Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 2 (P2-9709/02) | Year 2013 | May-Jun | (P2-9709/21) | Q#6

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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 for .

   ii.       Verify by calculation that this root lies between x= 0.7 and x=0.9.

  iii.       Show that this root also satisfies the equation

  iv.       Use the iterative formula

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 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.

Description: 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).

Description: 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

-2

-1.6

-0.8

0.0

0.7

1.6

2.4

3.2

4

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

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 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  0.7 and 0.9 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.
 

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 except where  or undefined;

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

12

13

14

15

It is evident that .

Hence, is a root of .

The root given correct to 2 decimal places is 0.76.

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