IBSL- Review Ex. Mark Schemes
1.
( a)
evidence of setting function to zero 2 e.g. f ( x x) = 0, 8 x = 2 x evidence of correct working − 8 ± 64 −4 e.g. 0 = 2 x(4 – x x), x-intercepts are at 4 and 0 (accept (4, 0) and (0, 0), or x x = 4, x = 0)
(b)
(i)
x = 2 (must be equation)
(ii)
x) substituting x = 2 into f ( x y = 8
(M1) A1
A1A1 N1N1 A1
N1
(M1) A1
N2 [7]
2.
(a)
13 5 6 WP =
A1A1A1
N3
Note: Award Award A1 for A1 for each correct element.
(b)
order. Note: The first two steps may be done in any order.
subtracting 26 12 10 – 2WP e.g.
(A1)
multiplying WP by by 2 26 10 12 e.g. 0 2 − 2 S =
(A1)
A1
N2 [6]
3.
( a)
(b)
evidence of expanding 4 3 2 2 3 4 2 2 e.g. 2 + 4(2 ) x x + 6(2 ) x x + 4(2) x x + x , (4 + 4 x + x )(4 + 4 x + x )
M1
(2 + x)4 = 16 + 32 x + 24 x2 + 8 x3 + x4
A2
finding coefficients 24 and 1 term is 25 x2
IB Questionbank Maths SL
N2
(A1)(A1) A1
N3 1
[6]
4.
( a)
tan θ =
3 3 do not accept x 4 4 3
(b)
(i)
A1
N1
4
sin θ = 5 , cos θ = 5 correct substitution 3 4 e.g. sin 2θ = 2 5 5
(A1)(A1) A1
24
sin 2θ = (ii)
A1
25
correct substitution 2 2 2 3 , 4 − 3 e.g. cos 2θ = 1 – 2 5 5 5
N3
A1
7
cos 2θ =
A1
25
N1 [7]
5.
( a)
interchanging x and y (seen anywhere) e.g. x =
(M1)
log y
(accept any base) evidence of correct manipulation 1
y , 3
x
y
e.g. 3 = –1 2 x f ( x x) = 3
(b)
–1 y > 0, f ( x x) > 0
(c)
METHOD 1
= x 2 , x =
A1
1 2
log3 y, 2 y = log3 x
finding g (2) (2) = log3 2 (seen anywhere) attempt to substitute log 2 –1 e.g. ( f f g )(2) )(2) = 3 ° evidence of using log or index rule
AG
N0
A1
N1
A1 (M1)
3
–1 e.g. ( f f g )(2) )(2) = 3
log 3 4
°
,3
(A1)
log 3 2 2
–1 f g )(2) = 4 ( f
A1
°
N1
METHOD 2
attempt to form composite (in any order) 2 log x –1 e.g. ( f f g )( x) = 3 )( x ° evidence of using log or index rule
(M1)
3
IB Questionbank Maths SL
(A1) 2
–1 e.g.( f f g )( x) = 3 )( x
°
log 3 x 2
,3
log 3 x 2
–1 2 f g )( x) = x ( f )( x
A1
–1 f ( f
A1
°
g )(2) = 4
°
N1 [7]
6.
( a)
(i)
p = 0.2
A1
N1
(ii)
q = 0.4
A1
N1
(iii)
r = 0.1
A1
N1
A2
N2
2
(b)
A│ B′) = P( A
3 0.2 Note: Award Award A1 for an unfinished answer such as 0.3 .
( c)
valid reason
R1
2 e.g. 3 ≠ 0.5, 0.35 ≠ 0.3 thus, A and B are not independent
AG
N0 [6]
7.
(a)
f ′( x) = x ′( x
2
– 2 x – 3 x) = 0 evidence of solving f ′(′( x 2 e.g. x – 2 x – 3 = 0 evidence of correct working 2 ± 16 2 e.g. ( x x + 1)( x x – 3), x = –1 (ignore x = 3) x) evidence of substituting their negative x-value into f ( x 1 e.g. 3 5
(−1) 3
A1A1A1 (M1) A1 (A1) (M1)
1
− (−1) 2 − 3(−1), − − 1 + 3
y = 3
3
A1
− 1, 5 3 coordinates are
(b)
(i)
(–3, –9)
IB Questionbank Maths SL
N3
A1
N1 3
(ii)
(1, –4)
A1A1
(iii) reflection gives (3, 9) 3 , 9 2 stretch gives
N2
(A1) A1A1
N3 [14]
8.
(a) (a)
any any cor corre rect ct equa equatition on in the the for form m r = a + t b (accept any parameter) − 8 2 − 5 + t 1 25 − 8 e.g. r =
A2
N2
Note: Award Award A1 for a + t b , A1 for L = a + t b , A0 for r = b + t a.
(b)
recognizing scalar product must be zero (seen anywhere)
e.g. a • b = 0
2 − 7 1 , − 2 evidence of choosing direction vectors − 8 k correct calculation of scalar product e.g. 2(–7) + 1(–2) – 8 k simplification that clearly leads to solution e.g. –16 – 8k , –16 – 8k =0 k = –2 ( c)
evidence of equating vectors − 3 2 5 − 7 − 3 + p 1 = 0 + q − 2 − 25 − 8 3 − 2 e.g. L = L , 1
(d)
R1
(A1)(A1) (A1) A1 AG
N0
(M1)
3
any two correct equations e.g. –3 + 2 p = 5 – 7q, –1 + p = –2q, –25 – 8 p = 3 –2q attempting to solve equations p = –3, q = 2) finding one correct parameter ( p the coordinates of A are (–9, –4, –1) (i) evidence of appropriate approach − 8 − 9 OA + AB = OB, AB = − 5 − − 4 25 − 1 e.g. 1 AB = − 1 26
IB Questionbank Maths SL
A1A1 (M1) A1 A1 (M1)
N3
A1
N2 4
(ii)
7 AC = 2 2 finding evidence of finding magnitude AC = 7 2 + 2 2 + 2 2
A1 (M1)
e.g.
AC
=
57
A1
N3 [18]
− 4.33 − 2 1.67 1 − 0.333 1.67 − 0.667 0 0.333 9.
13 5 − −2 3 3 1 = 5 1 − 3 3 2 1 0 − 3 3
(a)
–1 A =
(b)
evidence of attempting to solve equation –1 e.g. multiply by A (on left or right), setting up system of equations − 1 0 − 1 X = (accept x = 1, y = 0, z = –1)
A2
N2
(M1)
A2
N3 [5]
10.
( a)
common difference is 6
(b)
evidence of appropriate approach e.g. u = 1353 n correct working e.g. 1353 = 3 + (n – 1)6, n = 226
( c)
A1
1353 + 3
N1
(M1) A1
6
evidence of correct substitution 226(3 + 1353) 226 , 2 2 (2 × 3 + 225 × 6) e.g. S = 226 S = 153 228 (accept 153 000) 226
A1
N2
A1
A1
N1 [6]
IB Questionbank Maths SL
5
11.
(a)
Σ fx = 1(2) + 2(4) + ... + 7(4), Σ fx = 146 + 5 x (seen anywhere)
∑ fx f evidence of substituting into mean = ∑
correct equation 146 + 5 x e.g. 34 + x = 4.5, 146 + 5 x = 4.5(34 + x) x = 14 (b)
σ =
1.54
A1 (M1) A1 A1
N2
A2
N2 [6]
12.
(a)
(i)
(ii)
evidence of finding the amplitude 7+3 e.g. 2 , amplitude = 5 p = –5
(M1)
period = 8
(A1)
= 2π = π 8 4 q = 0.785 7−3 (iii) r = 2 r = 2 (b)
k = –3 (accept y = –3)
A1
N2
A1
N2
(A1) A1
N2
A1
N1 [7]
13.
(a)
(b)
correct substitution 2 2 e.g. 25 + 16 – 40cos x, 5 + 4 – 2 × 4 × 5 cos x AC = 41 − 40 cos x
AG
correct substitution
A1
AC sin x
=
4
,
1
sin 30 2
AC
= 4 sin x accept 4 sin x sin 30 AC = 8 sin x (c) (i) evidence of appropriate approach using AC e.g. 8 sin x = 41 − 40 cos x , sketch showing intersection correct solution 8.682..., 111.317... obtuse value 111.317... x = 111.32 to 2 dp (do not accept the radian answer 1.94) e.g.
