Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 2 (P2-9709/02) | Year 2013 | May-Jun | (P2-9709/23) | 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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