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.

desmos-graph (8).png

But we are looking for in the interval .

desmos-graph (10).png

Next we sketch the graph of .

desmos-graph (19).png

Sketching both graphs on the same axes, we get following.

desmos-graph (11).png

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.

1

2

3

4

5

6

7

8

9

10

It is evident that .

Hence, is a root of

The root given correct to 2 decimal places is 0.58.

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