IB Questionbank Maths SL
A1
A1 M1 (A1) (A1) A1
N1
N2 6
(ii)
(d)
(i)
(ii)
substituting value of x into either expression for AC e.g. AC = 8 sin 111.32 AC = 7.45
(M1)
evidence of choosing cosine rule a2 + c2 − b2 2ac e.g. cos B = correct substitution 4 2 + 4 2 − 7.45 2 2× 4× 4 e.g. , 7.452 = 32 – 32 cos y, cos y = –0.734... y = 137
(M1)
correct substitution into area formula
(A1)
A1
N2
A1 A1
N2
1
× 4 × 4 × sin 137, 8 sin 137 area = 5.42 e.g. 2
A1
N2 [14]
14.
(a)
q = –2, r = 4 or q = 4, r = –2
(b)
x = 1 (must be an equation)
(c)
substituting (0, –4) into the equation e.g. –4 = p(0 – (–2))(0 – 4), –4 = p(–4)(2) correct working towards solution e.g. –4 = –8 p 4 1 = 8 p = 2
A1A1
N2
A1
N1
(M1) (A1) A1
N2 [6]
15.
(a)
(b)
evidence of appropriate approach − 2 6 BC = BA + AC, − 3 − − 2 2 3 e.g. − 8 BC = − 1 − 1
(M1)
attempt to find the length of AB 2 2 2 AB = 6 + (−2) + 3 (= 36 + 4 + 9 =
(M1)
IB Questionbank Maths SL
A1 49
= 7)
N2
(A1)
7
6 − 2 7 3 1
unit vector is (c)
6 7 = − 2 7 3 7
recognizing that the dot product or cos θ being 0 implies perpendicular correct substitution in a scalar product formula − 12 + 6 + 6 e.g. (6) × (–2) + (–2) × (–3) + (3) × (2), cos θ = 7 × 17 correct calculation e.g. AB • AC = 0, cos θ = 0 therefore, they are perpendicular
A1
N2
(M1) A1
A1 AG
N0 [8]
16.
(a)
(b)
evidence of multiplying e.g. one correct element − 15 5 AB =
(M1)
A1A1
N3
METHOD 1
evidence of multiplying by A (on left or right) –1 e.g. AA X = AB, X = AB − 15 5 (accept x = – 15, y = 5) X =
(M1)
A1
N2
METHOD 2
attempt to set up a system of equations 4 x + 2 y − 3 x + y = −5, =5 e.g.
10
(M1)
10
− 15 5 (accept x = – 15, y = 5) X =
A1
N2 [5]
17.
(a)
π = cosπ 2
f
= –1
π (b) ( g f ) 2 = g (–1) (= 2(–1)2 – 1) ° =1
IB Questionbank Maths SL
(A1) A1
N2
(A1) A1
N2
8
(c)
( g f )( x) = 2(cos (2 x))2 – 1 (= 2 cos2(2 x) – 1) ° evidence of 2 cos2 θ – 1 = cos 2θ (seen anywhere) ( g f )( x) = cos 4 x ° k = 4
A1 (M1) A1
N2 [7]
18.
recognizing log a + log b = log ab (seen anywhere) 2 e.g. log ( x( x – 2)), x – 2 x 2
(A1)
recognizing loga b = x ⇔ a x = b (seen anywhere) 3 e.g. 2 = 8 correct simplification 3 2 e.g. x( x – 2) = 2 , x – 2 x – 8 evidence of correct approach to solve e.g. factorizing, quadratic formula correct working 2 ± 36 2 e.g. ( x – 4)( x + 2), x = 4
(A1) A1 (M1) A1 A2
N3 [7]
19.
(a)
(i)
(ii)
(b)
evidence of appropriate approach e.g. 9 + 25 + 35, 34 + 35 p = 69
(M1)
evidence of valid approach e.g. 109 – their value of p, 120 – (9 + 25 + 35 + 11) q = 40
(M1)
evidence of appropriate approach
∑ fx
e.g. substituting into
mean = 3.16 (c)
n
, division by 120
1.09
A1
A1
N2
N2
(M1)
A1
N2
A1
N1 [7]
20.
(a)
(b)
evidence of equation for u27 e.g. 263 = u + 26 × 11, u = u + (n – 1) × 11, 263 – (11 × 26) 1 27 1 u = –23 1
M1
(i)
A1
correct equation e.g. 516 = –23 + ( n – 1) × 11, 539 = ( n – 1) × 11 n = 50
IB Questionbank Maths SL
A1
A1
N1
N1 9
(ii)
correct substitution into sum formula 50(−23 + 516) 50( 2 × (−23) + 49 × 11) , S 50 = 2 2 e.g. S = 50 S = 12325 (accept 12300) 50
A1
A1
N1 [6]
21.
(a)
(b)
36 outcomes (seen anywhere, even in denominator) valid approach of listing ways to get sum of 5, showing at least two pairs e.g. (1, 4)(2, 3), (1, 4)(4, 1), (1, 4)(4, 1), (2, 3)(3, 2) , lattice diagram 4 1 = 36 P(prize) = 9
(A1) (M1)
recognizing binomial probability 8 1 3 8 5 8, 1 3 9 , binomial pdf, 9 9 e.g. B P(3 prizes) = 0.0426
(M1)
A1
A1
N3
N2 [5]
22.
evidence of substituting into binomial expansion 5 4 5 3 2 a b + a b + ... 2 5 1 e.g. a + identifying correct term for x4 evidence of calculating the factors, in any order 2 5 4 6 2 3 − 2 , 27 x , x 2 ; 10(3 x ) x 2 e.g.
(M1)
(M1) A1A1A1
Note: Award A1 for each correct factor.
term = 1080 x4
A1
N2
Note: Award M1M1A1A1A1A0 for 1080 with working shown. [6]
23.
(a)
IB Questionbank Maths SL
10
(b)
x = –1.32, x = 1.68 (accept x = –1.41, x = 1.39 if working in degrees)
(c)
–1.32 < x < 1.68 (accept –1.41 < x < 1.39 if working in degrees)
A1A1A1
N3
A1A1
N2
A2
N2 [7]
24.
(a)
(b)
appropriate approach e.g. 6 = 8θ ˆC AO = 0.75
(M1)
evidence of substitution into formula for area of triangle
(M1)
A1
N2
1 e.g. area = 2
× 8 × 8 × sin(0.75)
area = 21.8… evidence of substitution into formula for area of sector
(A1) (M1)
1 e.g. area = 2
× 64 × 0.75 area of sector = 24 evidence of substituting areas 1
r 2θ −
1
ab sin C
, area of sector – area of triangle area of shaded region = 2.19 cm 2 (c) attempt to set up an equation for area of sector e.g. 2
2
(A1) (M1) A1 (M1)
N4
A1
N2
1
× 82 × θ = 1.40625 (1.41 to 3 sf)
e.g. 45 = 2 ˆE CO
(d)
METHOD 1
attempting to find angle EOF e.g. π – 0.75 – 1.41 ˆF EO
= 0.985 (seen anywhere) evidence of choosing cosine rule correct substitution 2 2 e.g. EF = 8 + 8 − 2 × 8 × 8 × cos 0.985 IB Questionbank Maths SL
(M1) A1 (M1) A1
11
EF = 7.57 cm
A1
N3
METHOD 2
attempting to find angles that are needed e.g. angle EOF and angle OEF ˆ ˆ ˆF EO = 0.9853... and OEF (or OFE) = 1.078... evidence of choosing sine rule correct substitution EF e.g. sin0.985
=
(M1) A1 (M1) (A1)
8 sin 1.08
EF = 7.57 cm
A1
N3
METHOD 3
attempting to find angle EOF
(M1)
e.g. π – 0.75 – 1.41 ˆF EO
= 0.985 (seen anywhere) evidence of using half of triangle EOF
A1 (M1)
0.985 e.g. x = 8 sin
2
correct calculation e.g. x = 3.78 EF = 7.57 cm
A1 A1
N3 [15]
25.
