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Algebra (NON GDC) IB Questionbank Maths SL 41

A1. (a) attempt to find d (M1)

e.g. 2

13 uu −, 8 = 2 + 2d

d = 3 A1 N2 2

(b) correct substitution (A1) e.g. u20 = 2 + (20 –1)3, u20 = 3 × 20 –1 u20 = 59 A1 N2 2

(c) correct substitution (A1)

e.g. S20 = 220

(2 + 59), S20 = 220

(2 × 2 + 19 × 3)

S20 = 610 A1 N2 2 [6]


A2. (a) attempt to apply rules of logarithms (M1)

e.g. ln ab = bln a, ln ab = ln a + ln b

correct application of ln ab = bln a (seen anywhere) A1

e.g. 3ln x = ln x3

correct application of ln ab = ln a + ln b (seen anywhere) A1

e.g. ln 5x3 = ln 5 + ln x3

so ln 5x3 = ln 5 + 3ln x

g (x) = f (x) + ln5 (accept g (x) = 3ln x + ln 5) A1 N1 4

(b) transformation with correct name, direction, and value A3

e.g. translation by ⎟⎟⎠

⎞⎜⎜⎝

⎛5ln

0, shift up by ln 5, vertical translation of ln 5 3

[7]


A3. (a) evidence of expanding M1 e.g. 24 + 4(23)x + 6(22)x2 + 4(2)x3 + x4, (4 + 4x + x2)(4 + 4x + x2)

(2 + x)4 = 16 + 32x + 24x2 + 8x3 + x4 A2 N2

(b) finding coefficients 24 and 1 (A1)(A1)

term is 25x2 A1 N3 [6]


A4. (a) interchanging x and y (seen anywhere) (M1) e.g. x = ylog (accept any base) evidence of correct manipulation A1

e.g. 3x = 21,3, 2

1

== xxy y log3 y, 2y = log3 x

f–1(x) = 32x AG N0

(b) y > 0, f–1(x) > 0 A1 N1

(c) METHOD 1

finding g(2) = log3 2 (seen anywhere) A1

attempt to substitute (M1) e.g. (f–1 ° g)(2) = 2log33

evidence of using log or index rule (A1)

e.g. (f–1 ° g)(2) = 22log4log 33 3,3

(f–1 ° g)(2) = 4 A1 N1

METHOD 2

attempt to form composite (in any order) (M1) e.g. (f–1 ° g)(x) = x3log23

evidence of using log or index rule (A1)

e.g.(f–1 ° g)(x) = 2loglog 3

23 3,3 xx

(f–1 ° g)(x) = x2 A1

(f–1 ° g)(2) = 4 A1 N1 [7]


A5. recognizing log a + log b = log ab (seen anywhere) (A1) e.g. log2(x(x – 2)), x2 – 2x

recognizing loga b = x ⇔ ax = b (seen anywhere) (A1) e.g. 23 = 8

correct simplification A1 e.g. x(x – 2) = 23, x2 – 2x – 8

evidence of correct approach to solve (M1) e.g. factorizing, quadratic formula

correct working A1

e.g. (x – 4)(x + 2), 2362 ±

x = 4 A2 N3 [7]


A6. (a) r = ⎟⎠⎞⎜

⎝⎛=21

3216

A1 N1

(b) correct calculation or listing terms (A1)

e.g. 32 × 316

218,

21

⎟⎠⎞⎜

⎝⎛×⎟

⎠⎞⎜

⎝⎛

−

, 32, ... 4, 2, 1

u6 = 1 A1 N2

(c) evidence of correct substitution in S∞ A1

e.g.

2132,

211

32

−

S∞ = 64 A1 N1 [5]


A7. (a) correct substitution into the formula for the determinant (A1) e.g. det A = 9ex × e3x – ex × ex

det A = 9e4x – e2x A1 N2

(b) recognizing that no inverse implies det A = 0 R1 e.g. 9e4x – e2x = 0, ad – bc = 0

attempt to solve equation (M1)

e.g. e2x = 91

, e–2x = 9, e2x(9e2x – 1) = 0, 9e4x = e2x

rearranging to get correct log equation

e.g. 2x = )eln()e9ln(,9ln2,91ln 24 xxx ==− (A1)

isolating x A1

e.g. x 9,21,

31ln,9ln

21,

91ln

21 =−==−= baxx

x = –ln 3 (accept a = –1, b = 3) A1 N3 [7]


A8. (a) n = 10 A1 N1

(b) a = p, b = 2q (or a = 2q, b = p) A1A1 N1N1

(c) ⎟⎟⎠

⎞⎜⎜⎝

⎛510

p5(2q)5 A1A1A1 N3

[6]


A9. (a) 5 A1 N1

(b) METHOD 1

yxy

x

8log32log832

log 222 −=⎟⎟⎠

⎞⎜⎜⎝

⎛ (A1)

= x log2 32 – y log2 8 (A1)

log2 8 = 3 (A1)

p = 5, q = –3 (accept 5x – 3y) A1 N3

METHOD 2

y

x

y

x

)2()2(

832

3

5

= (A1)

= y

x

3

5

22

(A1)

= 25x–3y (A1)

log2 (25x–3y) = 5x – 3y p = 5, q = –3 (accept 5x – 3y) A1 N3

[5]


A10. (a) METHOD 1

recognizing that f(8) = 1 (M1) e.g. 1 = k log2 8

recognizing that log2 8 = 3 (A1) e.g. 1 = 3k

k = 31

A1 N2

METHOD 2

attempt to find the inverse of f(x) = k log2 x (M1)

e.g. x = k log2 y, y = kx

2 substituting 1 and 8 (M1)

e.g. 1 = k log2 8, k1

2 = 8

k = ⎟⎠⎞⎜

⎝⎛ =

31

8log1

2

k A1 N2

(b) METHOD 1

recognizing that f(x) = 32

(M1)

e.g. x2log31

32 =

log2 x = 2 (A1)

f–1 ⎟⎠⎞⎜

⎝⎛32

= 4 (accept x = 4) A2 N3

METHOD 2

attempt to find inverse of f(x) = 31

log2 x (M1)

e.g. interchanging x and y , substituting k = 31

into y = kx

2

correct inverse (A1) e.g. f–1(x) = 23x, 23x

f–1 ⎟⎠⎞⎜

⎝⎛32

= 4 A2 N3

[7]


