = x2 + x du = (2x + 1)dx

at x = 6 u = (6)2 + 6 = 36 + 6 = 42

at x = 3 u = (3)2 + 3 = 9 + 3 = 12

6

3

)2)((

252

2

−+−xxx

x dx = 42

12

u

1 du + 6

3

2

3

−x dx

= [ln | u | ]4212 + 3[ln | x – 2 | ]6

3

= ln 42 – ln 12 + 3[ ln 4 – ln 1 ]

= ln

12

42 + 3 ln

1

4

= ln

2

7 + 3 ln 4

= ln

2

7(64)

= ln 224

** Note: –3 is the maximum deduction when evaluating limits.

Blunders (–3) B1: Error in differentiation. B2: Error in indices. B3: Error in integration. B4: Not changing limits.


Attempts (3) A1: Incorrect substitution.

2011 L.17/20_MS 70/72 of 70 examsDEB

8(c) The diagram shows the graphs of the curves y = f (x) and y = g(x), where f (x) = 3cos x and g(x) = sin x.

(i) Find the values of x for which f (x) = g(x), where 0 ≤ x ≤ 2π. (5, Att. 2)

f (x) = g(x) 3 cos x = sin x

x

x

cos

sin = 3

tan x = 3

x = tan–1 3

x = 3

π, x = π +

3

π

= 3

4π

Blunders (–3) B1: Trigonometry. B2: One value only.

Slips (–1) S1: Numerical slip. S2: Trig value.

Attempts (2) A1: One value correct.

(ii) Hence, calculate the area of the shaded region. (5, 5, 5, Att. 2, 2, 2)

f (x) ∩ x axis 3 cos x = 0

cos x = 0

x = 2

π,

2

3π

g(x) ∩ x axis sin x = 0 x = 0, π, 2π

Area of shaded region

=

π

π

34

2

f (x) dx – π

π

34

g(x) dx

+

π

π3

g(x) dx –π

π

2

3

f (x) dx

x

yy f x( )�

y g x( )�

� 2�0

2011 L.17/20_MS 71/72 of 70 examsDEB

8(c) (continued)

π

π

34

2

f (x) dx = π

π

34

2

3 cos x dx

= 3 sin x 3

4

2

π

π

= 3 [sin

3

4π – sin

2

π]

= 3 [–sin

3

π – 1]

= 3 [–2

3 – 1]

= –2

3 – 3

= 2

323 −−

= 2

323 +

π

π

34

g(x) dx = π

π

34

sin x dx

= –cos x 3

4π

π

= –cos

3

4π + cos π

= cos

3

π – 1

= 2

1 – 1

= –2

1

= 2

1

π

π3

g(x) dx = π

π3

sin x dx

= –cos x π

π3

= –cos π + cos

3

π

= –(–1) + 2

1

= 1 + 2

1

= 2

3

π

π

2

3

f (x) dx = π

π

2

3

3 cos x dx

= 3 sin x 2

3

π

π

= 3 (sin

2

π – sin

3

π)

= 3 (1 – 2

3 )

= 3 – 2

3

Area of shaded region =

2

323 + –

2

1

+

2

3 – ( 3 –

2

3)

=

2

1323 −+

+

2

3 – 3 +

2

3

=

2

322 +

+

2

6 – 3

= 1 + 3 + 3 – 3 = 4

∗∗ First correct area (5, Att. 2) ∗∗ Second correct area (5, Att. 2) ∗∗ Finishes (5, Att. 2)

Blunders (–3) B1: Integration. B2: Limits of integration. B3: | value |. B4: Fails to substitute limits.

Slips (–1) S1: Numerical slip. S2: Trig. values.

Attempts (2, 2, 2) A1: Integration.

2011 L.17/20_MS 72/72 of 70 examsDEB

Notes:

2011 L.17/20_MS 17/72 of 70 examsDEB


Mathematics

Higher Level – Paper 2 Marking Scheme (300 marks)

Attempt Five questions from Section A and ONE question from Section B. General Instructions: See Page 2.

SECTION A

Attempt FIVE questions from this section.

1. (a) The following parametric equations define a circle:

x = 3 + 2 cos , y = –1 + 2 sin , where ∈ ℝ.

Find the Cartesian equation of the circle. (10, Att. 3)

x = 3 + 2 cos 2 cos = x – 3

cos = 2

3−x

y = –1 + 2 sin 2 sin = y + 1

sin = 2

1+y

cos2 + sin2 = 1

2

2

3

−x +

2

2

1

+y = 1

2

)3( 2−x +

2

)1( 2+y = 1

(x – 3)2 + (y + 1)2 = 2 Blunders (–3)

B1: Indices. B2: Not stating cos2 + sin2 = 1. B3: Not in required form.


Attempts (3) A1: cos in terms of x. A2: sin in terms of x.

examsDEB

2011 L.17/20_MS 18/72 of 70 examsDEB

1(b) Two circles, each of radius 2 units, intersect at the point (–1, –1). The centres of both circles lie on the line x + y + 4 = 0.

Find the equations of the two circles that satisfy these conditions. (5, 5, 5, 5, Att. 2, 2, 2, 2)

centres on the line x + y + 4 = 0 (–g, –f ) ∈ x + y + 4 = 0 (–g) + (–f ) + 4 = 0 –g – f = –4 g = 4 – f

(–1, –1) ∈ circle (–1, –1) ∈ x2 + y2 + 2gx + 2fy + c = 0 (–1)2 + (–1)2 + 2g(–1) + 2f (–1) + c = 0 1 + 1 – 2g – 2f + c = 0 –2g – 2f + c = –2

radius = 2

cfg −+ 22 = 2

g2 + f 2 – c = 4

g2 + f 2 – c = 4 –2g – 2f + c = –2 g2 + f 2 – 2g – 2f = 2

g2 + f 2 – 2g – 2f = 2 g = 4 – f

(4 – f )2 + f 2 – 2(4 – f ) – 2f = 2 16 – 8f + f 2 + f 2 – 8 + 2f – 2f = 2 2f 2 – 8f + 6 = 0 f 2 – 4f + 3 = 0 ( f – 3)( f – 1) = 0

f – 3 = 0 f – 1 = 0 f = 3 f = 1

g = 4 – f g = 4 – 3 g = 4 – 1 = 1 = 3

–2g – 2f + c = –2 –2(1) – 2(3) + c = –2 –2(3) – 2(1) + c = –2 –2 – 6 + c = –2 –6 – 2 + c = –2 –8 + c = –2 –8 + c = –2 c = –2 + 8 c = –2 + 8 = 6 = 6

Equations of the two circle: x2 + y2 + 2x + 6y + 6 = 0 x2 + y2 + 6x + 2y + 6 = 0

** 2 equations (5, Att. 2) ** 3rd equation (5, Att. 2) ** Solves correctly for f , g or c (5, Att. 2) ** Finish (5, Att. 2)

Blunders (–3) B1: One equation evaluated. B2: Error in squaring. B3: Error in factors. B4: Equation of one circle only.