(a)
(i)
(3, –4, 0)
(ii)
− 2 3 choosing velocity vector 1 finding magnitude of velocity vector (−2) 2 + 3 2 + 12 , 4 + 9 + 1 e.g. speed = 3.74 (
(b)
A1
14 )
(i)
substituting p = 7 B = (–11, 17, 7)
(ii)
METHOD 1
appropriate method to find e.g. AO + OB , A – B
IB Questionbank Maths SL
AB or BA
N1
(M1) (A1) A1
N2
(M1) A1
N2
(M1)
12
− 14 14 AB = 21 or BA = − 21 7 − 7 distance = 26.2
(A1)
(7 14 )
A1
N3
METHOD 2
evidence of applying distance is speed × time e.g. 3.74 × 7 distance = 26.2
(7 14 )
(M2) A1
N3
(M1) 2 2 2 2 2 2 (3 − ( −11)) + (−4 − 17) + (0 − 7) e.g. (3 – (–11)) + (–4 – 17) + (0 – 7) , AB2 = 686, AB = 686 (A1) distance AB = 26.2 ( 7 14 ) A1
N3
METHOD 3
attempt to find AB2, AB
(c)
− 2 – 1 3 and 2 1 a correct direction vectors −1 − 2 − 1 2 = a 2 + 5 , 3 • 2 1 a a = a + 8 substituting a +8 2 e.g. cos 40° = 14 a + 5 a = 3.21, a = –0.990
(A1)(A1)
(A1)(A1) M1
A1A1
N3 [16]
26.
(a)
0
(i)
g (0) = e
(ii)
METHOD 1
= –1
–2
(A1) A1
substituting answer from (i) e.g. ( f g )(0) = f (–1) ° correct substitution
f (–1) = 2(–1) f (–1) = 1
N2
(M1) 3
+3
(A1) A1
N3
METHOD 2
IB Questionbank Maths SL
13
attempt to find ( f g )( x) ° 3 x 3 x 3 e.g. ( f g )( x) = f (e – 2) = 2(e – 2) + 3 ° correct expression for ( f g )( x) ° 3 x 3 e.g. 2(e – 2) + 3 ( f g )(0) = 1 ° (b)
interchanging x and y (seen anywhere) 3 e.g. x = 2 y + 3 attempt to solve x − 3 3 e.g. y = 2 x − 3 3 –1 2 f ( x) =
(M1)
(A1) A1
N3
(M1) (M1)
A1
N3 [8]
27.
(a)
(b)
evidence of equating scalar product to 0 2 × 3 + 3 × (–1) + (–1) × p = 0 (6 – 3 – p = 0, 3 – p = 0) p = 3
(M1) A1 A1
evidence of substituting into magnitude formula 1 + q 2 + 25 e.g. , 1 + q2 + 25
(M1)
setting up a correct equation q 2 + 25 = 42 + 1 e.g. , 1 + q2 + 25 = 42, q2 = 16
A1
q = ±4
A1
N2
N2 [6]
28.
(a) (b)
A2 evidence of appropriate approach
IB Questionbank Maths SL
N2
(M1)
14
1 e.g. reference to any horizontal shift and/or stretch factor, x = 3 + 1, y = 2
P is (4, 1) (accept x = 4, y = 1)
×2 A1A1
N3 [5]
29.
evidence of substituting for cos2 x evidence of substituting into sin 2 x + cos2 x = 1 correct equation in terms of cos x (seen anywhere)
(M1) (M1) A1
2 2 e.g. 2cos x – 1 – 3 cos x – 3 = 1, 2 cos x – 3 cos x – 5 = 0
evidence of appropriate approach to solve e.g. factorizing, quadratic formula appropriate working
(M1) A1 3 ± 49
e.g. (2 cos x – 5)(cos x + 1) = 0, (2 x – 5)( x + 1), cos x =
4
correct solutions to the equation 5
5
e.g. cos x = 2 , cos x = –1, x = 2 , x = –1 x = π
(A1) A1
N4 [7]
30.
(a)
METHOD 1
recognizing that f (8) = 1 e.g. 1 = k log 8 2
(M1)
recognizing that log2 8 = 3 e.g. 1 = 3k
(A1)
1 k = 3
A1
N2
METHOD 2
attempt to find the inverse of f ( x) = k log2 x
(M1)
x k e.g. x = k log y, y = 2 2
substituting 1 and 8
(M1)
1 k e.g. 1 = k log 8, 2 2
1 k =
log 2 8
IB Questionbank Maths SL
=8
k = 1 3
A1
N2
15
(b)
METHOD 1
2
recognizing that f ( x) = 2
=
1
e.g. 3 3 log2 x = 2
(M1)
3
log 2 x
(A1)
2 –1 3 f = 4 (accept x = 4)
A2
N3
METHOD 2
1
attempt to find inverse of f ( x) =
3
log2 x
(M1) 1
e.g. interchanging x and y , substituting k = 3
correct inverse –1 3 x 3 x e.g. f ( x) = 2 , 2 2 –1 3 f =4
x
into y =
2 k
(A1)
A2
N3 [7]
31.
(a)
(b)
(i)
evidence of substituting into n( A ∪ B) = n( A) + n( B) – n( A ∩ B) e.g. 75 + 55 – 100, Venn diagram 30
(ii)
45
(i)
METHOD 1
evidence of using complement, Venn diagram e.g. 1 – p, 100 – 30 70 7 = 100 10
(M1) A1
N2
A1
N1
(M1) A1
N2
METHOD 2
attempt to find P(only one sport), Venn diagram 25
+
(M1)
45
e.g. 100 100 70 7
=
100 10
IB Questionbank Maths SL
A1
N2
16
45
9 = 70 14
(ii) (c)
(d)
A2
valid reason in words or symbols e. g. P( A ∩ B) = 0 if mutually exclusive, P( A ∩ B) if not mutually exclusive correct statement in words or symbols e.g. P( A ∩ B) = 0.3, P( A ∪ B) ≠ P( A) + P( B), P( A) + P( B) > 1, some students play both sports, sets intersect
(R1)
valid reason for independence e.g. P( A ∩ B) = P( A) × P( B), P( B│ A) = P( B) correct substitution
(R1)
30 e.g. 100
≠
75
55 30 , 100 100 55
×
≠
A1
A1A1
N2
N2
N3
75 100 [12]
32.
METHOD 1
substituting into formula for S 40 correct substitution 40(u1 + 106) 2 e.g. 1900 = u = –11 1
(M1) A1
substituting into formula for u40 or S 40 correct substitution e.g. 106 = –11 + 39 d , 1900 = 20(–22 + 39d ) d = 3
(M1) A1
A1
A1
N2
N2
METHOD 2
substituting into formula for S 40
(M1)
correct substitution e.g. 20(2u + 39d ) = 1900 1 substituting into formula for u40 correct substitution e.g. 106 = u + 39d 1 u = –11, d = 3 1
A1 (M1) A1 A1A1 N2N2 [6]
33.
3 2 − 3 − 2 1 2 –1 − 8 − 6 9 A (a) = (b)
evidence of subtracting matrices
IB Questionbank Maths SL
A2
N2
(M1) 17
e.g.
7 6 1
− 7 − 3 − 8 − 5 − 5 − 9
6 5 7
2 3 2
10 4 , 1 10 10 1
4 2 5
− 8 − 12 − 15 , D – C
evidence of multiplying on left by A –1 3 2 − 3 10 − 2 1 2 1 − 8 − 6 9 10 –1 –1 e.g. A AB, A ( D – C ),
2 1 4 B =
4 2 5
− 8 − 12 − 15
− 3 2 1
1 0 1
(M1)
A2
N3 6]
34.
(a)
choosing sine rule correct substitution AD e.g. sin0.8
=
(M1) A1
4 sin 0.3
AD = 9.71 (cm) (b)
A1
N2
METHOD 1
finding angle OAD = π – 1.1 = (2.04) (seen anywhere) choosing cosine rule correct substitution 2 2 2 e.g. OD = 9.71 + 4 – 2 × 9.71 × 4 × cos( π – 1.1) OD = 12.1 (cm)
(A1)
(M1) A1 A1
N3
METHOD 2
finding angle OAD = π – 1.1 = (2.04) (seen anywhere)
(A1)
choosing sine rule correct substitution
(M1) A1
OD e.g.
sin(π – 1.1)
=
9.71 sin 0.8
=
4 sin 0.3
OD = 12.1 (cm) (c)
(d)
A1
correct substitution into area of a sector formula 2 e.g. area = 0.5 × 4 × 0.8 area = 6.4 (cm2)
(A1)
substitution into area of triangle formula OAD correct substitution
(M1) A1
1 e.g. A = 2 IB Questionbank Maths SL
A1
N3
N2
1
× 4 × 12.1 × sin 0.8, A =
2
× 4 × 9.71 × sin 2.04, 18
1 A = 2
× 12.1 × 9.71 × sin 0.3 subtracting area of sector OABC from area of triangle OAD e.g. area ABCD = 17.3067 – 6.4 area ABCD = 10.9 (cm 2)
(M1) A1
N2 [13]
35.