A11. (a) d = 3 (A1) evidence of substitution into un = a + (n − 1) d (M1) e.g. u101 = 2 + 100 × 3 u101 = 302 A1 N3

(b) correct approach (M1) e.g. 152 = 2 + (n − 1) × 3 correct simplification (A1) e.g. 150 = (n − 1) × 3, 50 = n − 1, 152 = −1 + 3n n = 51 A1 N2

[6]


A12. evidence of using binomial expansion (M1)

e.g. selecting correct term, ...28

18 26708 +⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟⎠

⎞⎜⎜⎝

⎛+ bababa

evidence of calculating the factors, in any order A1A1A1

e.g. 56, ( )53

53

3

332

58

,3,32 −⎟

⎠⎞⎜

⎝⎛⎟⎟⎠

⎞⎜⎜⎝

⎛− x

−4032x3 (accept = −4030x 3 to 3 s.f.) A1 N2 [5]


A13. (a) u10 = 3(0.9)9 A1 N1

(b) recognizing r = 0.9 (A1) correct substitution A1

e.g. S = 9.01

3−

S = 1.03

(A1)

S = 30 A1 N3 [5]


A14. (a) (i) attempt to set up equations (M1) –37 = u1 + 20d and –3 = u1 + 3d A1 –34 = 17d d = –2 A1 N2

(ii) –3 = u1 – 6 ⇒ u1 = 3 A1 N1

(b) u10 = 3 + 9 × –2 = –15 (A1)

S10 = 210

(3 + (–15)) M1

= –60 A1 N2 [7]


A15. (a) u1 = 1, u2 = –1, u3 = –3 A1A1A1 N3

(b) Evidence of using appropriate formula M1

correct values S20 = 220

(2 × 1 + 19 × –2) (= 10(2 – 38)) A1

S20 = –360 A1 N1 [6]


A16. (a) loga 10 = loga (5 × 2) (M1)

= loga 5 + loga 2

= p + q A1 N2

(b) loga 8 = loga 23 (M1)

= 3 loga 2

= 3q A1 N2

(c) loga 2.5 = loga 25

(M1)

= loga 5 − loga 2

= p − q A1 N2 [6]


A17. (a) (i) logc 15 = logc 3 + logc 5 (A1)

= p + q A1 N2

(ii) logc 25 = 2 logc 5 (A1)

= 2q A1 N2

(b) METHOD 1

621

=d M1

d = 36 A1 N1

METHOD 2

For changing base M1

eg dd 1010

10

10 log6log2,21

log6log

==

d = 36 A1 N1 [6]


A18. (a) ln a3b = 3ln a + ln b (A1)(A1)

ln a3b = 3p + q A1 N3

(b) ln 21=

ba ln a − ln b (A1)(A1)

ln 21=

ba p − q A1 N3

[6]


A19. (a) 1 1 7u S= = (A1) (C1)

(b) 2 2 1 18 7u S u= − = −

11= (A1)

11 7d = − (M1)

4= (A1) (C3)

(c) 4 1 ( 1) 7 3(4)u u n d= + − = + (M1)

4 19u = (A1) (C2) [6]


A20. METHOD 1

Using binomial expansion (M1)

( ) ( ) ( ) ( )32233773

23

7313

373 +⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟⎠

⎞⎜⎜⎝

⎛+=+ (A1)

= 27 + 27 7 + 63 + 7 7 (A2)

( ) 73490733

+=+ (so p = 90 , q = 34 ) (A1)(A1)(C3)(C3)

METHOD 2

For multiplying (M1)

( ) ( ) ( )( )73776973732

+++=++ (A1)

= 27 + 9 7 + 18 7 + 42 + 21 + 7 7

( = 27 + 27 7 + 63 + 7 7 ) (A2)

( ) 73490733

+=+ (so p = 90 , q = 34 ) (A1)(A1)(C3)(C3) [6]


A21. METHOD 1

log10 ⎟⎟⎠

⎞⎜⎜⎝

⎛

zyx2

= log10 x – log10 y2 – log10 z (A1)(A1)(A1)

log10 y2 = 2 log10 y (A1)

log10 z = 21

log z (A1)

log10 ⎟⎟⎠

⎞⎜⎜⎝

⎛

zyx2

= log10 x – 2log y – 21

log z

= p – 2q – 21 r (A1) (C2)(C2)(C2)

METHOD 2

x = 10, y2 = 102p, 210r

z = (A1)(A1)(A1)

log10⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛=⎟

⎟⎠

⎞⎜⎜⎝

⎛

22102

1010

10log

rq

p

zyx (A1)

= log10 ⎟⎠⎞⎜

⎝⎛ −−=⎟⎟⎠

⎞⎜⎜⎝

⎛ −−

2210 2

2 rqprqp

(A2) (C2)(C2)(C2)

[6]


A22. 5 38(2) ( 3 )

3x

⎛ ⎞−⎜ ⎟

⎝ ⎠

8Accept

5⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠ (M1)(A1)(A1)(A1)

Term is 348384x− (A2) (C6) [6]


A23. METHOD 1 2log 2logx x= (A1)

1log log

2y y= (A1)

3log 3logz z= (A1)

12log log 3log

2x y z+ − (A1)(A1)

12 32

a b c+ − (A1) (C6)

METHOD 2

2 2 3 3210 , 10 , 10b

a cx y z= = = (A1)(A1)(A1)

2 2 2

10 103 3

10 10log log

10

ba

c

x yz

⎛ ⎞⎛ ⎞ ×⎜ ⎟=⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

(A1)

2 3

210log 10 2 3

2

ba c ba c+ −⎛ ⎞ ⎛ ⎞= = + −⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠

(A2)

[6]


A24. Selecting one term (may be implied) (M1)

⎟⎠⎞⎜

⎝⎛27

52(2x2)5 (A1)(A1)(A1)

= 16800x10 (A1)(A1) (C6) Note: Award C5 for 16800.

[6]


A25. (a) log5 x2 = 2 log5 x (M1) = 2y (A1) (C2)

(b) log5x1

= –log5 x (M1)

= –y (A1) (C2)

(c) log25 x = 25log

log

5

5 x (M1)

= y21

(A1) (C2)

[6]


A26. ... + 6 × 22(ax)2 + 4 × 2(ax)3 + (ax)4 (M1)(M1)(M1) = ...+ 24a2x2 +8a3x3 + a4x4 (A1)(A1)(A1) (C6)

Notes: Award C3 if brackets omitted, leading to 24ax2 + 8ax3+ ax4. Award C4 if correct expression with brackets as in first line of markscheme is given as final answer.

