Question 1
The table below shows values x_1, x_2, x_3, x_4 and x_5 for the equation x_{n+1}=\sqrt{(4+x_n)}.
1(a):
Use the table above to estimate a solution to x_{n+1}=\sqrt{(4+x_n )} , to 1 decimal place.
ANSWER: Simple Text Answer
Answer: 2.6
Workings:
The value for x_{n+1} increases towards 2.6 by a smaller amount each iteration.
Marks = 1
1(b):
Which pair of statements is true?
Statement 1: x_5 third decimal place is below 5 so answer rounded down to 2.5.
Statement 2: x_5 second decimal place is above 5 so answer rounded up to 2.6.
Statement 3: The solutions are approaching 2.6 with the increases getting smaller.
Statement 4: The solutions are approaching 2.5 with decreases getting smaller.
ANSWER: Multiple Choice (Type 1)
A: Statement 1 & Statement 3
B: Statement 1 & Statement 4
C: Statement 2 & Statement 3
D: Statement 2 & Statement 4
Answer: C
Workings:
Always round up when the following figure is 5 or more. The iterations keep increasing by smaller amounts so 2.60 won’t be reached.
Marks = 1
Question 2
Using x_{n+1}=\sqrt[3]{(6-4x_n )}
With x_0=0.5
2(a):
Find the value or x_1, x_2 and x_3.
Give your answers to 3 decimal places.
ANSWER: Multiple Answers (Type 1)
Answer: 1.597, -0.704, 2.066
Workings:
Substitute x_0=0.5 into the equation for x_{n+1} to get the value for x_1. Repeat to get iterations for x_2 and x_3.
Marks = 3
2(b):
Give the values of x correct to 3 decimal places.
ANSWER: Multiple Answers (Type 2)
Answer: 2.279 or -1.461
Workings:
Keep running iterations until the values for x remain at a constant value when rounded to 3d.p.
Marks = 1
Question 3
Using x_{n+1}=-3 - \dfrac{2}{x_n^2}
With x_0 = -3
3(a):
Find the values of x_1, x_2 and x_3
ANSWER: Multiple Answers (Type 1)
Answer: -3.22, -3.19,-3.20
Workings:
Substitute x_0=-3 into the equation for x_{n+1} to get the value for x_1. Repeat for x_2 and x_3.
Marks = 3
3(b):
Which statement explains the relationship between the values of x1, x2 and x3 and the equation
x^3 + 3x^2 + 2 = 0ANSWER: Multiple Choice (Type 1)
A: x_{n+1}=-3-\dfrac{2}{x_n^2} is the iterative form of x^3+3x^2+2=0, so x_1, x_2 and x_3 are estimations of the solution x^3+3x^2+2=0.
B: x^3+3x^2+2=0 is the iterative form of x_{n+1}=-3-\dfrac{2}{x_n^2}, so x_1, x_2 and x_3 are estimations of the solution x^3+3x^2+2=0.
Answer: A
Workings:
If we have an equation that equals 0, rearranging it to equal x puts it in iterative form. Each iteration then gives estimations for the solution to the original equation.
Marks = 2
Question 4
4(a):
Choose the correct iteration formula which can be found by rearranging the equation
10=x^2+xANSWER: Multiple Choice (Type 1)
A: x_{n+1}^2=10-x_n
B: x_{n+1}^2=\sqrt{10}-x_n
C: x_{n+1}=\sqrt{10-x_n}
D: x_{n+1}=\sqrt[3]{10-x_n}
Answer: C
Workings:
10-x=x^2
x_{n+1}=\sqrt{10-x_n}
Marks = 1
4(b):
Choose the correct iteration formula which can be found by rearranging the equation
x^3-5x+1=0ANSWER: Multiple Choice (Type 1)
A: x_{n+1}=\sqrt[3]{5x_n-1}
B: x_{n+1}=\sqrt[3]{\dfrac{x_n}{5}-\dfrac{1}{5}}
C: x_{n+1}=\dfrac{x_n}{5}-\dfrac{1}{5}
D: x_{n+1}=5+\dfrac{x_n}{5}
Answer: A
Workings:
x^3=5x-1
x_{n+1}=\sqrt[3]{5x_n-1}
Marks = 1
4(c):
Choose the correct iteration formula which can be found by rearranging the equation
x^3-10x^2-10=20ANSWER: Multiple Choice (Type 1)
A: x_{n+1}=\sqrt[3]{10x_n^2+30}
B: x_{n+1}=\sqrt[3]{\dfrac{x_n^2-30}{10}}
C: x_{n+1}=10+\dfrac{x_n^2}{10}
D: x_{n+1}=\sqrt[3]{10+\dfrac{x_n^2}{10}}
Answer: A
Workings:
x^3=30+10x^2
x_{n+1}=\sqrt[3]{10x_n^2+30}
Marks = 1
Question 5
3x^3+6x=4 has a solution between 0 and 1
5(a):
Find an appropriate iteration formula for the equation 3x^3+6x=4
ANSWER: Multiple Choice (Type 1)
A: x_{n+1}=\sqrt{\dfrac{4-6x_n}{3}}
B: x_{n+1}=\sqrt[3]{\dfrac{4-6x_n}{3}}
C: x_{n+1}=\sqrt[3]{\dfrac{3-4x_n}{6}}
D: x_{n+1}=\sqrt[3]{\dfrac{6-3x_n}{4}}
Answer: B
Workings:
3x^3=4-6x
x^3=\dfrac{4-6x}{3}
x_{n+1}=\sqrt[3]{\dfrac{4-6x}{3}}
Marks = 2
5(b):
Starting with x_0=0 use the iteration formula three times (x_3) to find an estimate for the solution to 3x^3+6x=4. Give answer to 2 d.p.
ANSWER: Simple Text Answer
Answer: 1.48
Workings:
Substitute x_0=0 into the equation for x_{n+1} This gives the value of x_1.
Repeat for x_2 and finally x_3 to get the solution.
Marks = 3
Question 6
A tank of water is slowly leaking.
One morning, the volume of water in the tank is V_A
The next morning, the volume of water in the tank is given by V_{A+1}=0.98V_A
On Monday morning, there was 50 litres in the tank.
What will the volume of water be on Friday morning?
Give your answer to the nearest whole litre.
ANSWER: Simple Text Answer
Answer: 46
Workings:
Monday = 50
Tuesday = 50\times 0.98 = 49
Wednesday = 49\times 0.98=48.02
Thursday = 48.02\times 0.98 = 47.0596
Friday = 47.0596\times 0.98 = 46.1184
This gives us 46 litres to the nearest litre.
Marks = 3
Question 7
The diagram shows a cube and a cuboid.
All the measurements are in cm.
The volume of the cube is 20cm^3 more than the volume of the cuboid.
Use an appropriate iteration formula to find x correct to 1 decimal place for
x^3 – x^2 – 10x = 20ANSWER: Simple Text Answer
Answer: 4.4
Workings:
Using the equation x^3-x^2-10x
We can start with a trial of values x=4 and x=5
Putting x=4 gives a solution of 8 which is too small andx=5 gives 50 which is too large
Putting x=4.5 gives 25.875, a result that is too large
Putting x=4.4 gives 21.24 which is still to large
x=4.3 gives 18.017 which is now too small
Putting x=4.35 gives 19.89 which is too small, so the solution must be 4.4 to 1 decimal place
Marks = 4