Slips (–1) S1: Numerical slip. S2: f , g and c found - no equations given.

Attempts (2, 2, 2, 2) A1: One equation only. A2: Eliminates unknown.

2011 L.17/20_MS 19/72 of 70 examsDEB

1(c) (i) R (3, 4) is the centre of a circle. P and Q are points on the circle such that | PQ | = 6 units and | ∠PRQ | = 90°. Find the equation of the circle. (10, Att. 3)

(3, 4) = centre of circle (x – 3)2 + (y – 4)2 = r2

| PQ | = 6 and | ∠PRQ | = 90° | PR |2 + | RQ |2 = 62 r2 + r2 = 62 2r2 = 36 r2 = 18 r = 18

(x – 3)2 + (y – 4)2 = ( 18)2

(x – 3)2 + (y – 4)2 = 18

Blunders (–3) B1: Pythagoras Theorem. B2: Indices. B3: r2 = 18 and stops.


Attempts (3) A1: (x – 3)2 + (y – 4)2 = r2 and stops.

(ii) The line x – y + k = 0 is a tangent to the circle. Find the two possible values of k. (10, Att. 3)

x – y + k = 0 is a tangent ⊥distance from (3, 4) to x – y + k = 0 = 18

22 )1(1

)4(1)3(1

−+

+− k = 18

11

43

++− k

= 18

2

1−k = 18

| k – 1 | = 2 18

= 36

| k – 1 | = 6 k – 1 = 6 k – 1 = –6 k = 6 + 1 k = –6 + 1 = 7 = –5

Blunders (–3) B1: Error in ⊥distance formula. B2: ⊥distance ≠ 18 . B3: Only one value found.


Attempts (3) A1: ⊥distance formula with some substitution. A2: x = y – k or y = x + k .

2011 L.17/20_MS 20/72 of 70 examsDEB

2. (a) →p = –6

→i + 8

→j .

Find the unit vector in the direction of →p ⊥. (10 Att. 3)

→p = –6

→i + 8

→j

→p ⊥ = –8

→i + 6

→j

| →p ⊥ | = 22 )6()8( +−

= 3664 +

= 100 = 10

Unit vector = 10

68→→

−− ji

= –5

4 →i –

5

3 →j

Blunders (–3) B1:

→p ⊥ error.

B2: | →p ⊥ | error.


Attempts (3) A1:

→p ⊥ correct.

A2: | →p | correct.

A3: Unit vector in the direction of →p .

2(b) →a =

→i – →j and

→b = 4

→i + 2

→j .

C is a point on [ AB ] such that | AC | : | CB | = 2 : 1.

(i) Express →c and

→AC in terms of

→i and

→j . (5, 5 Att. 2, 2)

→

AC = 3

2(

→AB )

= 3

2(

→b –

→a )

= 3

2(4

→i + 2

→j –

→i + →j )

= 3

2(3

→i + 3

→j )

= 2(→i +

→j )

= 2→i + 2

→j

→

AC = →c –

→a

→c =

→AC +

→a

= 2→i + 2

→j +

→i – →j

= 3→i + →j

Blunders (–3) B1: Incorrect ratio.

B2: →

AB ≠ →b –

→a .

B3: →

AC ≠ →c –

→a .


Attempts (2, 2)

A1: →

AB = →b –

→a .

A2: →

AC = →c –

→a .

2011 L.17/20_MS 21/72 of 70 examsDEB

2(b) (ii) Hence, find the measure of the angle between →c and

→AC ,

correct to the nearest degree. (10, Att. 3)

cos = ||||

.→→

→→

ACc

ACc

= |22||3|

)22)(3(→→→→

→→→→

++

++

jiji

jiji

= 2222 2213

26

++

+

= 4419

8

++

= 810

8

= 10

8

= 5

4

= 5

2

= cos–1

5

2

= 26.565051...° ≅ 27°

Blunders (–3) B1: Incorrect norm formula. B2: Incorrect vector multiplication. B3: Indices.

B4: cos–1

5

2 and stops.

Slips (–1) S1: Numerical slip. S2: Not to nearest degree. S3: Trig. value.

Attempts (3) A1: Correct formula with some substitution.

2011 L.17/20_MS 22/72 of 70 examsDEB

2(c) (i) →r = |

→y |

→x + |

→x |

→y and

→s = |

→y |

→x – |

→x |

→y for vectors

→x and

→y .

Prove that →r is perpendicular to

→s . (10, Att. 3)

→r = |

→y |

→x + |

→x |

→y

→s = |

→y |

→x – |

→x |

→y

If →r ⊥ →s

→r .

→s = [ |

→y |

→x + |

→x |

→y ][ |

→y |

→x – |

→x |

→y ]

= | →y |2

→x .

→x – |

→x |2

→y .

→y

= | →y |2.|

→x |2 – |

→x |2.|

→y |2

= 0

→r ⊥ →s

Blunders (–3)

B1: →r .

→s .

B2: →x .

→x ≠ |

→x |2.

B3: Fails to finish.


Attempts (3) A1:

→r .

→s = 0.

