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.

Next, we need to sketch graph of We know that 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.

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.

 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 1.76.