(a)
evidence of correct approach 3 2 PQ = OQ − OP, − 3 − − 1 8 5 e.g. 1 PQ = − 2 3
(b)
(i)
correct description
A1
AG R1
N0 N1
A2
N2
3 − 3 8 e.g. reference to being the position vector of a point on the line, a vector to the line, a point on the line. (ii)
any correct expression in the form r = a + t b
3 1 − 2 − 3 8 3 where a is , and b is a scalar multiple of 3 − 1 3 + 2 s r t s − + = − − 3 2 , 3 4 8 − 3 8 + 6 s e.g. r = (c)
(d)
one correct equation e.g. 3 + s = –1, –3 – 2 s = 5
(A1)
s = –4 p = –4
A1 A1
one correct equation e.g. –3 + t = –1, 9 – 2t = 5
A1
t = 2
A1
substituting t = 2 e.g. 2 + 2q = –4, 2q = –6 q = –3
IB Questionbank Maths SL
A1 AG
N2
N0
19
(e)
1 1 − 2 and − 2 choosing correct direction vectors 3 − 3 finding correct scalar product and magnitudes scalar product (1)(1) + (–2)(–2) + (–3)(3) (= –4) 2 2 2 2 3 2 magnitudes 1 + ( −2) + 3 = 14 , 1 + (−2) + (−3) = evidence of substituting into scalar product −4 e.g. cos θ = 3.741... × 3.741... θ = 1.86 radians (or 107°)
(A1)(A1) (A1)(A1)(A1) 14
M1 A1
N4 [17]
36.
(a)
evidence of addition e.g. at least two correct elements
4 1 A + B = (b)
2
0
A1
evidence of multiplication e.g. at least two correct elements − 3 − 6 − 9 3 −3 A =
(c)
(M1)
(M1)
A1
evidence of matrix multiplication (in correct order) 1( 3) + 2( − 2) 3( 3) + ( − 1) ( − 2) e.g. AB = − 1 2 11 − 1 AB =
N2
N2
(M1)
1( 0) + 2( 1)
3( 0) + ( − 1) ( 1) A2
N3 [7]
37.
(a)
(b)
(i)
sin 140° = p
A1
N1
(ii)
cos 70° = −q
A1
N1
METHOD 1
evidence of using sin2 θ + cos2 θ = 1 e.g. diagram,
cos 140° = ± cos 140° = −
1− p 2 1− p 2 1− p 2
(M1)
(seen anywhere) (A1) A1
N2
METHOD 2
IB Questionbank Maths SL
20
evidence of using cos2 θ = 2 cos2 θ − 1
(M1)
cos 140° = 2 cos2 70 − 1
(A1)
cos 140° = 2(− q)2 −1 (= 2q2 − 1) (c)
A1
N2
A1
N1
A1
N1
METHOD 1
sin 140°
tan 140° =
cos 140°
=−
p
1 − p 2
METHOD 2
p
tan 140° =
2q 2 −1
[6]
38.
(a)
d = 3
(A1) (M1)
evidence of substitution into un = a + (n − 1) d e.g. u = 2 + 100 × 3 101 u
(b)
101
= 302
A1
correct approach e.g. 152 = 2 + (n − 1) × 3 correct simplification e.g. 150 = (n − 1) × 3, 50 = n − 1, 152 = −1 + 3n n = 51
N3
(M1) (A1) A1
N2 [6]
39.
(a)
period = π
A1
N1
A1A1A1
N3
(b) y
4 3 2 1 0 – 1 – 2 – 3
π 2
π
3π 2
2π
x
– 4
Note: Award A1 for amplitude of 3, A1 for their period, A1 for a sine curve passing through (0, 0) and (0, 2 π ). IB Questionbank Maths SL
21
(c)
evidence of appropriate approach e.g. line y = 2 on graph, discussion of number of solutions in the domain 4 (solutions)
(M1)
A1
N2 [6]
40.
(a)
METHOD 1
ln ( x + 5) + ln 2 = ln (2( x + 5)) (= ln (2 x + 10)) interchanging x and y (seen anywhere) e.g. x = ln (2 y + 10) evidence of correct manipulation x
e.g. e
(A1) (M1) (A1)
= 2 y + 10
f −1 ( x ) =
e x
− 10 2
A1
N2
METHOD 2
y = ln ( x + 5) + ln 2 y − ln 2 = ln ( x + 5)
evidence of correct manipulation
(A1) (A1)
y − ln 2
e.g. e
= x + 5 interchanging x and y (seen anywhere) x − ln 2
e.g. e f
(b)
−1
(M1)
= y + 5
( x) = e x − ln 2 − 5
A1
N2
METHOD 1
evidence of composition in correct order e.g. ( g ◦ f ) ( x) = g (ln ( x + 5) + ln 2) = eln (2( x + 5)) = 2( x + 5) (g ◦ f ) ( x) = 2 x + 10
(M1)
A1A1
N2
METHOD 2
evidence of composition in correct order
(M1)
ln( x + 5) + ln 2 e.g. ( g ◦ f ) ( x) = e
= eln ( x + 5) × eln 2 = ( x + 5) 2 (g ◦ f ) ( x) = 2 x + 10
A1A1
N2 [7]
41.
(a)
f ( x) = 3( x
2
+ 2 x + 1) − 12
A1
= 3 x2 + 6 x + 3 − 12
A1
= 3 x2 + 6 x − 9
AG
IB Questionbank Maths SL
N0 22
(b)
(i)
vertex is (−1, −12)
(ii)
x = −1 (must
be an equation)
(iii) (0, − 9) (iv) evidence of solving f ( x) = 0 e.g. factorizing, formula, correct working e.g. 3( x + 3)( x − 1) = 0,
x =
A1A1
N2
A1
N1
A1 (M1)
N1
A1
− 6±
36 + 108 6
(−3, 0), (1, 0)
A1A1 N1N1
y
x
1
– 3
– 9
(c)
– 1 2
A1A1
N2
A1A1A1
N3
Notes: Award A1 for a parabola opening upward,
A1 for vertex and intercepts in approximately correct positions.
(d)
p − 1 = − 12 , q t = 3
(accept p = − 1, q = −12, t = 3)
[15]
3 42.
(a)
(i)
P( B) =
A1
N1
A1
N1
4
A1
N1
1 3 s = , t = 4 4
A1
N1
4 1
(ii) (b)
(c)
p =
(i)
P( R) =
4
3
P( X = 3) 1
= P (getting 1 and 2) = IB Questionbank Maths SL
4
3
×
4
A1 23
3
=
AG
16 1
(ii)
P( X = 2) =
1
×
4
4
or 1 − 3 16
3
+
4
N0
(A1)
13
=
16
A1
N2
A2 (M1)
N2
(d) (i) X
2
3
P( X = x)
13
3
16
16
(ii) evidence of using E( X ) = ∑ xP( X = x)
13 + 3 3 16 16 E( X ) = 2
= 2 3 16
35
= (e)
(A1)
16
A1
win $10 ⇒ scores 3 one time, 2 other time 13
N2
(M1)
3
P(3) × P(2) = 16 × 16 (seen anywhere) evidence of recognizing there are different ways of winning $10
A1 (M1)
13 × 3 , 16 16 e.g. P(3) × P(2) + P(2) × P(3), 2
36 256
+
3 256
+
36 256
+
3 256
= 39 128
78
P(win $10) =
256
A1
N3 [16]
43.
(a)
(b)
(i) (ii)
p = 65
for evidence of using sum is 125 (or 99 − p) q = 34
evidence of median position
A1 (M1) A1
N1 N2
(M1)
125 rd e.g. 63 student,
2
median is 17 (sit-ups) IB Questionbank Maths SL
A1
N2 24
∑ f ( x)
(c)
evidence of substituting into 125 15(11) + 16( 21) + 17( 33) + 18( 34) + 19( 18) + 20( 8) e. g .