(ii) Express the vector →i +

→j as the sum of two vectors, one of which

is parallel to →i + 2

→j and the other which is perpendicular to 3

→i +

→j . (10, Att. 3)

→i +

→j = m(

→i + 2

→j ) + n(

→i + 3

→j )

= (m – n)→i + (2m + 3n)

→j

m – n = 1 (×3) 2m + 3n = 1 (×1)

3m – 3n = 3 2m + 3n = 1 5m = 4

m = 5

4

m – n = 1

5

4 – n = 1

–n = 1 – 5

4

= 5

1

n = –5

1

→i +

→j =

5

4(

→i + 2

→j ) –

5

1(

→i + 3

→j )

Blunders (–3) B1: Like to like. B2: Only one value found.


Attempts (3) A1: One correct equation.

2011 L.17/20_MS 23/72 of 70 examsDEB

3. (a) A(–1, 2) and B(3, 1) are two points. Find the equation of the perpendicular bisector of [AB]. (10, Att. 3)

Midpoint AB =

++−

2

12,

2

31

=

2

3,

2

2

=

2

3,1

Slope AB = )1(3

21

−−−

= 13

1

+−

= –4

1

⊥slope = 4

Equation of ⊥bisector: y – y1 = m(x – x1)

y – 2

3 = 4(x – 1) ... full marks

2y – 3 = 8(x – 1) 2y – 3 = 8x – 8 8x – 2y = 8 – 3 8x – 2y = 5

Blunders (–3) B1: Incorrect midpoint. B2: Incorrect slope. B3: Incorrect ⊥slope. B4: Error in equation of line.


Attempts (3) A1: Correct midpoint or slope.

3(b) f is the transformation (x, y) → (x′, y′), where x′ = 3x + 2y and y′ = 3x + ty, t ∈ ℝ. l is the line 3x + y – 5 = 0.

(i) The equation of f (l), the image of l under f, is 2x – y – 5 = 0. Find the value of t. (5, Att. 2)

f (l): 2x – y – 5 = 0 2(3x + 2y) – (3x + ty) – 5 = 0 6x + 4y – 3x – ty – 5 = 0 3x + (4 – t)y – 5 = 0

l: 3x + y – 5 = 0 4 – t = 1 –t = 1 – 4 –t = –3 t = 3

Blunders (–3) B1: Like with like. B2: Fails to evaluate t.


Attempts (2) A1: x and y correctly substituted and stops.

2011 L.17/20_MS 24/72 of 70 examsDEB

3(b) (ii) k is the line 12x + 9y – 7 = 0. Find the measure of the acute angle between f (l) and f (k). (5, 5, 5, Att. 2, 2, 2)

k: 12x + 9y – 7 = 0

x′ = 3x + 2y (×–1) y′ = 3x + 3y (×1)

–x′ = –3x – 2y y′ = 3x + 3y –x′ + y′ = y

x′ = 3x + 2y (×3) y′ = 3x + 3y (×–2)

3x′ = 9x + 6y –2y′ = –6x – 6y 3x′ – 2y′ = 3x

x = 3

1(3x′ – 2y′)

f (k):

12 × 3

1(3x′ + 2y′) + 9(–x′ + y′) – 7 = 0

4(3x′ – 2y′) – 9x′ + 9y′ – 7 = 0 12x′ – 8y′ – 9x′ + 9y′ – 7 = 0 3x′ + y′ – 7 = 0

Slope of f (k) = 1

3−

= –3

Slope of f (l) = 1

2

−−

= 2

Angle between f (k) and f (l)

tan = ±21

21

1 mm

mm

+−

= ±)2)(3(1

23

−+−−

= ±61

5

−−

= ±5

5

−−

= ±1

is an acute angle tan = 1 = tan–1(1) = 45°

** Slopes (5, Att. 2) ** Formula filled in (5, Att. 2) ** Solution (5, Att. 2)

Blunders (–3) B1: Incorrect slopes. B2: Incorrect formula. B3: tan = 1 and stops.


Attempts (2, 2, 2) A1: One slope.

2011 L.17/20_MS 25/72 of 70 examsDEB

3(c) P and Q are two points on the line x – y + 1 = 0 such that | PO | = | OQ | = 5, where O is the origin.

(i) Find the perpendicular height of the triangle OPQ. (5, Att. 2)

⊥height = ⊥distance from (0,0) to x – y + 1 = 0

= 22 )1(1

1)0(1)0(1

−+

+−

= 11

100

++−

= 2

1

Blunders (–3)

B1: Indices. Slips (–1) S1: Numerical slip.

Attempts (2) A1: Formula with substitution.

(ii) Hence, find the area of the triangle OPQ. (10, 5, Att. 3, 2)

Area ΔOPQ = 2

1| PQ |.⊥height

= 2

1| PQ |.

2

1

Let 2

1| PQ | = x

x2 + 2

2

1

= 52

x2 + 2

1 = 25

x2 = 25 – 2

1

= 2

49

x = 2

49

x = 2

7

Area ΔOPQ = 2

7 ×

2

1

= 2

7

= 3.5 units2 ** x (10, Att. 3) ** Finish (Area ΔOPQ) (5, Att. 2)

Blunders (–3) B1: Pythagoras theorem. B2: Indices.

B3: x ≠ 2

1| PQ |.


Attempts (3, 2) A1: Area formula with substitution. A2: Pythagoras theorem with substitution.

2011 L.17/20_MS 26/72 of 70 examsDEB

4. (a) Evaluate 0

lim→θ

θθ

5sin

3sin. (10, Att. 3)

0

lim→θ

θθ

5sin

3sin =

0lim

→θ

θθ

3

3sin.

θθ5sin

5.

θθ

5

3

= 0

lim→θ

θθ

3

3sin ×

0lim

→θ

θθ5sin

5 ×

0lim

→θ

5

3

= 1 × 1 × 5

3

= 5

3

or

Let f () = sin 3 Let g() = sin 5

0

lim→θ

θθ

5sin

3sin =

0lim→θ

)(

)(

θθ

g

f

= )0(

)0(

g

f′′

= )0(cos5

)0(cos3

= 5

3

Blunders (–3) B1: sin 3 = 3sin . B2: sin 5 = 5sin . B3: Error in differentiation.


Attempts (3)

A1: Has θ

θ3

3sin or

θθ5sin

5 in solution.

A2: Correct differentiation.

2011 L.17/20_MS 27/72 of 70 examsDEB

4(b) x and y are acute angles such that sin x = 13

12 and cos y =

5

4.

