Past Papers’ Solutions  Cambridge International Examinations (CIE)  AS & A level  Mathematics 9709  Pure Mathematics 2 (P29709/02)  Year 2006  MayJun  (P29709/02)  Q#6
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Question
i. By sketching a suitable pair of graphs, show that there is only one value of x that is a root of the equation
ii. Verify, by calculation, that this root lies between 1 and 2.
iii. Show that, if a sequence of values given by the iterative formula
converges, then it converges to the root of the equation given in part (i).
iv. Use the iterative formula, with x_{1} = 1, to calculate the root correct to 2 decimal places.
with initial value x_{1 }= 1.8, to determine this 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.
Next, we need to sketch graph of .
Where
Some properties, which help to plot/sketch, of this graph are as follows.
· The graphs of all exponential functions contain the point .
· The domain is all real numbers .
· The range is only the positive .
· The graph is increasing.
· The graph is asymptotic to the xaxis as x approaches negative infinity
· The graph increases without bound as x approaches positive infinity
· The graph is continuous and smooth.
Therefore graph of is as shown below.
Therefore graph of is as shown below.
Therefore graph of is as shown below.
Functions and are reflections of each other in yaxis.
Hence, graph of is as shown below.
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 and 2 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.
If we can write the given equation and then transform it to , then both will have the same root.
We can show the desired result if we can show that given equation can be written as;
Therefore, if the given equation can be rewritten as , it is evident that roots of both will be same.
Taking logarithm of both sides;
Multiplication Rule;
Since for any ;
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 1.07.
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