125
mean = 17.4
(M1) ,
2176 125
A1
N2 [7]
44.
(a)
choosing sine rule
(M1) sin R
correct substitution sin R = 0.676148...
7
=
sin 75°
A1
10
ˆQ PR
= 42.5° (b) P = 180 − 75 − R P = 62.5 substitution into any correct formula 1 e.g. area ∆ PQR =
2
× 7 ×10 × sin
A1
N2
(A1) A1
(their P )
= 31.0 (cm2)
A1
N2 [6]
45.
(a)
(b)
(c)
evidence of appropriate approach 2π r e.g. 3π = 9 r =13.5 (cm) adding two radii plus 3π
M1
A1 (M1)
perimeter = 27+3π (cm) (= 36.4)
A1
evidence of appropriate approach 1 2π ×13.5 2 ×
M1
area = 20.25π (cm2) (= 63.6)
A1
e.g. 2
N1
N2
9
N1 [6]
46.
(a)
IB Questionbank Maths SL
25
y
15
10
5
–
π 3
π 3
0
x
– 5
A1A1A1
N3
Note: Award A1 for passing through (0, 0), A1 for correct shape, A1 for a range of approximately − 1 to 15.
(b)
evidence of attempt to solve f ( x) = 1 tan x = e.g. line on sketch, using
(M1)
sin x cos x
x = −0.207 x = 0.772
A1A1
N3 [6]
47.
(a)
evidence of binomial distribution (may be seen in parts (b) or (c)) e.g. np, 100 × 0.04 mean = 4
(M1) A1
N2
100 ( 0.04) 6 ( 0.96) 94 (b) P( X = 6) = 6 = 0.105
(A1) A1
N2
(c)
for evidence of appropriate approach e.g. complement, 1 − P( X = 0)
(M1)
P( X = 0) = (0.96)100 = 0.01687... P(X ≥ 1) = 0.983
(A1) A1
N2 [7]
48. pw = pi + 2 p j − 3 pk (seen anywhere)
attempt to find v + pw e.g. 3i + 4 j + k + p(i + 2 j − 3k ) collecting terms (3 + p)i + (4 + 2 p) j + (1 − 3 p) k attempt to find the dot product IB Questionbank Maths SL
(A1) (M1) A1 (M1) 26
e.g. 1(3 + p) + 2(4 + 2 p) − 3(1 − 3 p)
setting their dot product equal to 0 e.g. 1(3 + p) + 2(4 + 2 p) − 3(1 − 3 p) = 0 simplifying e.g. 3 + p + 8 + 4 p − 3 + 9 p = 0, 14 p + 8 = 0
(M1) A1
− 8 14 P = − 0.571
A1
N3 [7]
49.
(a)
(i)
evidence of approach →
e.g. AO
→
AB
→
→
+
OB
M1
=
AB,
B −A
− 4 6 = − 1
AG →
(ii)
→
N0
→
for choosing correct vectors, ( AO with AB , or OA
with
→
BA )
(A1)(A1) Using AO with BA will lead to ˆ π − 0.799. If they then say B AO = 0.799, this is a correct solution.
Note:
→ →
→
→
AO , AB
calculating AO • AB , e.g. d •d = (−1)(−4) + (2)(6) + (−3)(−1) (= 19) 1 2 d 1
= ( − 1) 2 + 2 2 + ( − 3) 2 ( =
14 ,
d 2
= ( − 4) 2 + 6 2 + ( − 1) 2 ( =
53
(A1)(A1)(A1)
)
)
evidence of using the formula to find the angle ( − 1) ( − 4) + ( 2) ( 6) + ( − 3) ( − 1) ( − 1) 2 + 2 2 + ( − 3) 2 ( − 4) 2 + 6 2 + ( − 1) 2 e.g. cos θ =
M1 ,
19 , 0.69751... 14 53 ˆO BA
(b)
= 0.799 radians (accept 45.8 °)
two correct answers e.g. (1, − 2, 3), (−3, 4, 2), (−7, 10, 1), (−11, 16, 0)
IB Questionbank Maths SL
A1
N3
A1A1 N2
27
(c)
(i)
1 − 3 − 2 + t 4 3 2 r =
(ii)
k 1 − 3 − k = − 2 + t 4 5 3 2 C on L2, so evidence of equating components e.g. 1 − 3t = k , − 2 + 4t = −k , 5 = 3 + 2 t one correct value t = 1, k = −2 (seen anywhere)
(d)
coordinates of C are ( −2, 2, 5) for setting up one (or more) correct equation using − 2 3 1 2 = − 8 + p − 2 5 0 − 1
A2
N2
(M1) (A1) (A1) A1
N3
(M1)
e.g. 3 + p = −2, −8 − 2 p = 2, − p = 5 p = − 5
A1
N2 [18]
50.
(a)
evidence of using ∑ f i = 100 k = 4
(b)
(i)
(M1) A1
evidence of median position
N2
(M1)
th e.g. 50 item, 26 + 10 + 20 = 56
median = 3 (ii)
A1
Q = and Q = 5 1 3
interquartile range = 4 (accept 1 to 5 or 5−1, etc.)
N2
(A1)(A1) A1
N3 [7]
51.
(a)
evidence of attempting to solve f ( x) = 0 evidence of correct working e.g.
( x +1) ( x − 2) ,
1± 9 2
intercepts are (−1, 0) and (2, 0) (accept x = −1, x = 2) (b)
evidence of appropriate method x + x 2 b x v = 1 , xv = − , 2 2a reference to symmetry e.g. x = 0.5 v
IB Questionbank Maths SL
(M1) A1
A1A1 N1N1 (M1)
A1
N2 28
[6]
52.
(a)
det M = − 4
1 − 1
(b)
−1 M =
− 4 − 2
1 1 − 1 4 4 = 1 1 2 − 2 2 −
Note: Award A1 for matrix.
(c)
−1 X = M
4 8
A1
N1
A1A1
N2
1 4 and A1 for the correct
− 1 − 1 4 X = − 1 8 − 2 2 4
M1
3 ( x = 3 , y = − 2) 2 − X =
A1A1
N0
Note: Award no marks for an algebraic solution of the system 2x + y = 4, 2x − y = 8. [6]
53.
(a)
evidence of choosing the formula cos2 A = 2 cos2 A − 1
(M1)
Note: If they choose another correct formula, do not award the M1 unless there is evidence 1 2 of finding sin A = 1− 9 .
correct substitution
A1
2
2
1 − 8 , cos 2 A = 2 × 1 − 1 9 3 e.g. cos 2 A = 3 cos 2 A = −
(b)
7
A1
9
N2
METHOD 1
evidence of using sin2 B + cos2 B = 1 2
e.g.
2 + cos 2 B =1, 3 ±
cos B =
IB Questionbank Maths SL
5 9
=±
(M1)
5 9
(seen anywhere),
5 3
(A1)
29
− cos B =
5 9
=−
5
3
A1
N2
METHOD 2
diagram
M1
e.g.
5
for finding third side equals cos B =
−
(A1)
5
A1
3
N2 [6]
54.
(a)
(i)
correct calculation 9 e.g. 20
+
5 20
−
2 20
(A1) 4 + 2 + 3+ 3
,
20 12
(ii)
P(male or tennis) = correct calculation 6 11 3 + 3 ÷ , e.g. 20
20
20
= 3 5
A1 (A1)
N2
A1
N2
11
6
P(not football | female) =
11
11
(b)
P(first not football) =
20
10
, P(second not football) =
20 110
P(neither football) =
380
A1
10
11
P(neither football) =
19
×
19
= 11 38
A1 A1
N1 [7]
55.
(a)
IB Questionbank Maths SL
30
y
4 3 2 1 – 2
0
– 1
1
3
2
4
x
– 1 – 2 – 3 – 4
M1A1
N2
Note: Award M1 for evidence of reflection in x-axis, A1 for correct vertex and all intercepts approximately correct.
(b)
(i)
g (−3) = f (0)
(A1)
f (0) = − 1.5
(ii)
− 3 0 etc translation (accept shift, slide, .) of
A1
N2
A1A1
N2 [6]
56.