[6]


A27. (a) 10 (A2) (C2)

(b) (3x2)3 61– ⎟⎠⎞⎜

⎝⎛x

[for correct exponents] (M1)(A1)

⎟⎟⎠

⎞⎜⎜⎝

⎛69

33 x6 ⎟⎠⎞⎜

⎝⎛ × 6

636

13 84or

1x

xx

(A1)

constant = 2268 (A1) (C4) [6]


A28. Statement (a) Is the statement true for all

real numbers x? (Yes/No) (b) If not true, example

A No x = –l (log10 0.1 = –1) (a) (A3) (C3)

B No x = 0 (cos 0 = 1) (b) (A3) (C3)

C Yes N/A

Notes: (a) Award (A1) for each correct answer. (b) Award (A) marks for statements A and B only if NO in column (a). Award (A2) for a correct counter example to statement A, (A1) for a correct counter example to statement B (ignore other incorrect examples). Special Case for statement C: Award (A1) if candidates write NO, and give a valid reason (eg

arctan 1 = 4π5

).

[6]


A29. Term involving x3 is ⎟⎟⎠

⎞⎜⎜⎝

⎛35

(2)2(–x)3 (A1)(A1)(A1)

⎟⎟⎠

⎞⎜⎜⎝

⎛35

= 10 (A1)

Therefore, term = –40x3 (A1) ⇒ The coefficient is –40 (A1) (C6)

[6]


A30. (3x + 2y)4 = (3x)4 + ⎟⎟⎠

⎞⎜⎜⎝

⎛14

(3x)2(2y) + ⎟⎟⎠

⎞⎜⎜⎝

⎛24

(3x)2(2y)2 + ⎟⎟⎠

⎞⎜⎜⎝

⎛34

(3x)(2y)3 +(2y)4 (A1)

= 81x4 + 216x3y + 216x2y2 + 96xy3 + 16y4 (A1)(A1)(A1) (C4) [4]


A31. (a) (1 + 1)4 = 24 = 1 + )1(34

)1(24

)1(14 32

⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟⎠

⎞⎜⎜⎝

⎛ + 14 (M1)

⇒ ⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟⎠

⎞⎜⎜⎝

⎛34

24

14

= 16 – 2

= 14 (A1) (C2)

(b) (1 + 1)9 = 1 + ⎟⎟⎠

⎞⎜⎜⎝

⎛++⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟⎠

⎞⎜⎜⎝

⎛89

...39

29

19

+ 1 (M1)

⇒ ⎟⎟⎠

⎞⎜⎜⎝

⎛++⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟⎠

⎞⎜⎜⎝

⎛89

...39

29

19

= 29 – 2

= 510 (A1) (C2) [4]


A32. log10

2

3 ⎟⎟⎠

⎞⎜⎜⎝

⎛QRP = 2log10 ⎟⎟⎠

⎞⎜⎜⎝

⎛3QRP (M1)

2log10 ⎟⎟⎠

⎞⎜⎜⎝

⎛3QRP = 2(log10P – log10(QR3)) (M1)

= 2(1og10P – log10Q – 3log10R) (M1) = 2(x – y – 3z) = 2x – 2y – 6z or 2(x – y – 3z) (A1)

[4]


A33. (a) log2 5 = 2log5log

a

a (M1)

= xy

(A1) (C2)

(b) loga 20 = loga 4 + loga 5 or loga 2 + loga 10 (M1) = 2 loga 2 + loga 5 = 2x + y (A1) (C2)

[4]


A34. S = ⎟⎠⎞⎜

⎝⎛−−

=−

321

32

11

ru

(M1)(A1)

= 53

32 × (A1)

= 52

(A1) (C4)

[4]


A35. (a + b)12

Coefficient of a5b7 is ⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟⎠

⎞⎜⎜⎝

⎛712

512

(M1)(A1)

= 792 (A2) (C4) [4]


A36. 17 + 27 + 37 + ... + 417 17 + (n – 1)10 = 417 (M1) 10(n – 1) = 400 n = 41 (A1)

S41 = 241

(2(17) + 40(10)) (M1)

= 41(17 + 200) = 8897 (A1)

OR

S41 = 241

(17 + 417) (M1)

= 241

(434)

= 8897 (A1) (C4) [4]


A37. 9x–1 = x2

31⎟⎠⎞⎜

⎝⎛

32x–2 = 3–2x (M1) (A1) 2x – 2 = –2x (A1)

x = 21

(A1) (C4)

[4]


A38. Required term is ⎟⎟⎠

⎞⎜⎜⎝

⎛58

(3x)5(–2)3 (A1)(A1)(A1)

Therefore the coefficient of x5 is 56 × 243 × –8 = –108864 (A1) (C4)

[4]


A39. S5 = 25

{2 + 32} (M1)(A1)(A1)

S5 = 85 (A1) OR a = 2, a + 4d = 32 (M1) ⇒ 4d = 30 d = 7.5 (A1)

S5 = 25

(4 + 4(7.5)) (M1)

= 25

(4 + 30)

S5 = 85 (A1) (C4) [4]


A40. (5a + b)7 = ...+ ⎟⎟⎠

⎞⎜⎜⎝

⎛47

(5a)3(b)4 + ... (M1)

= 43214567

××××××

× 53 × (a3b4) = 35 × 53 × a3b4 (M1)(A1)

So the coefficient is 4375 (A1) (C4) [4]


A41. 43x–1 = 1.5625 × 10–2 (3x – 1)log10 4 = log10 1.5625 – 2 (M1)

⇒ 3x – 1 = 4log

25625.1log

10

10 − (A1)

⇒ 3x – 1 = –3 (A1)

⇒ x = –32

(A1) (C4)

[4]

Algebra (GDC OPTIONAL) IB Questionbank Maths SL 52

A1. (a) evidence of choosing the formula for 20th term (M1)

e.g. u20 = u1 + 19d

correct equation A1

e.g. 19764,19764 −=+= dd

d = 3 A1 N2 3

(b) correct substitution into formula for un A1

e.g. 3709 = 7 + 3(n – 1), 3709 = 3n + 4

n = 1235 A1 N1 2 [5]


A2. (a) B, D A1A1 N2 2

(b) (i) f′(x) = 2

e2 xx −− A1A1 N2

Note: Award A1 for 2xe− and A1 for –2x.