Without evaluating x and y , find the value of

(i) cos 2x (5, Att. 2)

cos 2x = cos2 x – sin2 x = 1 – sin2 x – sin2 x = 1 – 2sin2 x

= 1 – 22

13

12

= 1 – 2

169

144

= 1 – 169

288

= 169

288169 −

= –169

119

Blunders (–3) B1: cos 2x ≠ 1 – 2sin2 x. B2: Indices.


Attempts (2) A1: cos 2x = 1 – 2sin2 x.

(ii) cos 2y (5, Att. 2)

cos 2y = cos2 y – sin2 y = cos2 y – (1 – cos2 y) = cos2 y – 1 + cos2 y = 2cos2 y – 1

= 22

5

4

– 1

= 2

25

16 – 1

= 25

32 – 1

= 25

2532 −

= 25

7

Blunders (–3) B1: cos 2y ≠ 2cos2 y – 1. B2: Indices.


Attempts (2) A1: cos 2y = 2cos2 y – 1.

2011 L.17/20_MS 28/72 of 70 examsDEB

4(b) (continued)

Explain why y < x . (10, Att. 3)

x and y are acute angles 0 < x < 90° 0 < 2x < 180°

cos 2x < 0 90° < 2x < 180° 45° < x < 90°

also 0 < y < 90° 0 < 2y < 180°

cos 2y > 0 0° < 2y < 90° 0° < y < 45°

x > 45° and y < 45° y < x

or

cos 2y > 0 2y < 90°

cos 2x < 0 2x > 90°

2y < 2x y < x

Blunders (–3) B1: 2y not less than 90°. B2: 2x not greater than 90°. B3: y not less than x stated.


Attempts (3) A1: 2y < 90° or 2x > 90° and stops.

2011 L.17/20_MS 29/72 of 70 examsDEB

4(c) Two circles, each of radius 5 cm, intersect at the points P and R, as shown.

OPQR is a square where O and Q are the centres of the two circles.

(i) Find the area of the sector OPR. (5, Att. 2)

OPQR is a square | ∠ROP | = 90°

Area of sector = °

°θ360

πr2

Area of sector OPR = °°

360

90 × π × 52

= 4

π × 25

= 4

25 π cm2 or 19.634954... cm2

Blunders (–3)

B1: ≠ 90° or 2

π.

B2: Sector formula incorrect.

Slips (–1) S1: Numerical slip. S2: Units of measurement omitted or incorrect.


(ii) Hence, or otherwise, find the area of the shaded region. (15, Att. 5)

Area of the shaded region = Area of sector OPR + area of sector PQR – area of rectangle OPQR = 2(area of sector OPR) – area of rectangle OPQR

= 2 × 4

25 π – (5 × 5)

= 2

25 π – 25

= 25(2

π – 1) cm2 or 14.269908... cm2

Blunders (–3) B1: Area of square. B2: Area of Δopr or Δpqr.

Slips (–1) S1: Numerical slip. S2: Units of measurement.

Attempts (5) A1: Area of square. A2: Area of Δ.

O

P

R

Q

2011 L.17/20_MS 30/72 of 70 examsDEB

5. (a) Find all the solutions of the equation

tan + tan2 = 0

in the domain 0° ≤ θ ≤ 180°. (10, Att. 3)

tan + tan2 = 0 tan (1 + tan ) = 0 tan = 0 = tan–1 0 = 0°, 180°

1 + tan = 0 tan = –1 = tan–1 –1 = 180° – 45° = 135°

Blunders (–3) B1: Indices. B2: Factors. B3: Missing or extra solution.

Slips (–1) S1: Numerical slip. S2: Trig. value.

Attempts (3) A1: Factors.

2011 L.17/20_MS 31/72 of 70 examsDEB

5(b) (i) Show, using the formula for sin (A + B), that

sin 2A = 2sin A cos A. (5, Att. 2)

sin 2A = sin(A + A ) = sin A cos A + cos A sin A = 2sin A cos A

Blunders (–3) B1: Incorrect formula. B2: Fails to show.



(ii) Hence, given that (cos x + sin x)2 = y2cos

1,

show that sin 2x = tan2 y. (15, Att. 5)

(cos x + sin x)2 = y2cos

1

cos2 x + 2cos x sin x + sin2 x = y2cos

1

(cos2 x + sin2 x) + 2cos x sin x = y2cos

1

cos2 x + sin2 x = 1 2cos x sin x = sin 2x

1 + sin 2x = y2cos

1

sin 2x = y2cos

1 – 1

= y

y2

2

cos

cos1 −

= y

y2

2

cos

sin

= tan2 y

Blunders (–3) B1: Squaring. B2: cos2 x + sin2 x ≠ 1. B3: Fails to show.


Attempts (5) A1: Squaring.

2011 L.17/20_MS 32/72 of 70 examsDEB

5(c) A, B and E are points on horizontal ground. D and C are points on a vertical wall directly

above A and B respectively. The angle of elevation of D from E is 40°. | AB | = 10 m and | AD | = 5 m. (i) Find | DE |, correct to one decimal place. (5, Att. 2)

sin 40° = ||

5

DE

| DE | = °40sin

5

= ...642787.0

5

= 7.778619... ≅ 7.8 m

Blunders (–3) B1: Trig. formula.

Slips (–1) S1: Numerical slip. S2: Trig. value. S3: Incorrectly or not rounded off. Attempts (2) A1: Finds | AE | and stops.

(ii) Given that | ∠CDE | = 60°, find | ∠DEC |,

correct to the nearest degree. (15, Att. 5)

Let | ∠DEC | = | ∠DCE | = 180° – (60° + ) = 120° –

| AB | = 10 m | DC | = 10 m

||sin

||

DEC

DC

∠ =

||sin

||

DCE

DE

∠

θsin

10 =

)120(sin

7.8

θ−°

θsin

10 =

)120sin(

8.7

θ−°

10 sin (120° – ) = 7.8 sin 10[sin 120°cos – cos 120°sin ] = 7.8 sin 10[(0.866025...)cos + (0.5)sin ] = 7.8 sin (8.660254...)cos + 5sin = 7.8 sin (8.660254...)cos = 7.8 sin – 5 sin = 2.8 sin

θθ

cos

sin =

8.2

...660254.8

tan = 3.092947... = tan–1 3.092947... = 72.083142... ≅ 72°

Blunders (–3) B1: Sine formula. B2: sin (A – B). B3: Fails to find value for .