(a)
(i)
evidence of combining vectors →
e.g. AB
→
AB
=
→
→
OB
OA
−
→
(or AD =
(M1) →
→
AO
OD
+
in part (ii))
2 − 4 = − 2
→
2 k − 5 − 2 =
(ii) (b) evidence of using perpendicularity ⇒ scalar product = 0 AD
2 2 e. g . − 4 • k − 5 = 0 − 2 − 2 4 − 4(k − 5) + 4 = 0 −4k + 28 = 0 (accept any correct equation clearly leading to k = 7) k = 7 IB Questionbank Maths SL
A1
N2
A1 (M1)
N1
A1 A1 AG
N0 31
(c)
→
AD
2 2 − 2 =
(A1)
1 1 → BC = − 1 evidence of correct approach →
→
→
OC = OB + BC , e.g.
→
OC
(d)
A1 (M1)
3 1 x − 3 1 1 + 1 , y −1 = 1 2 − 1 z − 2 − 1
4 2 1 =
A1
N3
METHOD 1 →
→
BA , BC
choosing appropriate vectors, finding the scalar product e.g. −2(1) + 4(1) + 2(−1), 2(1) + (−4)(1) + (−2)(−1) cos
ˆC AB
=0
(A1) M1 A1
N1
METHOD 2 →
→
BC
AD
parallel to
→
BC
→
⊥
ˆC AB
cos
(may show this on a diagram with points labelled) R1
AB
(may show this on a diagram with points labelled)
R1
= 90°
ˆC AB
=0
A1
N1 [13]
57.
(a)
evidence of using area of a triangle e.g.
A =
1 2
(M1)
× 2 × 2 × sinθ
A = 2 sin θ
IB Questionbank Maths SL
A1
N2
32
(b)
METHOD 1
ˆA PO
= π − θ 1
area ∆OPA =
2
(A1) 2 × 2 × sin ( π − θ )
(= 2 sin (π − θ ))
since sin (π − θ ) = sin θ then both triangles have the same area
A1 R1 AG
N0
R3 AG
N0
METHOD 2
triangle OPA has the same height and the same base as triangle OPB then both triangles have the same area 1
(c)
(d)
× π( 2 ) 2 ( = 2 π )
area semi-circle = 2 area ∆ APB = 2 sin θ + 2 sin θ
(= 4 sin θ )
A1 A1
S = area of semicircle − area ∆APB (= 2π − 4 sin θ )
M1
S = 2(π − 2 sin θ )
AG
N0
METHOD 1
attempt to differentiate dS e.g. dθ
(M1)
= − 4 cos θ
setting derivative equal to 0 correct equation e.g. −4 cos θ = 0, cos θ = 0, 4 cos θ = 0
(M1) A1
π θ
=
A1
2
N3
EITHER
evidence of using second derivative S ′′(θ ) = 4 sin θ
(M1) A1
π = 4 S ′′ 2
A1
π > 0 it is a minimum because S ′′ 2
R1
N0
OR
evidence of using first derivative IB Questionbank Maths SL
(M1) 33
π
, 2 S ′(θ ) < 0
for θ <
π
(may use diagram)
,
for θ > 2 S ′(θ ) > 0 (may use diagram) it is a minimum since the derivative goes from negative to positive
A1 A1 R1
N0
METHOD 2
2π − 4 sin θ is minimum when 4 sin θ is a maximum 4 sin θ is a maximum when sin θ = 1
R3 (A2)
π θ
(e)
=
A3
2
S is greatest when 4 sin θ
θ
is smallest (or equivalent)
= 0 (or π)
N3
(R1) A1
N2 [18]
58.
(a)
evidence of dividing two terms
−
e.g.
1800 3000
,
−
1800 1080
r = − 0.6
(b)
(M1)
A1
evidence of substituting into the formula for the 10th term
N2
(M1)
9 e.g. u = 3000(− 0.6) 10
u
(c)
10
= −30.2 (accept the exact value −30.233088)
evidence of substituting into the formula for the infinite sum e. g . S =
A1
N2
(M1)
3000 1.6
S = 1875
A1
N2 [6]
59.
evidence of using binomial expansion e.g. selecting correct term,
(M1)
8 7 8 6 2 a b + 2 a b + ... 1
a 8 b 0 +
evidence of calculating the factors, in any order
A1A1A1
8 2 3 , − 3 , x ( − 3) 5 3 5 3 e.g. 56, 3 23
5
−4032 x3 (accept = −4030 x 3 to 3 s.f.)
A1
N2 [5]
60.
(a)
intercepts when f ( x) = 0
IB Questionbank Maths SL
(M1) 34
(1.54, 0) (4.13, 0) (accept x = 1.54
x = 4.13)
A1A1
N3
A1A1A1
N3
A1
N1
(b) y
3 2 1 – 2 – 1 0 – 1
1
2
3
4
5
6
x
– 2 – 3 – 4 – 5 – 6 – 7 – 8 – 9 – 1 0
Note: Award A1 for passing through approximately (0, − 4), A1 for correct shape, A1 for a range of approximately − 9 to 2.3.
(c)
gradient is 2
[7]
61.
(a)
evidence of binomial distribution (seen anywhere)
(M1)
3, 1 4 e.g. X ~ B 3
mean =
(b)
4
( = 0.75)
3 1 2 3 2 4 4 P( X = 2) = P( X = 2) = 0.141
(c)
A1
= 9 64
evidence of appropriate approach e.g. complement, 1 − P( X = 0), adding probabilities P( X = 0) = (0.75)3
IB Questionbank Maths SL
= 0.422 , 27 64
N2
(A1) A1
N2
M1
(A1) 35
= 37 64
P( X ≥ 1) = 0.578
A1
N2 [7]
62.
evidence of equating vectors e.g. L = L 1 2 for any two correct equations e.g. 2 + s = 3 − t, 5 + 2 s = −3 + 3t , 3 + 3 s = 8 − 4t attempting to solve the equations finding one correct parameter ( s = −1, t = 2) the coordinates of T are (1, 3, 0)
(M1) A1A1 (M1) A1 A1
N3 [6]
63.
(a)
(i) 7 (ii) 1 (iii) 10
A1 A1 A1
N1 N1 N1
(b)
(i)
evidence of appropriate approach 18 − 2 A =
M1
A = 8
AG
N0
(ii)
C = 10
A2
N2
(iii)
METHOD 1
2
e.g.
period = 12 evidence of using B × period = 2π (accept 360°)
(A1) (M1)
2π e.g. 12 = B
B =
π 6
(accept 0.524 or 30)
A1
N3
METHOD 2
evidence of substituting e.g. 10 = 8 cos 3 B + 10 simplifying
(M1) (A1)
3 B = π 2 e.g. cos 3 B = 0 B =
π 6
IB Questionbank Maths SL
(accept 0.524 or 30)
A1
N3 36
(c)
correct answers e.g. t = 3.52, t = 10.5, between 03:31 and 10:29 (accept 10:30)
A1A1 N2 [11]
64.
(a)
(i)
n=5
(A1)
T = 280 × 1.12
5
T = 493
(ii)
A1 (A1)
evidence of doubling e.g. 560 setting up equation n
e.g. 280 × 1.12
A1
= 560, 1.12n = 2
n = 6.116...
in the year 2007 P =
(b)
(i)
2 560 000 10 + 90 e −0.1( 5)
P = 39 635.993... P = 39 636 P =
(ii)
not doubled valid reason for their answer e.g. P < 51200 (i)
correct value e. g .
(ii)
25600 280
N3
(A1) (A1) A1
N3
P T
A1 A1 R1
N0
A2
N2
, 91.4 , 640 : 7
setting up an inequality (accept an equation, or reversed inequality) e. g .
(A1) A1
2 560 000 10 + 90 e −0.1( 7 )
P = 46 806.997...
(c)
N2
< 70 ,
2 560 000
(10 + 90e −0.1 ) 280 ×1.12
finding the value 9.31.... after 10 years
n
n
M1
< 70 (A1) A1
N2 [17]
3 65.