(ii) finding the derivative of –2x, i.e. –2 (A1)

evidence of choosing the product rule (M1)

e.g. 22

e22e2 xx xx −− −×−− 22

e4e2 2 xx x −− +− A1

f ′′(x) = (4x2 – 2)2

e x− AG N0 5

(c) valid reasoning R1

e.g. f ′′(x) = 0

attempting to solve the equation (M1)

e.g. (4x2 – 2) = 0, sketch of f ′′(x)

p = 0.707 ⎟⎟⎠

⎞⎜⎜⎝

⎛−=−=⎟⎟⎠

⎞⎜⎜⎝

⎛=

21707.0,

21 q A1A1 N3 4

(d) evidence of using second derivative to test values on either side of POI M1

e.g. finding values, reference to graph of f′′, sign table

correct working A1A1

e.g. finding any two correct values either side of POI,

checking sign of f ′′ on either side of POI

reference to sign change of f ′′(x) R1 N0 4 [15]


A3. (a) 12 terms A1 N1 1

(b) evidence of binomial expansion (M1)

e.g. rrn barn −⎟⎟⎠

⎞⎜⎜⎝

⎛, an attempt to expand, Pascal’s triangle

evidence of choosing correct term (A1)

e.g. 10th term, r = 9, ⎟⎟⎠

⎞⎜⎜⎝

⎛911

(x)2 (2)9

correct working A1

e.g. ⎟⎟⎠

⎞⎜⎜⎝

⎛911

(x)2 (2)9, 55× 29

28160x2 A1 N2 4 [5]


A4. (a) common difference is 6 A1 N1

(b) evidence of appropriate approach (M1) e.g. un = 1353

correct working A1

e.g. 1353 = 3 + (n – 1)6, 631353+

n = 226 A1 N2

(c) evidence of correct substitution A1

e.g. S226 = 2226,

2)13533(226 +

(2 × 3 + 225 × 6)

S226 = 153 228 (accept 153 000) A1 N1 [6]


A5. (a) evidence of equation for u27 M1 e.g. 263 = u1 + 26 × 11, u27 = u1 + (n – 1) × 11, 263 – (11 × 26) u1 = –23 A1 N1

(b) (i) correct equation A1 e.g. 516 = –23 + (n – 1) × 11, 539 = (n – 1) × 11 n = 50 A1 N1

(ii) correct substitution into sum formula A1

e.g. S50 = 2

)1149)23(2(50,

2)51623(50

50×+−×

=+− S

S50 = 12325 (accept 12300) A1 N1 [6]


A6. evidence of substituting into binomial expansion (M1)

e.g. a5 + ...25

15 234 +⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟⎠

⎞⎜⎜⎝

⎛baba

identifying correct term for x4 (M1) evidence of calculating the factors, in any order A1A1A1

e.g. 2

322

6 2)3(10;4,27,25

⎟⎠⎞⎜

⎝⎛ −

⎟⎟⎠

⎞⎜⎜⎝

⎛x

xx

x

Note: Award A1 for each correct factor.

term = 1080x4 A1 N2

Note: Award M1M1A1A1A1A0 for 1080 with working shown. [6]


A7. (a) d = 2 A1 N1

(b) (i) 5 + 2n = 115 (A1)

n = 55 A1 N2

(ii) u1 = 7 (may be seen in above) (A1)

correct substitution into formula for sum of arithmetic series (A1)

e.g. S55 = ∑=

++=+55

155 )25()),2(54)7(2(

255),1157(

255

k

kS

S55 = 3355 (accept 3360) A1 N3 [6]


A8. (a) attempt to substitute into sum formula for AP (accept term formula) (M1)

e.g. S20 = { } ⎟⎠⎞⎜

⎝⎛ +−+− )7(

220

or ,19)7(22

2020ud

setting up correct equation using sum formula A1

e.g. 220 {2(–7) + 19d} = 620 A1 N2

(b) correct substitution u78 = –7 + 77(4) (A1) = 301 A1 N2

[5]


A9. (a) evidence of substituting into formula for nth term of GP (M1)

e.g. u4 = 3

811 r

setting up correct equation 31

811 3 =r A1

r = 3 A1 N2

(b) METHOD 1

setting up an inequality (accept an equation) M1

e.g. 64813;402

)31(811

;402

)13(811

>>−

−>

−n

nn

evidence of solving M1 e.g. graph, taking logs

n > 7.9888... (A1) n = 8 A1 N2

METHOD 2

if n = 7, sum = 13.49...; if n = 8, sum = 40.49... A2

n = 8 (is the smallest value) A2 N2 [7]


A10. (a) attempt to expand (M1) (x + h)3 = x3 + 3x2h + 3xh2 + h3 A1 N2

(b) evidence of substituting x + h (M1) correct substitution A1

e.g. f′(x) = h

xxhxhxh

)14(1)(4)(lim

33

0

+−−++−+→

simplifying A1

e.g. h

xxhxhxhhxx )1414433( 32223 −+−+−−+++

factoring out h A1

e.g. h

hxhxh )433( 22 −++

f′(x) = 3x2 – 4 AG N0

(c) f′(1) = –1 (A1) setting up an appropriate equation M1 e.g. 3x2 – 4 = –1

at Q, x = –1, y = 4 (Q is (–1, 4)) A1A1 N3

(d) recognizing that f is decreasing when f′(x) < 0 R1

correct values for p and q (but do not accept p = 1.15, q = –1.15) A1A1 N1N1

e.g. p = –1.15, q = 1.15; 32± ; an interval such as –1.15 ≤ x ≤ 1.15

(e) f′(x) ≥ –4, y ≥ –4, [–4, ∞[ A2 N2 [15]


A11. (a) 76547

4

22222 +++=∑=r

r (accept 16 + 32 + 64 + 128) A1 N1

(b) (i) METHOD 1

recognizing a GP (M1) u1 = 24, r = 2, n = 27 (A1) correct substitution into formula for sum (A1)

e.g. S27 = 12)12(2 274

−−

S27 = 2147483632 A1 N4

METHOD 2

recognizing ∑∑∑===

−=3

1

30

1

30

4 rrr

(M1)

recognizing GP with u1 = 2, r = 2, n = 30 (A1) correct substitution into formula for sum

S30 = 12)12(2 30

−− (A1)

= 2147483646

∑=

30

4

2r

r = 2147483646 – (2 + 4 + 8)

