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.

desmos-graph (10).png

Therefore graph of is as shown below.

desmos-graph (12).png

Therefore graph of is as shown below.

desmos-graph (13).png

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

Hence, graph of is as shown below.

desmos-graph (14).png

Sketching both graphs on the same axes, we get following.

desmos-graph (11).png

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.

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It is evident that .

Hence, is a root of .

The root given correct to 2 decimal places is 1.07.

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