IB Math – Standard Level Year 2: May ‘ 10, Paper 1, TZ 2
Alei - Desert Academy 2011-12
IB Practice Exam: 10 Paper 1 Zone 2 – 90 min, Calculator Allowed
Name: __________________________________________Date: _________________Class: _________
1.
Let f(x) = p(x – q)(x – r). Part of the graph of f is shown below.
The graph passes through the points (–2, 0), (0, –4) and (4, 0).
(a) Write down the value of q and of r.
(2)
(b)
Write down the equation of the axis of symmetry.
(c)
Find the value of p.
(1)
(3)
(Total 6 marks)
2.
 6 
 − 2
Let AB =  − 2  and AC =  − 3  .
 2 
 3 
 
 
(a)
Find BC.
(2)
(b)
Find a unit vector in the direction of AB .
(3)
(c)
Show that AB is perpendicular to AB .
(3)
(Total 8 marks)
3.
1 − 2
 and B =
Let A = 
3 4 
(a) Find AB.
 − 5
  .
 5 
(3)
(b)
–1
Solve A X = B.
(2)
(Total 5 marks)
4.
Let f(x) = cos 2x and g(x) = 2x2 – 1.
π
(a) Find f   .
2
(2)
(b)
π
Find (g ° f)   .
2
(c)
Given that (g ° f)(x) can be written as cos (kx), find the value of k, k ∈
(2)
.
(3)
(Total 7 marks)
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IB Math – Standard Level Year 2: May ‘ 10, Paper 1, TZ 2
Alei - Desert Academy 2011-12
5.
Let f(x) = kx4. The point P(1, k) lies on the curve of f. At P, the normal to the curve is parallel to y =
1
− x . Find the value of k.
8
6.
Solve log2x + log2(x – 2) = 3, for x > 2.
7.
A function f is defined for –4 ≤ x ≤ 3. The graph of f is given below.
(Total 6 marks)
(Total 7 marks)
The graph has a local maximum when x = 0, and local minima when x = –3, x = 2.
(a) Write down the x-intercepts of the graph of the derivative function, f′.
(b)
Write down all values of x for which f′(x) is positive.
(c)
At point D on the graph of f, the x-coordinate is –0.5. Explain why f′′(x) < 0 at D.
(2)
(2)
(2)
(Total 6 marks)
8.
Consider the function f with second derivative f′′(x) = 3x – 1. The graph of f has a minimum point at
 4 358 
A(2, 4) and a maximum point at B  − ,
.
 3 27 
(a) Use the second derivative to justify that B is a maximum.
(3)
(b)
3
Given that f′ = x 2 – x + p, show that p = –4.
2
(c)
Find f(x).
(4)
(7)
(Total 14 marks)
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IB Math – Standard Level Year 2: May ‘ 10, Paper 1, TZ 2
9.
Alei - Desert Academy 2011-12
José travels to school on a bus. On any day, the probability that José will miss the bus is
If he misses his bus, the probability that he will be late for school is
1
.
3
7
.
8
3
.
8
Let E be the event “he misses his bus” and F the event “he is late for school”.
The information above is shown on the following tree diagram.
If he does not miss his bus, the probability that he will be late is
(a)
Find
(i)
P(E ∩ F);
(ii) P(F).
(b)
Find the probability that
(i)
José misses his bus and is not late for school;
(ii) José missed his bus, given that he is late for school.
(4)
(5)
The cost for each day that José catches the bus is 3 euros. José goes to school on Monday and
Tuesday.
(c) Copy and complete the probability distribution table.
X (cost in euros)
P (X)
0
1
9
3
6
(3)
(d)
Find the expected cost for José for both days.
(2)
(Total 14 marks)
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IB Math – Standard Level Year 2: May ‘ 10, Paper 1, TZ 2
10.
Alei - Desert Academy 2011-12
Let f(x) = 6 + 6sinx. Part of the graph of f is shown below.
The shaded region is enclosed by the curve of f, the x-axis, and the y-axis.
(a) Solve for 0 ≤ x < 2π.
(i)
6 + 6sin x = 6;
(ii) 6 + 6 sin x = 0.
(5)
(b)
Write down the exact value of the x-intercept of f, for 0 ≤ x < 2.
(c)
The area of the shaded region is k. Find the value of k, giving your answer in terms of π.
(1)
(6)
π

Let g(x) = 6 + 6sin  x −  . The graph of f is transformed to the graph of g.
2

(d) Give a full geometric description of this transformation.
(2)
(e)
Given that
∫
p+
p
3π
2
g ( x)dx = k and 0 ≤ p < 2π, write down the two values of p.
(3)
(Total 17 marks)
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IB Math – Standard Level Year 2: May ’10 Paper 1, TZ 2: MarkScheme
Alei - Desert Academy 2011-12
IB Practice Exam: 10 Paper 1 Zone 2 – MarkScheme
1.
(a)
(b)
(c)
q = –2, r = 4 or q = 4, r = –2
x = 1 (must be an equation)
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 = –8p
4 1
p = = 
8  2
A1A1
N2
A1N1
(M1)
(A1)
A1N2
[6]
2.
(a)
evidence of appropriate approach
 − 2  6 
   
e.g. BC = BA + AC,  − 3  −  − 2 
 2   3 
   
(M1)
 − 8
 
BC =  − 1 
 −1
 
(b)
A1N2
attempt to find the length of AB
AB = 6 + (−2) + 3
2
(c)
2
2
(M1)
(= 36 + 4 + 9 = 49 = 7 )
(A1)
  6 
 

 6 
  7 
1 
 =  − 2 
unit vector is  − 2 
  7 
7 
  3 
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
A1N2
(M1)
A1
A1
e.g. AB • AC = 0, cos θ = 0
therefore, they are perpendicular
AGN0
[8]
3.
(a)
(b)
evidence of multiplying
e.g. one correct element
 − 15 