= 2147483632 A1 N4

(ii) valid reason (e.g. infinite GP, diverging series), and r ≥ 1 (accept r > 1)R1R1 N2 [7]


A12. METHOD 1

substituting into formula for S40 (M1) correct substitution A1

e.g. 1900 = 2

)106(40 1 +u

u1 = –11 A1 N2

substituting into formula for u40 or S40 (M1) correct substitution A1 e.g. 106 = –11 + 39d, 1900 = 20(–22 + 39d) d = 3 A1 N2

METHOD 2

substituting into formula for S40 (M1)

correct substitution A1 e.g. 20(2u1 + 39d) = 1900 substituting into formula for u40 (M1) correct substitution A1 e.g. 106 = u1 + 39d u1 = –11, d = 3 A1A1 N2N2

[6]


A13. (a) evidence of dividing two terms (M1)

e.g. 10801800,

30001800 −−

r = − 0.6 A1 N2

(b) evidence of substituting into the formula for the 10th term (M1)

e.g. u10 = 3000(− 0.6)9

u10 = −30.2 (accept the exact value −30.233088) A1 N2

(c) evidence of substituting into the formula for the infinite sum (M1)

6.13000.. =Sge

S = 1875 A1 N2 [6]


A14. (a) evidence of expanding M1 e.g. (x – 2)4 = x4 + 4x3(–2) + 6x2(–2)2 + 4x(–2)3 + (–2)4

(x – 2)4 = x4 – 8x3 + 24x2 – 32x + 16 A2 N2

(b) finding coefficients, 3 × 24 (= 72),4 × (–8)(= –32) (A1)(A1) term is 40x3 A1 N3

[6]


A15. (a) Recognizing an AP (M1) u1 = 15 d = 2 n = 20 (A1) substituting into u20 = 15 + (20 – 1) × 2 M1 = 53 (that is, 53 seats in the 20th row) A1 N2

(b) Substituting into S20 = 220 (2(15) + (20 – 1)2) (or into

220 (15 + 53)) M1

= 680 (that is, 680 seats in total) A1 N2 [6]


A16. (a) 5000(1.063)n A1 N1

(b) Value = $ 5000(1.063)5 (= $ 6786.3511...) = $ 6790 to 3 s.f. (accept $ 6786, or $ 6786.35) A1 N1

(c) (i) 5000(1.063)n > 10 000 or (1.063)n > 2 A1 N1

(ii) Attempting to solve the inequality nlog(1.063) > log2 (M1) n > 11.345 (A1) 12 years A1 N3

Note: Candidates are likely to use TABLE or LIST on a GDC to find n. A good way of communicating this is suggested below.

Let y = 1.063x (M1) When x = 11, y = 1.9582, when x = 12, y = 2.0816 (A1) x = 12 i.e. 12 years A1 N3

[6]


A17. (a) 51 (0.2) A1 N1

(b) (i) 9

10 5125 ⎟⎠⎞⎜

⎝⎛=u (M1)

= 0.0000128 ⎟⎟⎠

⎞⎜⎜⎝

⎛×⎟

⎠⎞⎜

⎝⎛ −

781251,1028.1,

51 5

7

A1 N2

(ii) 1

5125

−

⎟⎠⎞⎜

⎝⎛=

n

nu A1 N1

(c) For attempting to use infinite sum formula for a GP ⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

⎟⎠⎞⎜

⎝⎛−511

25 (M1)

S = ( )fs3to3.3125.314125 == A1 N2

[6]


A18. (a) 7 terms A1 N1

(b) A valid approach (M1)

Correct term chosen ( ) ( )333 336

xx −⎟⎟⎠

⎞⎜⎜⎝

⎛ A1

Calculating ( ) 273,2036 3 −=−=⎟⎟⎠

⎞⎜⎜⎝

⎛ (A1)(A1)

Term is −540x12 A1 N3 [6]


A19. Identifying the required term (seen anywhere) M1

eg 22810

×⎟⎟⎠

⎞⎜⎜⎝

⎛

⎟⎟⎠

⎞⎜⎜⎝

⎛810

= 45 (A1)

4y2, 2 × 2, 4 (A2)

a = 180 A2 N4 [6]


A20. (a) For taking three ratios of consecutive terms (M1)

( )3162486

54162

1854 === A1

hence geometric AG N0

(b) (i) r = 3 (A1)

un = 18 × 3n − 1 A1 N2

(ii) For a valid attempt to solve 18 × 3 n − 1 = 1062882 (M1)

eg trial and error, logs

n = 11 A1 N2 [6]


A21. (a) 3, 6, 9 A1 N1

(b) (i) Evidence of using the sum of an AP M1

eg ( ) 312032220 ×−+×

∑=

=20

1

6303n

n A1 N1

(ii) METHOD 1

Correct calculation for ∑=

100

1

3n

n (A1)

eg ( ) 15150,399322100 ×+×

Evidence of subtraction (M1)

eg 15150 − 630

145203100

21

=∑=n

n A1 N2

METHOD 2

Recognising that first term is 63, the number of terms is 80 (A1)(A1)

eg ( ) ( )379126280,30063

280 ×++

145203100

21

=∑=n

n A1 N2

[6]


A22. (a) For finding second, third and fourth terms correctly (A1)(A1)(A1)

Second term ,e1e

14 3 ⎟

⎠⎞⎜

⎝⎛

⎟⎟⎠

⎞⎜⎜⎝

⎛ third term ,

e1e

14 2

2 ⎟⎠⎞⎜

⎝⎛

⎟⎟⎠

⎞⎜⎜⎝

⎛

fourth term 3

e1e

14

⎟⎠⎞⎜

⎝⎛

⎟⎟⎠

⎞⎜⎜⎝

⎛

For finding first and last terms, and adding them to their three terms (A1)

+⎟⎠⎞⎜

⎝⎛

⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟

⎠⎞⎜

⎝⎛

⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟

⎠⎞⎜

⎝⎛

⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟

⎠⎞⎜

⎝⎛ +

32234

4

e1e

34

e1e

24

e1e

14

e04

e1e

4

e1

44

⎟⎠⎞⎜

⎝⎛⎟⎟⎠

⎞⎜⎜⎝

⎛

432

2344

e1

e1e4

e1e6

e1e4e

e1e ⎟

⎠⎞⎜

⎝⎛+⎟

⎠⎞⎜

⎝⎛+⎟

⎠⎞⎜

⎝⎛+⎟

⎠⎞⎜

⎝⎛+=⎟

⎠⎞⎜

⎝⎛ +

⎟⎠⎞⎜

⎝⎛ ++++= 42

24

e1

e46e4e N4

(b) 432

2344

e1

e1e4

e1e6

e1e4e

e1e ⎟

⎠⎞⎜

⎝⎛+⎟

⎠⎞⎜

⎝⎛−⎟

⎠⎞⎜

⎝⎛+⎟

⎠⎞⎜

⎝⎛−=⎟

⎠⎞⎜

⎝⎛ −

⎟⎠⎞⎜

⎝⎛ +−+−= 42

24

e1

e46e4e (A1)