Slips (–1) S1: Numerical slip. S2: Trig. value. S3: Incorrectly or not rounded off. Attempts (5) A1: Sine formula with substitution. A2: sin (A – B) with substitution.

40�

CD

5 m

10 m

A B

60�

E

2011 L.17/20_MS 33/72 of 70 examsDEB

6. (a) The table below shows the marks awarded to two students in the Listening, Reading and Written sections of a language examination.

The overall result is the weighted mean of the marks awarded in the above sections. The table also shows the weight assigned to each section.

The two students achieve the same overall result.

Examination Listening Reading Written

Student 1 46 80 72

Student 2 40 64 82

Weight 3 2 w

Find the value of w. (10, Att. 3)

w

w

++++

23

72)2(80)3(46 =

w

w

++++

23

82)2(64)3(40

138 + 160 + 72w = 120 + 128 + 82w 298 + 72w = 248 + 82w 82w – 72w = 298 – 248 10w = 50 w = 5

Blunders (–3) B1: Sum of weights incorrect. B2: Incorrect denominator. B3: Add instead of multiplying each term.


Attempts (3) A1: Sum of weights.

2011 L.17/20_MS 34/72 of 70 examsDEB

6(b) Two unbiased dice, each with faces from 1 to 6, are thrown. Find the probability of getting

(i) two “fives” (5, Att. 2)

P(2 × “fives”) = 6

1 ×

6

1

= 36

1

Blunders (–3) B1: Incorrect possible outcomes. B2: Incorrect favourable outcomes.

B3: 6

1 +

6

1.


Attempts (2) A1: Correct possible outcomes. A2: Correct favourable outcomes.

(ii) two prime numbers (5, Att. 2)

P(2 out of 2, 3, 5) = 6

3 ×

6

3

= 36

9 or

4

1

Blunders (–3) B1: Incorrect possible outcomes. B2: Incorrect favourable outcomes.

B3: 6

3 +

6

3.



(iii) one number that is a factor of the other number. (10, Att. 3)

1 2 3 4 5 6 1 (1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6) 2 (2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6) 3 (3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6) 4 (4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6) 5 (5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6) 6 (6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6)

P(factors) = 36

22 or

18

11

Blunders (–3) B1: Incorrect possible outcomes. B2: Incorrect favourable outcomes. B3: Incomplete list.



2011 L.17/20_MS 35/72 of 70 examsDEB

6(c) (i) Solve the difference equation un + 2 – 4un + 1 + un = 0, where n ≥ 0, given that u0 = 0 and u1 = 2 3 . (5, 5, 5, Att. 2, 2, 2)

un + 2 – 4un + 1 + un = 0 x2 – 4x + 1 = 0

x = a

acbb

2

42 −±−

= )1(2

)1)(1(4)4()4( 2 −−±−−

= 2

4164 −±

= 2

124 ±

= 2 ± 3

un = l(2 + 3 )n + m(2 – 3 )n

u0 = 0 l(2 + 3 )0 + m(2 – 3 )0 = 0

l + m = 0

u1 = 2 3

l(2 + 3 )1 + m(2 – 3 )1 = 2 3

(2 + 3 )l + (2 – 3 )m = 2 3

(2 + 3 )l + (2 – 3 )m = 2 3 (×1)

l + m = 0 (×–2 – 3 )

(2 + 3 )l + (2 – 3 )m = 2 3

(–2 – 3 )l + (–2 – 3 )m = 0

–2 3m = 2 3

m = –1 l + m = 0 l + (–1) = 0 l = 1

un = l(2 + 3 )n + m(2 – 3 )n

un = (2 + 3 )n – (2 – 3 )n

** Characteristic equation (5, Att. 2) ** u0 and u1 (5, Att. 2) ** Finish (un) (5, Att. 2)

Blunders (–3) B1: Error in characteristic equation. B2: Error in quadratic formula. B3: Incorrect use of initial conditions. B4: Roots in decimal form.


Attempts (2, 2, 2) A1: un = ln + mn and stops.

2011 L.17/20_MS 37/72 of 70 examsDEB

6(c) (ii) Verify that your solution gives the correct value for u2. (5, Att. 2)

un = (2 + 3 )n – (2 – 3 )n

u2 = (2 + 3 )2 – (2 – 3 )2

= 4 + 4 3 – (4 – 4 3 + 3)

= 8 3

un + 2 – 4un + 1 + un = 0 u2 – 4u1 + u0 = 0 u2 – 4(2 3 ) + 0 = 0

u2 – 8 3 + 0 = 0

u2 = 8 3 ... verified

Blunders (–3) B1: Only one u2 found. B2: Indices. B3: Not verified.


Attempts (2) A1: u2 found.

2011 L.17/20_MS 37/72 of 70 examsDEB

7. (a) (i) How many five-letter arrangements can be formed from the letters of the word ABSOLUTE if each letter is used only once in each arrangement? (5, Att. 2)

No. of arrangements = 8 × 7 × 6 × 5 × 4 = 6,720

or = 8P5 = 6,720

Blunders (–3) B1: 8C5. B2: 8!

Attempts (2) A1: 8P5.

Worthless (0) W1: 8 + 7 + 6 + 5 + 4.

(ii) How many of these arrangements contain two consonants and three vowels? (5, Att. 2)

No. of arrangements = 4 × 3 × 4 × 3 ×2 = 288

or = 4P2 × 4P3 = 12 × 24 = 288

Blunders (–3) B1: 4C2. B1: 4C3. B2: 4P2 + 4P3.

Attempts (2) A1: 4Pn or 4P2 or 4P3.

Worthless (0) W1: 8 + 7 + 6 + 5 + 4.

2011 L.17/20_MS 38/72 of 70 examsDEB

7(b) There are 10 white tickets and x red tickets in a box. A ticket is drawn at random and not replaced. A second ticket is then drawn.