AB = 
 5 
METHOD 1
evidence of multiplying by A (on left or right)
e.g. AA–1 X = AB, X = AB
 − 15 
 (accept x = – 15, y = 5)
X = 
 5 
METHOD 2
attempt to set up a system of equations
4x + 2 y
− 3x + y
e.g.
= −5,
=5
10
10
(M1)
A1A1
N3
(M1)
A1N2
(M1)
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IB Math – Standard Level Year 2: May ’10 Paper 1, TZ 2: MarkScheme
Alei - Desert Academy 2011-12
 − 15 
 (accept x = – 15, y = 5)
X = 
 5 
A1N2
[5]
4.
(a)
(b)
(c)
π
f   = cosπ
2
= –1
π
(g ° f)   = g(–1) (= 2(–1)2 – 1)
2
=1
(g ° f)(x) = 2(cos (2x))2 – 1 (= 2 cos2(2x) – 1)
(A1)
A1N2
(A1)
A1N2
A1
evidence of 2 cos2 θ – 1 = cos 2θ (seen anywhere)
(g ° f)(x) = cos 4x
k=4
(M1)
A1N2
[7]
5.
gradient of tangent = 8 (seen anywhere)
f′(x) = 4kx3 (seen anywhere)
recognizing the gradient of the tangent is the derivative
setting the derivative equal to 8
e.g. 4kx3 = 8, kx3 = 2
substituting x = 1 (seen anywhere)
k=2
(A1)
A1
(M1)
(A1)
recognizing log a + log b = log ab (seen anywhere)
e.g. log2(x(x – 2)), x2 – 2x
recognizing loga b = x ⇔ ax = b (seen anywhere)
e.g. 23 = 8
correct simplification
e.g. x(x – 2) = 23, x2 – 2x – 8
evidence of correct approach to solve
e.g. factorizing, quadratic formula
correct working
2 ± 36
e.g. (x – 4)(x + 2),
2
x=4
(A1)
(M1)
A1N4
[6]
6.
(A1)
A1
(M1)
A1
A2N3
[7]
7.
(a)
(b)
x-intercepts at –3, 0, 2
–3 < x < 0, 2 < x < 3
(c)
correct reasoning
e.g. the graph of f is concave-down (accept convex), the first
derivative is decreasing
therefore the second derivative is negative
A2
N2
A1A1
N2
R2
AG
[6]
8.
(a)
substituting into the second derivative
 4
e.g. 3 ×  −  − 1
 3
 4
f″  −  = –5
 3
M1
A1
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IB Math – Standard Level Year 2: May ’10 Paper 1, TZ 2: MarkScheme
(b)
Alei - Desert Academy 2011-12
since the second derivative is negative, B is a maximum
setting f′(x) equal to zero
4

evidence of substituting x = 2  or x = − 
3

e.g. f′(2)
correct substitution
R1
N0
(M1)
(M1)
A1
2
3
3  4  4
( 2) 2 − 2 + p,  −  −  −  + p
2
2  3  3
correct simplification
8 4
e.g. 6 – 2 + p = 0, + + p = 0, 4 + p = 0
3 3
p = –4
evidence of integration
1
1
f(x) = x 3 − x 2 − 4 x + c
2
2
e.g.
(c)
A1
AGN0
(M1)
A1A1A
1
 4 358 
substituting (2, 4) or  − ,
 into their expression
 3 27 
correct equation
1
1
1
1
e.g. × 2 3 − × 2 2 − 4 × 2 + c = 4, × 8 − × 4 – 4 × 2 + c = 4, 4 − 2 − 8 + c = 4
2
2
2
2
1 3 1 2
f ( x) = x − x − 4 x + 10
2
2
(M1)
A1
A1N4
[14]
9.
(a)
(i)
(ii)
(b)
(i)
(ii)
7
24
evidence of multiplying along the branches
2 5 1 7
e.g. × , ×
3 8 3 8
adding probabilities of two mutually exclusive paths
1 7  2 3 1 1  2 5
e.g.  ×  +  × ,  ×  +  × 
3 8  3 8 3 8  3 8
13
P(F) =
24
1 1
×
3 8
1
24
recognizing this is P(E│F)
7 13
e.g.
÷
24 24
168  7 
= 
312  13 
A1
N1
(M1)
(M1)
A1N2
(A1)
A1
(M1)
A2N3
(c)
X (cost in euros)
0
3
6
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IB Math – Standard Level Year 2: May ’10 Paper 1, TZ 2: MarkScheme
1
9
P (X)
(d)
4
9
Alei - Desert Academy 2011-12
4
9
A2A1
N3
(M1)
correct substitution into E(X) formula
1
4
4 12 24
e.g. 0 × + 3 × + 6 × , +
9
9
9 9
9
E(X) = 4 (euros)
A1N2
[14]
10.
(a)
(i)
sin x = 0
x = 0, x = π
sin x = –1
3π
x=
2
(ii)
(b)
(c)
3π
2
evidence of using anti-differentiation
e.g.
3π
2
0
∫
(e)
N2
A1
A1N1
A1N1
(M1)
(6 + 6 sin x)dx
correct integral 6x – 6 cos x (seen anywhere)
correct substitution
 3π 
 3π 
e.g. 6  − 6cos  − (−6 cos 0), 9π – 0 + 6
 2 
 2 
k = 9π + 6
(d)
A1
A1A1
π
translation of  2 
 
0
recognizing that the area under g is the same as the shaded region in f
π
p= ,p=0
2
A1A1
(A1)
A1A1
N3
A1A1
N2
(M1)
A1A1
N3
[17]
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