Adding gives 2e4 + 12 + 4e2

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞⎜

⎝⎛⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟

⎠⎞⎜

⎝⎛

⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟⎠

⎞⎜⎜⎝

⎛ 4224

e1

44

2e1e

24

2e04

2accept A1 N2

[6]


A23. (a) For taking an appropriate ratio of consecutive terms (M1)

r = 32 A1 N2

(b) For attempting to use the formula for the nth term of a GP (M1)

u15 = 1.39 A1 N2

(c) For attempting to use infinite sum formula for a GP (M1)

S = 1215 A1 N2 [6]


A24. (a) (i) r = −2 A1 N1

(ii) u15 = −3 (−2)14 (A1)

= −49152 (accept −49200) A1 N2

(b) (i) 2, 6, 18 A1 N1

(ii) r = 3 A1 N1

(c) Setting up equation (or a sketch) M1

182

31

++=

−+

xx

xx (or correct sketch with relevant information) A1

x2 + 2x + 1 = 2x2 + 2x − 24 (A1)

x2 = 25

x = 5 or x = −5

x = −5 A1 N2 Notes: If “trial and error” is used, work must be documented with several trials shown. Award full marks for a correct answer with this approach. If the work is not documented, award N2 for a correct answer.

(d) (i) r = 21 A1 N1

(ii) For attempting to use infinite sum formula for a GP (M1)

S =

211

8

−

−

S = −16 A1 N2 Note: Award M0A0 if candidates use a value of r where r > 1, or r < −1.

[12]


A25. (a) (i) S4 = 20 A1 N1

(ii) u1 = 2, d = 2 (A1)

Attempting to use formula for Sn M1

S100 = 10100 A1 N2

(b) (i) M2 = ⎟⎟⎠

⎞⎜⎜⎝

⎛1041

A2 N2

(ii) For writing M3 as M2 × M or M × M2 ⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛1041

1021

or M1

M3 = ⎟⎟⎠

⎞⎜⎜⎝

⎛++++10002401

A2

M3 = ⎟⎟⎠

⎞⎜⎜⎝

⎛1061

AG N0

(c) (i) M4 = ⎟⎟⎠

⎞⎜⎜⎝

⎛1081

A1 N1

(ii) T4 = ⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟⎠

⎞⎜⎜⎝

⎛1081

1061

1041

1021

(M1)

= ⎟⎟⎠

⎞⎜⎜⎝

⎛40204

A1A1 N3

(d) T100 = ⎟⎟⎠

⎞⎜⎜⎝

⎛++⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟⎠

⎞⎜⎜⎝

⎛10

2001...

1041

1021

(M1)

= ⎟⎟⎠

⎞⎜⎜⎝

⎛1000

10100100 A1A1 N3

[16]


A26. Note: Throughout this question, the first and last terms are interchangeable.

(a) For recognizing the arithmetic sequence (M1) u1 = 1, n = 20, u20 = 20 (u1 = 1, n = 20, d = 1) (A1) Evidence of using sum of an AP M1

S20 = ( ) ( ))11912220

(or220201 ×+×=+ S A1

S20 = 210 AG N0

(b) Let there be n cans in bottom row

Evidence of using Sn = 3240 (M1)

eg ( ) ( )( ) ( )( )( ) 3240112

2,324012

2,3240

21 =−−+=−+=+ nnnnnnn

n2 + n − 6480 = 0 A1

n = 80 or n = −81 (A1)

n = 80 A1 N2

(c) (i) Evidence of using S = ( )2

1 nn+ (M1)

2S = n2 + n A1

n2 + n − 2S = 0 AG N0

(ii) METHOD 1

Substituting S = 2100

eg n2 + n − 4200 = 0, 2100 = ( )2

1 nn+ A1

EITHER

n = 64.3, n = −65.3 A1

Any valid reason which includes reference to integer being needed, R1

and pointing out that integer not possible here. R1 N1

eg n must be a (positive) integer, this equation does not have integer solutions.

OR

Discriminant = 16 801 A1

Valid reason which includes reference to integer being needed, R1

and pointing out that integer not possible here. R1 N1

eg this discriminant is not a perfect square, therefore no integer solution as needed.


A26. continued

METHOD 2 Trial and error S64 = 2080, S65 = 2145 A1A1

Any valid reason which includes reference to integer being needed, R1 and pointing out that integer not possible here. R1 N1

[14]


A27. (a) Recognizing an AP (M1) u1 =15 d = 2 n = 20 (A1) 4 substituting into u20 = 15 + (20 –1) × 2 M1 = 53 (that is, 53 seats in the 20th row) A1

(b) Substituting into S20= 220 (2(15) + (20–1)2) (or into

220 (15 + 53)) M1

= 680 (that is, 680 seats in total) A1 2 [6]


A28. (a) 5000(1.063)n A1 1

(b) Value = $5000(1.063)5 (= $6786.3511...) = $6790 to 3 sf (Accept $6786, or $6786.35) A1 1

(c) (i) 5000(1.063)n > 10000 or (1.063)n > 2 A1 1

(ii) Attempting to solve the inequality «log (1.063) > log 2 (M1) n > 11.345... (A1) 12 years A1 3

Note: Candidates are likely to use TABLE or LIST on a GDC to find n. A good way of communicating this is suggested below.