(i) Find the probability that the first ticket drawn is red. (5, Att. 2)

P(red) = x

x

+10

Blunders (–3) B1: Incorrect favourable outcomes. B2: Incorrect possible outcomes.


Attempts (2) A1: Correct favourable outcomes. A2: Correct possible outcomes.

(ii) Given that the first ticket drawn is red, find the probability that the second ticket drawn is also red. (5, Att. 2)

P(given red, red) = )1(10

1

−+−x

x

= x

x

+−

9

1


B3: x

x

+10.

x

x

+−

9

1.



(iii) If the probability that the first two tickets drawn are both red is 21

2,

find the value of x. (10, Att. 3)

x

x

+10.

x

x

+−

9

1 =

21

2

21x(x – 1) = 2(10 + x)(9 + x) 21x2 – 21x = 2(x2 + 19x + 90) = 2x2 + 38x + 180 19x2 – 59x – 180 = 0 (19x + 36)(x – 5) = 0 19x + 36 = 0 x – 5 = 0 19x = –36 x = 5

x = –19

36

no solution x = 5

Blunders (–3)

B1: x

x

+10 +

x

x

+−

9

1.

B2: Quadratic equation. B3: Factors/quadratic formula. B4: Solution ∉ N.


Attempts (3) A1: Quadratic equation.

2011 L.17/20_MS 39/72 of 70 examsDEB

7(c) The mean of the real numbers p and q is x and the standard deviation is σ.

(i) Express x and σ in terms of p and q. (10, Att. 3)

x = mean

= 2

qp +

σ = standard deviation

σ2 =

+−+

+−

22

222

1 qpq

qpp

=

−−+

−− 22

2

2

2

2

2

1 qpqqpp

=

+−+

− 22

222

1 qpqp

= 8

2

8

2 2222 qpqpqpqp +−++−

= 8

2(p – q)2

= 4

)( 2qp −

σ = 4

)( 2qp −

= 2

qp −

Blunders (–3) B1: Error in mean. B2: Error in standard deviation.


Attempts (3) A1: Correct mean.

(ii) Hence, show that pq

1[x 2 – σ2] simplifies to a constant. (10, Att. 3)

pq

1[x 2 – σ2] =

pq

1

−−

+ 22

22

qpqp

= pq

1

+−−++4

2

4

2 2222 qpqpqpqp

= pq

1

−+−++4

22 2222 qpqpqpqp

= pq

1

4

4 pq

= 1

Blunders (–3) B1: Incorrect squaring. B2: Not simplified to a constant.


Attempts (3) A1: x 2 – σ2 expressed in terms of p and q.

2011 L.17/20_MS 40/72 of 70 examsDEB

SECTION B

Attempt ONE question from this section.

8. (a) Derive the Maclaurin series for f (x) = sin x up to and including the term containing x5. (10, Att. 3)

f (x) = sin x f (0) = sin (0) = 0

f ′(x) = cos x f ′(0) = cos (0) = 1

f ′′(x) = –sin x f ′′(0) = –sin (0) = 0

f ′′′(x) = –cos x f ′′′(0) = –cos (0) = –1

f ′′′′(x) = sin x f ′′′′(0) = sin (0) = 0

f ′′′′′(x) = cos x f ′′′′′(0) = cos (0) = 1

f (x) = f (0) + !1

)0( xf ′ +

!2

)0( 2xf ′′ +

!3

)0( 3xf ′′′ +

!4

)0( 4xf ′′′′ + ...

sin x = 0 + !1

)1( x +

!2

)0( 2x +

!3

)1( 3x− +

!4

)0( 4x +

!5

)1( 5x +...

= 0 + x + 0 – 6

1x3 + 0 +

120

1x5 +...

= x – 6

1x3 +

120

1x5 + ... or

!1

x –

!3

3x +

!5

5x + ...

Blunders (–3) B1: Incorrect differentiation. B2: Incorrect evaluation of f ′′(0). B3: Each term not derived. B4: Omitted term.

Slips (–1) S1: Numerical slip. S2: Incorrect trig. value.

Attempts (3) A1: Differentiation. A2: f (0) evaluated.

2011 L.17/20_MS 41/72 of 70 examsDEB

8(b) (i) Use integration by parts to find x2 sin x dx. (10, Att. 3)

u dv = uv – v du

Let u = x2

dx

du = 2x

du = 2x dx

Let dv = sin x dx

dv = sin x dx

v = –cos x

x2 sin x dx = –x2 cos x + cos x 2x dx

Let u = 2x

dx

du = 2

du = 2 dx

Let dv = cos x dx

dv = cos x dx

v = sin x

x2 sin x dx = –x2 cos x + 2x sin x – 2 sin x dx

= –x2 cos x + 2x sin x + 2 cos x + c

Blunders (–3) B1: Incorrect differentiation or integration. B2: Constant of integration omitted. B3: Incorrect ‘parts’ formula.


Attempts (3) A1: Correct assigning to ‘parts’ formula. A2: Correct differentiation or integration.

2011 L.17/20_MS 42/72 of 70 examsDEB

8(b) (ii) Use the ratio test to test the series ∞

= 1n 1

)2(2 +−

n

x n

for convergence and specify clearly

the range of values of x for which the series converges. (10, Att. 3)

un = 1

)2(2 +−

n

x n

un + 1 = 1)1(

)2(2

1

++− +

n

x n

= 112

)2(2

1

+++− +

nn

x n

= 22

)2(2

1

++− +

nn

x n

∞→n

lim

n

n

u

u 1+ =

∞→nlim

1

)2(

22

)2(

2

2

1

+−

++− +

n

x

nn

x

n

n

= ∞→n

lim n

n

x

n

nn

x

)2(

1

22

)2( 2

2

1

−+×

++− +

= | x – 2 | ∞→n

lim

22

12

2

+++nn

n

= | x – 2 | ∞→n

lim

2

2

221

11

nn

n

++

+

= | x – 2 | × 001

01

+++

= | x – 2 | × | 1 | = | x – 2 |

∞

= 1n 1

)2(2 +−

n

x n

is convergent for

| x – 2 | < 1 –1 < x – 2 < 1 –1 + 2 < x < 1 + 2 1 < x < 3

Blunders (–3) B1: Indices. B2: Error in un + 1. B3: Error in evaluating limit. B4: Evaluate limit and stops.