(a)
4
IB Questionbank Maths SL
A1
N1
37
(b)
P( A ∪ B) = P( A) + P( B) − P( A ∩ B)
(M1)
P( A ∩ B) = P( A) + P( B) − P( A ∪ B) 2
=
5
3
7
4
8
A1
(0.275)
A1
+ −
11
=
40
P( A ∩ B) P( B)
(c)
P( A B) = 11
=
11 = 40 3 4
(0.367)
30
N2
A1 A1
N1 [6]
66.
(a)
(b)
(i)
m = 165
A1
(ii)
Lower quartile (1st quarter) = 160
(A1)
Upper quartile (3rd quarter) = 170 IQR = 10
(A1) A1
Recognize the need to use the 40th percentile, or 48th student eg a horizontal line through (0, 48) a = 163
N1
N3
(M1) A1
N2 [6]
67.
(a)
loga 10 = loga (5 × 2)
(M1)
= loga 5 + loga 2
(b)
= p + q
A1
loga 8 = loga 23
(M1)
N2
= 3 loga 2 = 3q
A1
N2
5
(c)
loga 2.5 = loga
2
(M1)
= loga 5 − loga 2 = p −
q
A1
N2 [6]
IB Questionbank Maths SL
38
Note:
68.
(a)
Throughout this question, do not accept methods which involve finding θ .
Evidence of correct approach BC eg sin θ = AB
, BC = 3 2 − 2 2
A1
=
5
5
sin θ =
(b)
AG
3
Evidence of using sin 2θ = 2 sin θ cos θ 5 2 3 3 =
N0
(M1)
2
A1
4 5
= (c)
AG
9
Evidence of using an appropriate formula for cos 2θ 4 eg
9
−
5 9
cos 2θ =
, 2×
−
4 9
− 1, 1 − 2 ×
5 9
,
N0
M1
1 − 80 81
1
A2
9
N2 [6]
69.
(a)
f −1 ( x ) = ln x
(b)
(i)
A1
Attempt to form composite ( f ◦ g ) ( x) = f (ln (1 + 2 x)) ( f ◦ g ) ( x) = eln (1 + 2 x) = (= 1 + 2 x)
(ii)
Simplifying y = eIn(1 + 2 x) to y = 1 + 2 x (may be seen in part (i) or later) Interchanging x and y (may happen any time) eg x = 1 + 2 y x − 1 = 2 y
N1
(M1) A1
N2
(A1) M1
x − 1
( f ◦ g )−1 ( x) =
2
A1
N2 [6]
70.
(a)
(i) (ii)
0 −
A1
N1
A1
N1
1 2
IB Questionbank Maths SL
39
y
5 4 3 2 1 – 5
–4
–3
–2
0
–1
1
– 1
3
2
4
5
x
– 2 – 3 – 4
b)
– 5
A2
N2
A2
N2
(c) y
5 4 3 2 1 – 5
–4
–3
–2
0
–1 – 1
1
2
3
4
5
x
– 2 – 3 – 4 – 5
[6]
Notes:Candidates may have differing answers due to using approximate answers from previous parts or using answers from the GDC. Some leeway is provided to accommodate this.
71.
(a)
METHOD 1
Evidence of using the cosine rule a2
+ b2 − c2
eg cos C =
2ab
, a2
(M1)
= b 2 + c 2 − 2bc cos A
Correct substitution 32 + 2 2 − 4 2
ˆ eg cos AOP
cos
=
2 × 3× 2
ˆP , 4 2 = 3 2 + 2 2 − 2 × 3 × 2 cos AO
ˆP AO
= −0.25
ˆP AO
= 26π 45 (radians) = 1.82
IB Questionbank Maths SL
A1
A1
N2 40
METHOD 2
Area of AOBP = 5.81 (from part (d)) Area of triangle AOP = 2.905
(M1)
2.9050 = 0.5 × 2 × 3 × sin AOˆP
(b)
A1
ˆP AO
= 1.32 or 1.82
ˆP AO
= 26π 45 (radians) = 1.82
ˆB AO
= 2(π − 1.82)
A1
(= 2π − 3.64)
(A1)
= 38π 45 (radians) = 2.64 (c)
(i)
Appropriate method of finding area 1 eg area = 2
A1
Area of sector PAEB =
(M1)
2
× 4 2 ×1.63
= 13.0 (cm2) (accept the exact value 13.04) 1
Area of sector OADB =
2
× 3 2 × 2.64
= 11.9 (cm2) (d)
(i) (ii)
N2
θr 2 1
(ii)
N2
Area AOBE = Area PAEB − Area AOBP (= 13.0 − 5.81) = 7.19 (accept 7.23 from the exact answer for PAEB) Area shaded = Area OADB − Area AOBE (= 11.9 − 7.19) = 4.71 (accept answers between 4.63 and 4.72)
A1 A1
N2
A1 A1 M1 A1 M1 A1
N1
N1 N1 [14]
IB Questionbank Maths SL
41
S e c o n d d ie in p a ir F ir s t d i e in p a ir
1 6
four
fo u r 1 6
5 6 1 6
5 6
72.
not four four
not fo u r 5 6
(a)
not four
A1A1A1
N3
Note: Award A1 for each pair of complementary probabilities.
(b)
P( E ) =
1 5
5 1
6 6
6 6
× + ×
10
=
(c)
36
= 5 + 5 36 36
= 5 or 0.278 18
Evidence of recognizing the binomial distribution
eg X ~
B 5 ,
A1
N3
(M1)
or p = 5 , q = 13 18 18 18 5
5 5 3 13 2 3 18 18 (or other evidence of correct setup) P( X = 3) = = 0.112 (d)
(A2)
(A1) A1
N3
METHOD 1
Evidence of using the complement eg P( X ≥ 3) = 1 − P( X ≤ 2) Correct value 1 − 0.865 = 0.135
M1 (A1) A1
N2
METHOD 2
Evidence of adding correct probabilities eg P( X ≥ 3) = P( X = 3) + P( X = 4) + P( X = 5) IB Questionbank Maths SL
M1 42
Correct values 0.1118 + 0.02150 + 0.001654 = 0.135
(A1) A1
N2 [12]
73.
(a)
7 terms
A1
(b)
A valid approach
N1
(M1)
6 3 3 ( x ) ( − 3 x ) 3 Correct term chosen 3
A1
6 = 20 , ( − 3) 3 = − 27 Calculating 3
(A1)(A1)
Term is −540 x12
A1
N3 [6]
74.
(a)
Two correct factors eg y
2
A1A1
+ y − 12 = ( y + 4)( y − 3), (2 x)2 + (2 x) − 12 = (2 x + 4)(2 x − 3)
a = 4, b = −3 (or a = −3, b = 4)
(b)
N2
2 x − 3 = 0
(M1)
2 x = 3 ln 3 x =
ln 2
log 2 3 , log 3 etc. log 2
A1
N2
A1 R1
N1
EITHER
Considering 2 x + 4 = 0 (2 x = −4) (may be seen earlier) Valid reason x
eg this equation has no real solution, 2 x-axis
> 0, graph does not cross the
OR
Considering graph of y = 22 x + 2 x − 12 (asymptote does not need to be indicated)
A1
There is only one point of intersection of the graph with x-axis.
R1
N1 [6]
IB Questionbank Maths SL
43
Note: In this question, accept any correct vector notation, including row vectors eg (1, − 2, 3).
75.
(a)
PQ
(i)
OQ
OP
(M1)
= i − 2 j + 3k
A1
=
−
N2
→
PQ r = OP + s = −5i + 11 j −8k + s(i − 2 j + 3k )
(ii)
(M1) A1
= (−5 + s) i + (11 − 2 s) j + (−8 + 3 s) k
(b)
If (2, y1, z 1) lies on L1 then −5 + s = 2
(M1)
s = 7
A1 A1A1
y
(c)
AG
1
= −3, z 1 = 13
Evidence of correct approach eg (−5 + s)i + (11 − 2 s) j + (−8 + 3 s) k = 2i + 9 j +13k + t (i +2 j + 3k ) At least two correct equations eg −5 + s = 2 + t , 11 − 2 s = 9 + 2t , −8 + 3 s = 13 + 3t Attempting to solve their equations One correct parameter ( s = 4, t = −3)
N0
N3
(M1) A1A1 (M1) A1
→
OT
(d)
= −i + 3 j + 4k
A2
Direction vector for L1 is d 1 = i − 2 j + 3k Note:
d1
cos θ = θ
6 14 14
= 64.6°
(A1)
Award A1FT for their vector from (a)(i).