Let y = 1.063x (M1) When x = 11, y = 1.9582, when x = 12, y = 2.0816 (A1) x = 12 ie 12 years A1 3

[6]


A29. (a) 6 terms (A1) (C1)

(b) ( )23 2510, ( 2) 8,

3x

⎛ ⎞= − = −⎜ ⎟

⎝ ⎠ (A1)(A1)(A1)

fourth term is 480x− (A1)

for extracting the coefficient 80A = − (A1) (C5) [6]


A30. METHOD 1

2 39 3 , 27 3= = (A1)(A1)

expressing as a power of 3, 2 2 3 1(3 ) (3 )x x−= (M1)

4 3 33 3x x−= (A1)

4 3 3x x= − (A1)

7 3x =

37

x⇒ = (A1) (C6)

METHOD 2

( )2 log9 1 log27x x= − (M1)(A1)(A1)

2 log27 31 log9 2xx

⎛ ⎞= =⎜ ⎟− ⎝ ⎠ (A1)

4 3 3x x= − (A1) 7 3x =

37

x⇒ = (A1) (C6)

Notes: Candidates may use a graphical method. Award (M1)(A1)(A1) for a sketch, (A1) for showing the point of

intersection, (A1) for 0.4285…., and (A1) for 37

.

[6]


A31. (a) 3 3 3log log ( 5) log5x

x xx

⎛ ⎞− − = ⎜ ⎟−⎝ ⎠ (A1)

5x

Ax

=−

(A1) (C2)

Note: If candidates have an incorrect or no answer to part (a) award (A1)(A0)

if ⎟⎠⎞⎜

⎝⎛

− 5log

xx

seen in part (b).

(b) EITHER

3log 15xx

⎛ ⎞ =⎜ ⎟−⎝ ⎠

( )13 35xx

= =−

(M1)(A1)(A1)

3 15x x= −

2 15x− = −

152

x = (A1) (C4)

OR

10

10

log5 1

log 3

xx

⎛ ⎞⎜ ⎟−⎝ ⎠ = (M1)(A1)

10 10log log 35xx

⎛ ⎞ =⎜ ⎟−⎝ ⎠ (A1)

7.5x = (A1) (C4) [6]


A32. For using u3 = u1r2 = 8 (M1)

8 = 18r2 (A1)

r2 = ⎟⎠⎞⎜

⎝⎛=94

188

r = ±32 (A1)(A1)

,11

ruS−

=∞

)8.10(554,54 ==∞S (A1)(A1)(C3)(C3)

[6]


A33. (a) (i) Neither

(ii) Geometric series

(iii) Arithmetic series

(iv) Neither (C3) Note: Award (A1) for geometric correct, (A1) for arithmetic correct and (A1) for both “neither”. These may be implied by blanks only if GP and AP correct.

(b) (Series (ii) is a GP with a sum to infinity)

Common ratio 43

(A1)

S∞ = ⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

−=

−431

11 ra

(M1)

= 4 (A1) (C3) Note: Do not allow ft from an incorrect series.

[6]


A34. ⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛710

accept)(2310 37 ax (A1)(A1)(A1)

⎟⎟⎠

⎞⎜⎜⎝

⎛310

= 120 (A1)

120 × 27 a3 = 414 720 (M1) a3 = 27 a = 3 (A1) (C6)

Note: Award (A1)(A1)(A0) for ⎟⎟⎠

⎞⎜⎜⎝

⎛310

27 ax3. If this leads to the

answer a = 27, do not award the final (A1). [6]


A35. (a) (i) $11400, $11800 (A1) 1

(ii) Total salary 10 (2 11000 9 400)2

= × + × (A1)

= $128000 (A1) (N2) 2

(b) (i) $10700, $11449 (A1)(A1)

(ii) 10th year salary 910 000(1.07)= (A1)

= $18384.59 or $18400 or $18385 (A1) (N2) 4

(c) EITHER

Scheme A ( )A 2 11000 ( 1)4002nS n= × + − (A1)

Scheme B B10 000(1.07 1)

1.07 1

n

S −=−

(A1)

Solving B AS S> (accept B AS S= , giving 6.33n = ) (may be implied) (M1)

Minimum value of n is 7 years. (A1) (N2)

OR

Using trial and error (M1)

Arturo Bill

6 years $72 000 $71532.91

7 years $85 400 $86 540.21

(A1)(A1) Note: Award (A1) for both values for 6 years, and (A1) for both values for 7 years.

Therefore, minimum number of years is 7. (A1) (N2) 4 [11]

A36. Arithmetic sequence d = 3 (may be implied) (M1)(A1)


n = 1250 (A2)

S = 2

1250 (3 + 3750) ⎟⎠⎞⎜

⎝⎛ ×+= 3) 1249 (6

21250 or S (M1)

= 2 345 625 (A1) (C6) [6]


A37. x f Σf

4 2 2

5 5 7

6 4 11

7 3 14

8 4 18

10 2 20

12 1 21

(a) m = 6 (A2) (C2)

(b) Q1 = 5 (A2) (C2)

(c) Q3 = 8 (A1) IQR = 8 – 5 (M1) = 3 (accept 5 – 8 or [5, 8]) (C2)

[6]


A38. Arithmetic sequence (M1) a = 200 d = 30 (A1)

(a) Distance in final week = 200 + 51 × 30 (M1) = 1730 m (A1) (C3)

(b) Total distance = 252 [2.200 + 51.30] (M1)

= 50180 m (A1) (C3) Note: Penalize once for absence of units ie award A0 the first time units are omitted, A1 the next time.

[6]


A39. (a) (i) Area B = 161 , area C =

641 (A1)(A1)

(ii) 41

161641

41

41161

== (Ratio is the same.) (M1)(R1)

(iii) Common ratio = 41

(A1) 5

(b) (i) Total area (S2) = 165

161

41 =+ = (= 0.3125) (0.313, 3 sf) (A1)

(ii) Required area = S8 =

411

411

41 8

−

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞⎜

⎝⎛−

(M1)

= 0.333328 2(471...) (A1) = 0.333328 (6 sf) (A1) 4

Note: Accept result of adding together eight areas correctly.

