Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 2 (P2-9709/02) | Year 2002 | Oct-Nov | (P2-9709/02) | Q#4

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  Question

     i.       By sketching a suitable pair of graphs, 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 1 and 1.5.

  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 3 significant figures, showing the value of each approximation that  you calculate.

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 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 in the interval   graphically.

Therefore, first we sketch  over the interval .

We know that graph of  is as shown below.

desmos-graph (8).png

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).

desmos-graph (9).png

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

100

11.11

4.0

2.04

1.23

0.83

0.59

0.44

We can plot these points to sketch the following graph of  over the interval .

desmos-graph (10).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  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 1 and 1.5. 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.
 

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 𝑥=, it is evident that roots of  both will be same.

We know that    provided that , therefore;

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.

0

1

2

3

4

5

6

7

8

9

10

11

It is evident that .

Hence,  is a root of .

The root given correct to 3 significant figures is 1.07.

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