Attempts (3) A1: Correct un + 1. A2: Correct statement of test with some substitution.

2011 L.17/20_MS 43/72 of 70 examsDEB

8(c) A ladder, of length l m, standing on level ground leans against a vertical wall, touching a fence. The fence is 2 m high and located 2 m from the wall.

The distance from the foot of the ladder to the fence is x m. The distance from the ground to the top of the ladder is y m.

(i) Using similar triangles, or otherwise,

show that y = x

x )2(2 +. (5, Att. 2)

2

y =

x

x 2+

y = x

x )2(2 +

Blunders (–3) B1: Ratio incorrect. B2: x in terms of y.


Attempts (2) A1: x in terms of y.

(ii) Hence, find l in terms of x. (5, Att. 2)

l2 = y2 + (x + 2)2

= 2

)2(2

+

x

x + (x + 2)2

= 2

2

)2(4

x

x + + (x + 2)2

= 2

2)2(

x

x +(4 + x2)

l = 2

22 )4()2(

x

xx ++

=

+

x

x 242 +x

Blunders (–3) B1: Pythagoras theorem incorrect. B2: Fails to show.


Attempts (2) A1: Pythagoras theorem.

2

yl

2x

2011 L.17/20_MS 44/72 of 70 examsDEB

8(c) (iii) Hence, find the shortest possible ladder that can reach the wall. (10, Att. 3)

l =

+

x

x 242 +x

dx

dl =

+

x

x 2

2

1 21

)4( 2 −+x (2x) + 42 +x

+−

2

)1)(2()1(

x

xx

= 4

22 +

+

x

x + 42 +x

−−

2

2

x

xx

= 4

22 +

+

x

x + 42 +x

−

2

2

x

= 4

)4(2)2(22

22

+

+−+

xx

xxx

= 4

82222

223

+

−−+

xx

xxx

= 4

822

3

+

−

xx

x

dx

dl = 0

4

822

3

+

−

xx

x = 0

x3 – 8 = 0 x3 = 8 x = 2

l =

+

x

x 242 +x

=

+

2

22422 +

= 2

444 +

= 2 8 m or 4 2 m

Blunders (–3) B1: Differentiation. B2: Indices. B3: No x value. B4: x found but no l value.


Attempts (3) A1: Correct differentiation and stops.

A2: dx

dl = 0 and stops.

2011 L.17/20_MS 45/72 of 70 examsDEB

9. (a) Four unbiased coins are tossed.

(i) Find the probability of getting three heads and one tail. (5, Att. 2)

p = 2

1

q = 2

1

P(3 head & 1 tail) = 2

1 ×

2

1 ×

2

1 ×

2

1 × 4

= 16

1 × 4

= 4

1

Blunders (–3) B1: p or q incorrect. B2: ‘× 4’ omitted.


Attempts (2) A1: p and q correct.

(ii) The four coins are tossed eight times. Find the probability of getting three heads and one tail exactly three times. (5, Att. 2)

p = 4

1

q = 4

3

P(3 head & 1 tail, 3 times) = 8C3 3

4

1

× 5

4

3

= )1)(2)(3(

)6)(7)(8(

64

1

042,1

243

= 128,400

648,81

= 336,8

701,1

Blunders (–3) B1: p or q incorrect. B2: Binomial.


Attempts (2) A1: p and q correct.

2011 L.17/20_MS 46/72 of 70 examsDEB

9(b) Five out of twelve items produced by a machine are found to be defective. Two items are chosen at random. What is the probability that

(i) both items are defective, (5, Att. 2)

P(both defective) = 12

5 ×

11

4

= 132

20

= 33

5


B3: 12

5 +

11

4.



(ii) at least one item is defective, (5, Att. 2)

P(at least one defective) = 1 – P(none)

= 1 – 12

7 ×

11

6

= 1 – 22

7

= 22

15

Blunders (–3) B1: Probability rule (1 – ...) B2: Incorrect favourable outcomes. B3: Incorrect possible outcomes.



(iii) at most one item is defective? (10, Att. 3)

P(at most one defective) = P(none defective) + P(one defective)

= 1 – 22

15 +

×

11

4

12

5 × 2

= 22

7 +

33

10

= 66

21 +

66

20

= 66

41

Blunders (–3) B1: Probability rule (1 – ...) B2: Incorrect probability. B3: ‘× 2’ omitted.


Attempts (3) A1: One correct probability.

2011 L.17/20_MS 47/72 of 70 examsDEB

9(c) A joke shop advertised a biased die which would not give the expected number of “sixes” as a fair die. To test the hypothesis, the die was rolled 500 times and the number of “sixes” observed is x.

Between what limits should x lie in order that the hypothesis not be rejected at the 5% level of significance? (5, 5, 5, 5, Att. 2, 2, 2, 2)

p = 6

1

q = 6

5

The null hypothesis: = np

= 6

500

= 3

250

σ = npq

= 6

5

6

1500 ××

= 36

500,2

= 6

50

= 3

25

x:

–1.96 <

3

253

250−x < 1.96

–16.33 3... < x – 83.333... < 16.333... 67 < x < 99.666... 67 < x < 99

** Correct (5, Att. 2) ** Correct σ (5, Att. 2)

** –1.96 < σ

μ−x < 1.96 (5, Att. 2)

** Solution (5, Att. 2)

Blunders (–3) B1: Incorrect value of p or q. B2: Incorrect formula for . B3: Incorrect formula for σ. B4: Incorrect limits. B5: Incorrect solution. B6: P(z ≠ 0.8413).


Attempts (2, 2, 2, 2) A1: Correct value of p or q. A2: Correct formula for . A4: Correct formula for σ. A5: Correct formula for limits | 1.96 |.

2011 L.17/20_MS 48/72 of 70 examsDEB

10. (a) For each of the following, give a reason why it is not a group:

(i) the set of all even natural numbers under subtraction. (5, Att. 2)

Not a group Reason: Not closed 2 – 4 = –2 ∉ even natural numbers

(ii) the set of all even natural numbers under multiplication (5, Att. 2)

Not a group Reason: No identity element 2 × any ≠ 2

(iii) the set of all odd natural numbers under subtraction (5, Att. 2)

Not a group Reason: Not closed 7 – 5 = 2 ∉ odd natural numbers

(iv) the set of all odd natural numbers under multiplication. (5, Att. 2)

Not a group Reason: No inverse 3 × any ≠ 1

Blunders (–3) B1: Closure omitted. B2: Identity omitted. B3: No Inverse omitted. B4: No example.

