Past Papers’ Solutions  Cambridge International Examinations (CIE)  AS & A level  Mathematics 9709  Pure Mathematics 2 (P29709/02)  Year 2014  MayJun  (P29709/21)  Q#4
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Question
i. By sketching a suitable pair of graphs, show that the equation
has exactly one real root.
ii. Show by calculation that the root lies between 2.0 and 2.5.
iii. Use the iterative formula to find the root correct to 3 decimal places. Give the result of each iteration to 5 decimal places.
Solution
i.
We are required to show that there is only one root of the following equation by sketching.
Root of an equation is the xcoordinate of a point of intersection of the graphs of and .
Therefore, first we sketch .
We know that graph of is as shown below. Or we can plot it as follows.
We can calculate a couple of values to be accurate in sketching the graph.

0 
0.1 
0.5 
1.0 
1.5 
2.0 
2.5 
3.0 
3.5 

– 
6.90 
2.08 
0.0 
1.22 
2.08 
2.75 
3.29 
3.76 
We can plot these points to sketch the following graph of
Next, we need to sketch graph of .
It can be seen that it is a quadratic equation which can be rearranged as;
We can calculate a couple of values to be accurate in sketching the graph.

0 
1 
2 
3 
4 

15 
14 
7 
12 
49 
We can plot these points to sketch the following graph of
Sketching both graphs on the same axes, we get following.
It can be seen that the two graphs of and intersect each other at only a single point, therefore, the equation has a single roots and at a single value of .
ii.
We are required to verify by calculation that the only root of equation lies between 2.0 and 2.5. We need to use signchange rule.
To use the signchange method we need to write the given equation as .
Therefore;
If the function is continuous in an interval of its domain, and if and have opposite signs, then has at least one root between and .
We can find the signs of at and as follows;
Since and have opposite signs for function , the function has root between and .
iii.
We are also given the iterative formula as;
If the sequence given by the inductive definition , with some initial value , converges to a limit , then is the root of the equation .
Therefore, if , then is a root of .
We have already found in (ii) through signchange rule that root of the given equation lies between and .
Therefore, for iteration method we use;
We use as initial value.



0 


1 


2 


3 


4 


5 


6 


7 


It is evident that .
Hence, is a root of .
The root given correct to 3 decimal places is 2.319.
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