# 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.

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).

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 0.7 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.