Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 2 (P2-9709/02) | Year 2005 | Oct-Nov | (P2-9709/02) | Q#5

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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 this root also satisfies the equation

  iv. Use the iterative formula

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.

desmos-graph (8).png

Next, we need to sketch graph of We know that graph of is as shown below.

desmos-graph (10).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.
 

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.

Taking anti-logarithm of both sides;

Since ;

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 1.76.

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