Past Papers’ Solutions  Cambridge International Examinations (CIE)  AS & A level  Mathematics 9709  Pure Mathematics 2 (P29709/02)  Year 2007  MayJun  (P29709/02)  Q#5
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Question
i. By sketching a suitable pair of graphs, show that the equation
Where x is in radians, has only one root in the interval .
ii. Verify by calculation that this root lies between 1.0 and 1.2.
iii. Show that this root also satisfies the equation
iv. Use the iterative formula
with initial value x_{1 }= 1.1, to calculate the root correct to 2 decimal places. Give the result of each iteration to 4 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 .
But we are required to show that there is only one root of the following equation graphically.
Therefore, first we sketch .
We know that graph of is as shown below.
But we are looking for in the interval .
Next we sketch the 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 1.0 and 1.22 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 provided that ;
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.



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It is evident that .
Hence, is a root of .
The root given correct to 2 decimal places is 1.04.
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