Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 2 (P2-9709/02) | Year 2008 | Oct-Nov | (P2-9709/02) | Q#7
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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 0.5 and 1.
iii. Show that, if a sequence of values given by the iterative formula
converges, then it converges to the root of the equation in part (i).
iv. Use this iterative formula, with initial value x1 = 0.6, 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 x-coordinate 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 0.5 and 1.0 radians. We need to use sign-change rule.
To use the sign-change 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 achieve this by showing that given equation can also be written as;
Therefore, if the given equation can be rewritten as
, it is evident that roots of both will be same.
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 sign-change 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 0.58.
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