EOY not calculus no calculator
[48 marks]
Let X be normally distributed with mean 100 cm and standard deviation 5 cm.
1a. On the diagram below, shade the region representing P(X > 105) .
[2 marks]
Markscheme
A1A1
N2
Note: Award A1 for vertical line to right of mean, A1 for shading to right of their vertical line.
1b. Given that P(X < d) = P(X > 105) , find the value of d .
[2 marks]
Markscheme
evidence of recognizing symmetry
(M1)
e.g. 105 is one standard deviation above the mean so d is one standard deviation below the mean, shading the corresponding part,
105 − 100 = 100 − d
d = 95
A1
N2
[2 marks]
1c. Given that P(X > 105) = 0.16 (correct to two significant figures), find P(d < X < 105) .
Markscheme
evidence of using complement
(M1)
e.g. 1 − 0.32 , 1 − p
P(d < X < 105) = 0.68
[2 marks]
f(x) =
1
3
−
2
− 3x
A1
N2
[2 marks]
Let f(x) = 12 x3 − x2 − 3x . Part of the graph of f is shown below.
There is a maximum point at A and a minimum point at B(3, − 9) .
2.
Write down the coordinates of
(i)
[6 marks]
the image of B after reflection in the y-axis;
(ii) the image of B after translation by the vector (
−2
);
5
(iii) the image of B after reflection in the x-axis followed by a horizontal stretch with scale factor 12 .
Markscheme
(i) (−3, − 9)
A1
(ii) (1, − 4)
A1A1
N1
N2
(iii) reflection gives (3, 9)
stretch gives ( 32 , 9)
(A1)
A1A1
N3
[6 marks]
Let f(x) = Aekx + 3 . Part of the graph of f is shown below.
The y-intercept is at (0, 13) .
3.
Show that A = 10 .
[2 marks]
Markscheme
substituting (0, 13) into function
e.g. 13 =
Ae0
13 = A + 3
A = 10
AG
M1
+3
A1
N0
[2 marks]
The diagram below shows part of the graph of f(x) = acos(b(x − c)) − 1 , where a > 0 .
The point P ( π4 ,2) is a maximum point and the point Q ( 3π
,−4) is a minimum point.
4
4a. Find the value of a .
[2 marks]
Markscheme
evidence of valid approach
e.g.
max y value−min y value
2
a=3
A1
(M1)
, distance from y = −1
N2
[2 marks]
4b. (i)
Show that the period of f is π .
(ii) Hence, find the value of b .
[4 marks]
Markscheme
(i) evidence of valid approach
(M1)
e.g. finding difference in x-coordinates,
evidence of doubling
π
2
A1
e.g. 2 × ( π2 )
period = π
AG
N0
(ii) evidence of valid approach
e.g. b =
2π
π
b=2
A1
(M1)
N2
[4 marks]
4c. Given that 0 < c < π , write down the value of c .
[1 mark]
Markscheme
c = π4
A1
N1
[1 mark]
The diagram below shows a circle with centre O and radius 8 cm.
The points A, B, C, D, E and F are on the circle, and [AF] is a diameter. The length of arc ABC is 6 cm.
5.
Find the size of angle AOC in radians .
Markscheme
appropriate approach
(M1)
e.g. 6 = 8θ
AÔC = 0.75
A1
[2 marks]
f(x) = ln(
4
+ 1)
N2
[2 marks]
Consider f(x) = ln(x4 + 1) .
6.
Find the value of f(0) .
[2 marks]
Markscheme
substitute 0 into f
(M1)
eg ln(0 + 1) , ln 1
f(0) = 0
A1 N2
[2 marks]
Let f(x) = 2x − 1 and g(x) = 3x2 + 2 .
7a. Find f −1 (x) .
[3 marks]
Markscheme
interchanging x and y (seen anywhere)
(M1)
e.g. x = 2y − 1
correct manipulation
(A1)
e.g. x + 1 = 2y
f −1 (x) = x+1
2
A1
N2
[3 marks]
7b. Find (f ∘ g)(1) .
[3 marks]
Markscheme
METHOD 1
attempt to find or g(1) or f(1)
g(1) = 5
(A1)
f(5) = 9
A1
(M1)
N2
[3 marks]
METHOD 2
attempt to form composite (in any order)
(M1)
e.g. 2(3x2 + 2) − 1 , 3(2x − 1)2 + 2
(f ∘ g)(1) = 2(3 × 12 + 2) − 1 (= 6 × 12 + 3)
(f ∘ g)(1) = 9
[3 marks]
A1
N2
(A1)
A Ferris wheel with diameter 122 metres rotates clockwise at a constant speed. The wheel completes 2.4 rotations every hour. The bottom of
the wheel is 13 metres above the ground.
A seat starts at the bottom of the wheel.
8a. Find the maximum height above the ground of the seat.
[2 marks]
Markscheme
valid approach
(M1)
eg 13 + diameter , 13 + 122
maximum height = 135 (m)
A1 N2
[2 marks]
After t minutes, the height h metres above the ground of the seat is given by
h = 74 + acos bt.
8b. (b)
(i)
Show that the period of h is 25 minutes.
(ii)
Write down the exact value of b .
(c)
Find the value of a .
(d)
Sketch the graph of h , for 0 ≤ t ≤ 50 .
[9 marks]
Markscheme
(a)
(i)
60
period = 2.4
period = 25 minutes
(ii)
b=
2π
25
A1
AG
(= 0.08π)
N0
A1
N1
[2 marks]
(b)
METHOD 1
valid approach
(M1)
eg max − 74 , |a| = 135−13
, 74 − 13
2
|a| = 61 (accept a = 61 )
a = −61
A1
(A1)
N2
METHOD 2
attempt to substitute valid point into equation for h
(M1)
eg 135 = 74 + a cos ( 2π×12.5
)
25
correct equation
(A1)
eg 135 = 74 + a cos(π) , 13 = 74 + a
a = −61
A1
N2
[3 marks]
(c)
A1A1A1A1
N4
Note: Award A1 for approximately correct domain, A1 for approximately correct range,
A1 for approximately correct sinusoidal shape with 2 cycles.
Only if this last A1 awarded, award A1 for max/min in approximately correct positions.
[4 marks]
Total [9 marks]
The diagram below shows part of the graph of a function f .
The graph has a maximum at A(1, 5) and a minimum at B(3, −1) .
The function f can be written in the form f(x) = p sin(qx) + r . Find the value of
9.
(a)
p
(b)
q
(c)
r.
[6 marks]
Markscheme
(a)
valid approach to find p
eg amplitude =
p=3
A1
max−min
2
(M1)
,p=6
N2
[2 marks]
(b)
valid approach to find q
eg period = 4 , q =
q = π2
A1
(M1)
2π
period
N2
[2 marks]
(c)
valid approach to find r
eg axis =
r=2
max+min
2
A1
(M1)
, sketch of horizontal axis, f(0)
N2
[2 marks]
Total [6 marks]
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