# Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 2 (P2-9709/02) | Year 2006 | May-Jun | (P2-9709/02) | Q#6

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Question

i. By sketching a suitable pair of graphs, show that there is only one value of x that is a root of the  equation

ii. Verify, by calculation, that this root lies between 1 and 2.

iii. Show that, if a sequence of values given by the iterative formula

converges, then it converges to the root of the equation given in part (i).

iv. Use the iterative formula, with x1 = 1, to calculate the root correct to 2 decimal places.

with initial value x1 = 1.8, 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 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 graphically.

Therefore, first we sketch .

We know that graph of is as shown below.

Next, we need to sketch graph of .

Where

Some properties, which help to plot/sketch, of this graph are as follows.

· The graphs of all exponential functions contain the point .

· The domain is all real numbers .

· The range is only the positive .

· The graph is increasing.

· The graph is asymptotic to the x-axis as x approaches negative infinity

· The graph increases without bound as x approaches positive infinity

· The graph is continuous and smooth.

Therefore graph of is as shown below.

Therefore graph of is as shown below.

Therefore graph of is as shown below.

Functions  and  are reflections of each other in y-axis.

Hence, graph of is as shown below.

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 1 and 2 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.

If we can write the given equation  and then transform it to , then both will have the  same root.

We can show the desired result if we can show that given equation  can be written as;

Therefore, if the given equation can be rewritten as  , it is evident  that roots of both will be same.

Taking logarithm of both sides;

Multiplication Rule;

Since  for any ;

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.

 1 2 3 4 5 6 7 8 9 10 11 12 13 14

It is evident that .

Hence, is a root of .

The root given correct to 2 decimal places is 1.07.