# Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 2 (P2-9709/02) | Year 2014 | May-Jun | (P2-9709/21) | Q#4

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Question

i.       By sketching a suitable pair of graphs, show that the equation

has exactly one real root.

ii.       Show by calculation that the root lies between 2.0 and 2.5.

iii.       Use the iterative formula to find the root correct to 3 decimal places. Give  the result of each iteration to 5 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 .

Therefore, first we sketch .

We know that graph of is as shown below. Or we can plot it as follows.

We can calculate a couple of values to be accurate in sketching the graph.

 0 0.1 0.5 1 1.5 2 2.5 3 3.5 – -6.9 -2.08 0 1.22 2.08 2.75 3.29 3.76

We can plot these points to sketch the following graph of

Next, we need to sketch graph of .

It can be seen that it is a quadratic equation which can be rearranged as;

We can calculate a couple of values to be accurate in sketching the graph.

 0 1 2 3 4 15 14 7 -12 -49

We can plot these points to sketch the following graph of

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,  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  2.0 and 2.5. 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 also given the iterative formula as;

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

It is evident that .

Hence, is a root of .

The root given correct to 3 decimal places is 2.319.