# Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 2 (P2-9709/02) | Year 2008 | May-Jun | (P2-9709/02) | Q#8

Hits: 53

Question

The constant , where , is such that

i.       Find an equation satisfied by , and show that it can be written in the form ii.
Verify, by calculation, that the equation ) has a root between 3 and 3.5.

iii.       Use the iterative formula with a1 = 3.2, to calculate the value of a correct to 2 decimal places. Give the result of each iteration  to 4 decimal places.

Solution

i.

We are given; Rule for integration of is:  Rule for integration of is: For ;    Since , , therefore;      Since , therefore; ii.

We are required to verify by calculation that the root of equation lies between 3  and 3.5. 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.

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 (i) 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  It is evident that .

Hence, is a root of .

The root given correct to 2 decimal places is 3.26.