(c) Sum to infinity =

411

41

− (A1)

= 31

(A1) 2

[11]


A40. (a) u4 = ul + 3d or 16 = –2 +3d (M1)

d = ( )32––16

(M1)

= 6 (A1) (C3)

(b) un = ul + (n – 1)6 or 11998 = –2 + (n – l)6 (M1)

n = 16

211998 ++ (A1)

= 2001 (A1) (C3) [6]


A41. log27 (x(x – 0.4)) = l (M1)(A1) x2 – 0.4x = 27 (M1) x = 5.4 or x = –5 (G2) x = 5.4 (A1) (C6)

Note: Award (C5) for giving both roots. [6]


A42. (a) Ashley AP 12 + 14 + 16 + ... to 15 terms (M1)

S15 = 215

[2(12) + 14(2)] (M1)

= 15 × 26 = 390 hours (A1) 3

(b) Billie GP 12, 12(1.1), 12(1.1)2… (M1)

(i) In week 3, 12(1.1)2 (A1) = 14.52 hours (AG)

(ii) S15 = ( )

1–1.1]1–1.1[12 15

(M1)

= 381 hours (3 sf) (A1) 4

(c) 12 (1.1)n–1 > 50 (M1)

(1.1)n–1 > 1250

(A1)

(n – 1) ln 1.1 > ln1250

n – 1 > 1.1ln1250ln

(A1)

n – 1 > 14.97 n > 15.97 ⇒ Week 16 (A1)

OR 12(1.1)n–1 > 50 (M1) By trial and error 12(1.1)14 = 45.6, 12(1.1)15 = 50.1 (A1) ⇒ n – l = 15 (A1) ⇒ n = 16 (Week 16) (A1) 4

[11]


A43. (a) (i) PQ = 22 AQAP + (M1)

= 22 22 + = ( )24 = 2 2 cm (A1)(AG)

(ii) Area of PQRS = (2 2 )(2 2 ) = 8 cm2 (A1) 3

(b) (i) Side of third square = ( ) ( )2222 + = √4 = 2 cm

Area of third square = 4 cm2 (A1)

(ii) 48

32

816

21

rd

nd

nd

st

== (M1)

⇒ Geometric progression, r = 21

84

168 == (A1) 3

(c) (i) u11 = u1r10 = 1610

21⎟⎠⎞⎜

⎝⎛ =

102416 (M1)

= 641 ( = 0.015625 = 0.0156, 3 sf) (A1)

(ii) S∞ = r

u–11 =

21–1

16 (M1)

= 32 (A1) 4 [10]


A44. (a) u1 = 7, d = 2.5 (M1) u41 = u1 + (n – 1)d = 7 + (41 – 1)2.5 = 107 (A1) (C2)

(b) S101 = 2n

[2u1 + (n – 1)d]

= 2101 [2(7) + (101 – 1)2.5] (M1)

= 2

)264(101

= 13332 (A1) (C2) [4]


A45. METHOD 1

log9 81 + log9 ⎟⎠⎞⎜

⎝⎛91 + log9 3 = 2 – 1 +

21

(M1)

⇒ 23

= log9 x (A1)

⇒ x = 23

9 (M1) ⇒ x = 27 (A1) (C4)

METHOD 2

log 81 + log9 ⎟⎠⎞⎜

⎝⎛91 + log9 3 = log9 ⎥

⎦

⎤⎢⎣

⎡⎟⎠⎞⎜

⎝⎛ 39181 (M2)

= log9 27 (A1) ⇒ x = 27 (A1) (C4)

[4]


A46. (a) r = 23

160240

240360 == = 1.5 (A1) 1

(b) 2002 is the 13th year. (M1) u13 = 160(1.5)13–1 (M1) = 20759 (Accept 20760 or 20800.) (A1) 3

(c) 5000 = 160(1.5)n–1

1605000 = (1.5)n–1 (M1)

log ⎟⎠⎞⎜

⎝⎛1605000 = (n – 1)log1.5 (M1)

n – 1 = 5.1log

1605000log ⎟

⎠⎞⎜

⎝⎛

= 8.49 (A1)

⇒ n = 9.49 ⇒ 10th year ⇒ 1999 (A1)

OR

Using a gdc with u1 = 160, uk+1 = 23

uk, u9 = 4100, u10 = 6150 (M2)

1999 (G2) 4

(d) S13 = 160 ⎥⎦

⎤⎢⎣

⎡−−15.115.1 13

(M1)

= 61958 (Accept 61960 or 62000.) (A1) 2

(e) Nearly everyone would have bought a portable telephone so there would be fewer people left wanting to buy one. (R1)

OR

Sales would saturate. (R1) 1 [11]


A47. (a) a1 = 1000, an = 1000 + (n – 1)250 = 10000 (M1)

n = 250100000010 −

+ 1 = 37.

She runs 10 km on the 37th day. (A1)

(b) S37 = 237 (1000 + 10000) (M1)

She has run a total of 203.5 km (A1) [4]


A48. The constant term will be the term independent of the variable x. (R1) 9

2

3

26

289

9

22...2

39

...292⎟⎠⎞⎜

⎝⎛ −++⎟

⎠⎞⎜

⎝⎛ −

⎟⎟⎠

⎞⎜⎜⎝

⎛++⎟

⎠⎞⎜

⎝⎛ −+=⎟

⎠⎞⎜

⎝⎛ −

xxx

xxx

xx (M1)

⎟⎠⎞⎜

⎝⎛ −=⎟

⎠⎞⎜

⎝⎛ −

⎟⎟⎠

⎞⎜⎜⎝

⎛6

63

26 8842

39

xx

xx (A1)

= –672 (A1) [4]


A49. a = 5 a + 3d = 40 (may be implied) (M1)

d = 335 (A1)

T2 = 5 + 335 (A1)

= 1632

or 350 or 16.7 (3 sf) (A1) (C4)

[4]


A50. (a) Plan A: 1000, 1080, 1160... Plan B: 1000, 1000(1.06), 1000(1.06)2… 2nd month: $1060, 3rd month: $1123.60 (A1)(A1) 2

(b) For Plan A, T12 = a + 11d = 1000 + 11(80) (M1) = $1880 (A1)

For Plan B, T12 = 1000(1.06)11 (M1) = $1898 (to the nearest dollar) (A1) 4

(c) (i) For Plan A, S12 = 212 [2000 + 11(80)] (M1)

= 6(2880) = $17280 (to the nearest dollar) (A1)

(ii) For Plan B, S12 = 106.1

)106.1(1000 12

−− (M1)

= $16870 (to the nearest dollar) (A1) 4 [10]


A51. (a) $1000 × 1.07510 = $2061 (nearest dollar) (A1) (C1)

(b) 1000(1.07510 + 1.0759 + ... + 1.075) (M1)

= 1075.1

)1075.1)(075.1(1000 10

−− (M1)

= $15208 (nearest dollar) (A1) (C3) [4]

Algebra (GDC) IB Questionbank Maths SL 2

A1. (a) combining 2 terms (A1)

e.g. log3 8x – log3 4, log321 x + log3 4

expression which clearly leads to answer given A1

e.g. 24

log,38

log 33xx

f(x) = log3 2x AG N0 2

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