Direction vector for L2 is d 2 = i − 2 j + 3k d • d = 6, 1 2
N4
14 ,
=
d2 =
14 ,
= 6 = 3 14 7 (= 1.13radians)
(A1) (A1)(A1)(A1) A1 A1
N4
Note: Award marks as per the markscheme if their (correct) direction vectors give d • d = − 6, leading to θ = 115° 1
2
(= 2.01 radians). IB Questionbank Maths SL
44
[22]
76.
(a)
METHOD 1
f (3) =
7
(A1) A1
( g ◦ f ) (3) = 7
N2
METHOD 2
x + 4
2
( g ◦ f ) ( x) = (= x + 4) ( g ◦ f ) (3) = 7 (b) For interchanging x and y (seen anywhere) Evidence of correct manipulation x=
eg
(c)
y + 4 , x 2
(A1) A1 (M1) A1
N2
= y+4
−1 2 f ( x) = x − 4
A1
N2
x ≥ 0
A1
N1 [6]
46 77.
(a)
97
13
(b)
51
59
(c)
97
( = 0.474)
( = 0.255)
( = 0.608)
A1A1
N2
A1A1
N2
A2
N2 [6]
78.
(a)
u • v = 8 + 3 + p
For equating scalar product equal to zero 8 + 3 + p = 0 p = −11
(b)
u
=
2 2 + 3 2 + ( − 1)
q 14 =14
IB Questionbank Maths SL
2
=
14 , 3.74
(A1) (M1) A1
N3
(M1) A1
45
q=
( = 3.74)
14
A1
N2 [6]
79.
(a)
For using perimeter = r + r + arc length 20 = 2r + r θ
θ=
20 − 2r
AG
r 1
(b)
(M1) A1
20 − 2r ( = 10r − r 2 ) r
r 2
Finding A = 2 For setting up equation in r Correct simplified equation, or sketch
(A1) M1
2 2 eg 10r – r = 25, r – 10r + 25 = 0
(A1) A1
r = 5 cm
N0
N2 [6]
80.
(a)
speed = =
(b)
3 2 + 4 2 + 10 2
125
=
(M1)
5 5 , 11.2, (metres per minute)
a b c Let the velocity vector be Finding a velocity vector 3 − 5 a 3 16 10 b 16 39 23 c 39 eg = + 2 , −
(i)
− 5 10 23
3 3 − 5 4 2 4 10 3 7 10 23 8 At Q, + t = + t Setting up one correct equation eg 3 + 3t = − 5 + 4t , 2 + 4 t = 10 + 3t , 7 + 10t = 23 + 8t
IB Questionbank Maths SL
N2
A2
4 3 Dividing by 2 to give 8 x − 5 4 y 10 3 z 23 8 = + t
(c)
A1
A1
AG
N0
(M1) A1
46
t = 8
(ii)
Correct answer eg after 8 minutes, 13:08 Substituting for t x 3 3 x y 2 4 y z 7 10 z = + 8 , or =
(A1) A1 N3 (M1)
− 5 4 10 3 23 8 + 8
27 34 87 x = 27, y = 34, z = 87 or (27, 34, 87), or
(d)
3 4 10 For choosing both direction vectors d 1 = and d 2 = d • d = 104, 1 2
d1
=
125 ,
d2 =
89
A1
4 3 8
N2
(A1) (A1)(A1)(A1)
104
cos θ = 125 89 = 0.98601... θ = 0.167 (radians) (accept θ = 9.59°)
A1 A1
N3 [17]
81.
(a)
METHOD 1
Using the discriminant = 0 (q2 − 4(4)(25) = 0) 2
= 400 q = 20, q = −20 q
M1 A1A1
N2
METHOD 2
Using factorizing: (2 x − 5)(2 x − 5) and/or (2 x + 5) (2 x + 5) q = 20, q = −20
M1 A1A1
N2
(b)
x = 2.5
A1
N1
(c)
(0, 25)
A1A1
N2 [6]
82.
(a)
IB Questionbank Maths SL
47
y
6 5
2
4 3 2
1 A (– 1, 1 )
– 5
– 4
– 3
– 2
– 1
3
1 0
1
2
3
4
5
x
– 1 – 2 – 3
Notes:
A1A1A1
N3
A1
N1
A1A1
N2
Award A1 for left end point in circle 1,
A1 for maximum point in circle 2, A1 for right end point in circle 3.
(b)
y = 1 (must be an equation)
(c)
(0, 3)
[6]
83.
(a)
253250
(accept 253000)
(b)
1972 → 2002 is 30 years, increase of 1.3% → 1.013 Evidence of any appropriate approach Correct substitution 250000 × 1.01330 368000 (accept 368318)
A1
N1
(A1)(A1) (M1) A1 A1
N3 [6]
84.
Identifying the required term (seen anywhere)
M1
10 2 ×2 8 eg 10 8
= 45
4 y2, 2 × 2, 4 a = 180
(A1) (A2) A2
N4 [6]
IB Questionbank Maths SL
48
85.
(a)
(b)
Attempting to multiply matrices 3 1+ x 2 1 x − 1 3 + x 2 − 2 x = 9 + x + 8 = 17 + x 3 1 4 2
(M1)
A1A1
Setting up equation
N3
M1
1+ x 2 20 2 + 2 x 2 20 1+ x 2 10 = 2 , 34 + 2 x , 17 + x 17 + x = 28 = 14 28 eg 2 + 2 x 2
= 20 34 + 2 x = 28
1 + x 2 =10 17 + x =14
(A1) A1
x = −3
N2 6]
86.
(a)
A = 18, B = 19, C = 23, D = 31, E = 36
(b)
IQR = 12
A1A1A1A1A1
N5
A1
N1 [6]
87.
(a)
Evidence of choosing cosine rule
(M1)
= b2 + c2 − 2bc cos A Correct substitution eg a
2
eg (AD)
2
A1
= 7.12 + 9.22 − 2(7.1) (9.2) cos 60°
(AD)2 = 69.73 AD = 8.35 (cm) (b)
(A1) A1
180° − 162° = 18° Evidence of choosing sine rule Correct substitution
N2
(A1) (M1) A1
8.35
DE
= sin 110° DE = 2.75 (cm)
eg sin 18°
(c)
A1
Setting up equation
N2
(M1) 1
1
eg 2 ab sin C = 5.68, 2 bh = 5.68
Correct substitution 1
A1 1
ˆ eg 5.68 = 2 (3.2) (7.1) sin DBC , 2 × 3.2 × h = 5.68, (h = 3.55)
sin
ˆC DB
= 0.5
IB Questionbank Maths SL
(A1) 49
ˆC DB
30° and/or 150°
A1
ˆ ˆ (d) Finding A B C (60° + D B C) Using appropriate formula
(A1) (M1)
N2
= (AB)2 + (BC)2, (AC)2 = (AB)2 + (BC)2 − 2 (AB) (BC) cos ABC eg (AC)
2
Correct substitution (allow FT on their seen
ˆC AB )
= 9.22 + 3.22 AC = 9.74 (cm)
eg (AC)
(e)
2
A1 A1
For finding area of triangle ABD
N3
(M1)
1
Correct substitution Area = 2 × 9.2 × 7.1 sin 60° = 28.28... Area of ABCD = 28.28... + 5.68 = 34.0 (cm2)
A1 A1 (M1) A1
N3 [21]
88.
(a)
P( F ∪ S ) = 1 − 0.14 (= 0.86) Choosing an appropriate formula eg P( A ∪ B) = P( A) + P( B) − P( A ∩ B) Correct substitution eg P( F ∩ S ) = 0.93 − 0.86 P( F ∩ S ) = 0.07
(A1) (M1)
A1 AG
N0
Notes: There are several valid approaches. Award (A1)(M1)A1 for relevant working using any appropriate strategy eg formula, Venn Diagram, or table.
Award no marks for the incorrect solution
P( F ∩ S ) = 1 − P( F ) + P(S ) = 1 − 0.93 = 0.07 (b)
Using conditional probability
(M1)
P ( F ∩ S ) = P ( ) S eg P( F | S ) 0.07
P( F | S ) = 0.62 = 0.113 (c)
F and S are not independent
(A1) A1
N3
A1
N1
EITHER
IB Questionbank Maths SL
50
