Past Papers’ Solutions  Cambridge International Examinations (CIE)  AS & A level  Mathematics 9709  Pure Mathematics 2 (P29709/02)  Year 2004  OctNov  (P29709/02)  Q#6
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Question
i. By sketching a suitable pair ofgraphs, show that there is only one value of x in the interval that is a root of the equation
ii. Verify by calculation that this root lies between 0.8 and 0.9 radians.
iii. Show that this value of x is also a root of the equation
iv. Use the iterative formula
to determine this root correct to 2 decimal places, showing the value of each iteration.
Solution
i.
We are required to show that there is only one root of the following equation in the interval by sketching.
Root of an equation is the xcoordinate of a point of intersection of the graphs of and .
But we are required to show that there is only one root of the following equation in the interval graphically.
Therefore, first we sketch over the interval .
We know that graph of is as shown below.
But we are looking for over the interval only.
Therefore, we consider only part of the above graph (of ) as shown below (in blue color).
Next, we need to sketch graph of over the interval .
We can calculate a couple of values to be accurate in sketching the graph.

0 
0.1 
0.3 
0.5 
0.7 
0.9 
1.1 
1.3 
1.5 


0 
0.1 
0.3 
0.5 
0.7 
0.9 
1.1 
1.3 
1.5 

We can plot these points to sketch the following graph of over the interval .
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 during interval , 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 0.8 and
0.9 radians. 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 required to show that root of equation is also a root of the equation
If we can write the given equation and then transform it to , then both will have the same root.
Therefore, if the given equation can be rewritten as , it is evident that roots of both will be same.
Since except where or undefined;
Since given equation can be rewritten as , the root of will also be root of .
iv.
If we can write the given equation and transform it to , then we can find the root of the equation by iteration method using sequence defined as.
As demonstrated in (iii) given equation can be rewritten as , therefore, iteration method can be used to find the root of the given equation using sequence defined by;
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 


8 


9 


10 


11 


12 


13 


14 


It is evident that .
Hence, is a root of .
The root given correct to 2 decimal places is 0.86.
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