Iteration (Edexcel GCSE Maths: Higher)

Exam Questions

1 hour16 questions
1a
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3 marks

Using  x subscript n plus 1 end subscript space equals space minus 2 space minus space fraction numerator 4 over denominator x subscript n superscript 2 end fraction

with  x subscript 0 space equals space minus 2.5

find the values of x subscript 1x subscript 2 and x subscript 3

1b
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2 marks

Explain the relationship between the values of x subscript 1 space comma space x subscript 2 and x subscript 3 and the equation space x cubed space plus space 2 x squared space plus 4 space equals 0

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2
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2 marks

A sequence of numbers is formed by the iterative process

u subscript n plus 1 end subscript equals fraction numerator 4 over denominator u subscript n minus 1 end fraction         u subscript 1 equals 9

Work out the values of u subscript 2 and u subscript 3

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3a
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2 marks

An approximate solution to an equation is found using the iterative formula

x subscript n plus 1 end subscriptfraction numerator open parentheses x subscript n close parentheses cubed space minus space 2 over denominator 10 end fraction   with x subscript 1 space equals space minus 1

Work out the values of x subscript 2 and x subscript 3

x subscript 2 = ......................

x subscript 3 = ......................

3b
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1 mark

Work out the solution to 5 decimal places.

x = ......................

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4a
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2 marks

An approximate solution to an equation is found using this iterative process.

   x subscript n plus 1 end subscript space equals space fraction numerator open parentheses x subscript n close parentheses cubed space minus space 3 over denominator 8 end fraction space and x subscript 1 space equals space minus 1

Work out the values of space x subscript 2and x subscript 3

space x subscript 2   = ........................

x subscript 3   = .........................

4b
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1 mark

Work out the solution to 6 decimal places.

x = .......................

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5
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3 marks

Use the formula x subscript n plus 1 end subscript space equals space open parentheses x subscript n close parentheses cubed over 30 plus 2 with x1 = 2 to calculate x2 and x3.

Round your answers correct to 4 decimal places.

x2 = ............. and x3 = ................

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1a
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2 marks

Show that the equation x cubed space plus space x space equals space 7 has a solution between 1 and 2

1b
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1 mark

Show that the equation space x to the power of 3 space end exponent plus space x space equals 7 can be rearranged to give x space equals space cube root of 7 minus x end root

1c
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3 marks

Starting with x subscript 0 space equals space 2 comma
use the iteration formula x subscript n plus 1 end subscript space equals space cube root of 7 minus x subscript n end root three times to find an estimate for a solution of  x cubed space plus space x space equals space 7

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2a
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2 marks

Show that the equation x cubed space plus space 4 x equals space 1 has a solution between x space equals space 0 and x space equals space 1

2b
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1 mark

Show that the equation x cubed space plus space 4 x equals space 1 can be arranged to give  x space equals space 1 fourth minus x cubed over 4

2c
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3 marks

Starting with space x subscript 0 space equals space 0 , use the iteration formula x subscript n plus 1 end subscript space equals space fraction numerator 1 over denominator 4 space end fraction space minus fraction numerator x subscript n superscript 3 over denominator 4 end fraction spacetwice , to find an estimate for the solution of  x cubed space plus space 4 x space equals 1

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3a
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2 marks

Show that the equation 3 x squared space minus space x cubed space plus space 3 space equals space 0 can be rearranged to give 

   x space equals space 3 space plus space 3 over x squared

3b
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3 marks

Using 

      x subscript n plus 1 end subscript space equals space 3 space plus space fraction numerator 3 over denominator x subscript n superscript 2 end fraction with x subscript 0 space equals space 3.2 comma

find the values of x subscript 1 space comma space x subscript 2 and x subscript 3

3c
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1 mark

Explain what the values of x subscript 1 space comma space x subscript 2 space comma space and  x subscript 3 represent.

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4a
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2 marks

A sphere has radius r cm

An approximate value of r can be found using the iterative formula

r subscript n plus 1 end subscript equals square root of 239 over r subscript n end root

The starting value is  r subscript 1 equals 7

Work out the values of  r subscript 2 and r subscript 3

r subscript 1 equals.....................
r subscript 2 equals.....................

4b
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1 mark

Continue the iteration to work out the radius to 1 decimal place.

................................................. cm

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5
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3 marks

x subscript n plus 1 end subscript equals cube root of 3 x subscript n plus 7 end root

Use a starting value of x subscript 1 space equals space 2 to work out a solution to x space equals space cube root of 3 x plus 7 end root
Give your answer to 3 decimal places.

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1a
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2 marks

Show that the equation x cubed space plus space 7 x space minus space 5 space equals space 0 has a solution between x space equals space 0 and x space equals space 1

1b
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2 marks

Show that the equation x cubed space plus space 7 x space minus 5 space equals 0 can be arranged to give x space equals space fraction numerator 5 over denominator x squared space plus 7 end fraction

1c
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3 marks

Starting with x subscript 0 space equals space 1, use the iteration formula x subscript n plus 1 end subscript space equals space fraction numerator 5 over denominator x subscript n superscript 2 space plus space 7 end fraction three times to find an estimate for the solution of x cubed space plus space 7 x space long dash space 5 space equals space 0

1d
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2 marks

By substituting your answer to part (c) into x cubed space plus space 7 x space long dash space 5 comma comment on the accuracy of your estimate for the solution to x cubed space plus space 7 x minus space 5 space equals space 0

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2
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3 marks

The number of bees in a beehive at the start of year n is P subscript n.
The number of bees in the beehive at the start of the following year is given by 

P subscript n plus 1 end subscript space equals space 1.05 open parentheses P subscript n space minus space 250 close parentheses

At the start of 2015 there were 9500 bees in the beehive.

How many bees will there be in the beehive at the start of 2018?

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3
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3 marks

The number of slugs in a garden space t spacedays from now is p subscript t where

   p subscript 0 space equals space 100
p subscript t plus 1 end subscript space equals 1.06 p subscript t

Work out the number of slugs in the garden 3 days from now.

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4a
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3 marks

The number of rabbits on a farm at the end of month n is P subscript n
The number of rabbits at the end of the next month is given by P subscript n plus 1 end subscript equals 1.2 space P subscript n space minus space 50

At the end of March there are 200 rabbits on the farm.

Work out how many rabbits there will be on the farm at the end of June.

4b
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1 mark

Considering your results in part (a), suggest what will happen to the number of rabbits on the farm after a long time.

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5
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3 marks

At the start of year n, the number of animals in a population is P subscript n At the start of the following year, the number of animals in the population is P subscript n plus 1 end subscript, where

P subscript n plus 1 end subscript equals k P subscript n

At the start of 2017 the number of animals in the population was 4000
At the start of 2019 the number of animals in the population was 3610
Find the value of the constant k.

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6a
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3 marks

Use the iteration formula  x subscript n plus 1 end subscript equals cube root of 10 minus 2 x subscript n end root  to find the values of x subscript 1, x subscript 2 and x subscript 3
Start with x subscript 0 equals 2

x subscript 1 equals space......     
x subscript 2 equals space......     
x subscript 3 equals space..... .     

6b
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1 mark

The values of x subscript 1, x subscript 2 and x subscript 3 found part (a) are estimates of the solution of an equation of the form x cubed plus a x plus b equals 0 where a and b are integers.

Find the value of a and the value of b.

a equals space......     
b equals space......     

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