Attempts (2, 2, 2, 2) A1: Closure or example. A2: Identity or example. A3: No inverse or example.

2011 L.17/20_MS 49/72 of 70 examsDEB

(b) S = {1, 2, 4, 7, 8, 11, 13, 14} is a group under multiplication modulo 15, which is assumed associative.

(i) Find the subgroup P of S that is generated by the element 2. (10, Att. 3)

21 = 2 22 = 4 23 = 8 24 = 16 = 1

R = {1, 2, 4, 8}

× 1 2 4 8

1 1 2 4 8

2 2 4 8 1

4 4 8 1 2

8 8 1 2 4

(i) Closed: yes (ii) Identity: yes: 1 (iii) Inverse: 1 → 1 2 → 8 4 → 4 8 → 2

Blunders (–3) B1: Not closed. B2: Incorrect entry.


Attempts (3) A1: Incomplete table.

(ii) Show that Q = {1, 4, 11, 14} is a subgroup of S under multiplication modulo 15. (10, Att. 3)

Q = {1, 4, 11, 14}

× 1 4 11 14

1 1 4 11 14

4 4 1 14 11

11 11 14 1 4

14 14 11 4 1

(i) Closed : yes (ii) Identity: yes: 1 (iii) Inverse: 1 → 1 4 → 4 11 → 11 14 → 14



Attempts (3) A1: Incomplete table.

2011 L.17/20_MS 50/72 of 70 examsDEB

10(b) (iii) Given that R = {1, 4, 7, 13} is a subgroup of S under multiplication modulo 15, show that R is isomorphic to P but not to Q.

Write down one such isomorphism from R to P. (5, 5, Att. 2, 2)

R = {1, 4, 7, 13}

× 1 4 7 13

1 1 4 7 13

4 4 1 13 7

7 7 13 4 1

13 13 7 1 4

Order P Order Q Order R

1 1 1 1 1 1

2 4 4 2 4 2

4 2 11 2 7 4

8 4 14 2 13 4 Possible R → P 1 → 1 1 → 1 2 → 4 2 → 4 7 → 2 7 → 8 13 → 8 13 → 2



Attempts (2, 2) A1: Incomplete table.

2011 L.17/20_MS 51/72 of 70 examsDEB

11. (a) Find the equation of the ellipse with one focus at (2, 0) and where (–4, 0) and (4, 0) are the co-ordinates of the endpoints of the major axis. (10, Att. 3)

Vertices are (–4, 0) and (4, 0) centre of ellipse is (0, 0)

Vertex (4, 0) a2 = 16

Focus (2, 0) a2 – b2 = 20 b2 = 4

∴ Equation of ellipse

16

2x +

4

2y = 1

Blunders (–3) B1: Value of a or b incorrect.


Attempts (3) A1: Standard equation of ellipse.

2011 L.17/20_MS 52/72 of 70 examsDEB

11(b) Tangents from an external point P to a circle touch the circle at Q and R. Prove that QR is the polar of P. (20, Att. 6)

c Q

P

R

OS

Given PQ and PR are tangents to the circle c and QR is a chord The centre of the circle c is O Let OP cut QR at the point S

Proof ΔOPQ and ΔORQ are congruent triangles | ∠OSQ | = | ∠OSR | = 90°

Consider ΔOPQ and ΔOSQ | ∠OQP | = | ∠OSQ | ... 90° | ∠QOP | = | ∠QOS | ... common angle ∴ ΔOPQ and ΔOSQ are equiangular

∴ ||

||

OQ

OP =

||

||

OS

OQ

| OP |.| OS | = | OQ |2 | OP |.| OS | = (radius)2 QR is the polar of P

Blunders (–3) B1: Congruent triangles incorrect. B2: Equiangular triangles incorrect. B3: Ratio incorrect. B4: ≠ r2.


Attempts (6) A1: Congruent triangles. A2: Diagram with information.

2011 L.17/20_MS 53/72 of 70 examsDEB

11(c) f is a similarity transformation. (i) l and m are two perpendicular lines. Prove that f (l) is perpendicular to f (m). (10, Att. 3)

P Q

fR

l

m

P�

Q�

R�

f m( )

f l( )

Let |det f | = k2

Area ΔP′Q′R′ = 2

1| P′Q′ |.| QR | sin

= k2 × area ΔPQR

k2 × area ΔPQR = k2 × 2

1| PQ |.| QR | sin 90°

= 2

1| P′Q′ |.| Q′R′ | sin

= 90° ∴ f (l) ⊥ f (m )

Blunders (–3) B1: Incorrect area. B2: Incorrect det. B3: ≠ 90°.


Attempts (3) A1: Diagram of merit.

2011 L.17/20_MS 54/72 of 70 examsDEB

11(c) (ii) Hence, prove that the circumcentre of a triangle is mapped onto the circumcentre of the image triangle under f. (10, Att. 3)

P

Q

R

b

a

f

G

P� Q�

R�

G�

f b( )

f a( )

Let G be the circumcentre of ΔABC and let a be the mediator of [ AB ] and b be the mediator of [ BC ].

f (A) = A′, f (B) = B′, f (C) = C′

Under f, a similarity transformation, perpendicularity and midpoint are invariant

a is the mediator of [ AB ] f (a) is the mediator of [ A′B′ ]

Similarily, f (b) is the mediator of [ B′C′ ] G′ is the circumcentre of ΔA′B′C′ as G → G′ the circumcentre is invariant under a similarity transformation

Blunders (–3) B1: ⊥ incorrect. B2: Midpoint invariant. B3: θ ≠ 90°.


Attempts (3) A1: Diagram of merit.

2011 L.17/20_MS 55/72 Page 68 of 70 examsDEB

2011 L.17/20_MS 56/72 Page 69 of 70 examsDEB

Mathematics Higher Level

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