# Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 2 (P2-9709/02) | Year 2013 | Oct-Nov | (P2-9709/22) | Q#7

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Question

The diagram shows the curve . The shaded region R is bounded by the curve and by  the lines x = 0, y = 0 and x = a, where a is positive. The area of R is equal to .

i.
Find an equation satisfied by a, and show that the equation can be written in the form

ii.
Verify by calculation that the equation  has a root between 0.2 and 0.3.

iii.
Use the iterative formula  to determine this root correct to 2 decimal places. Give

the result of each iteration to 4 decimal places.

Solution

i.

To find the area of region under the curve , we need to integrate the curve from point to   along x-axis.

It is evident from the diagram that area R of shaded region is area under the curve along x-axis from x=0 to x=a.

Therefore;

Rule for integration of  is:

Rule for integration of  is:

Rule for integration of , or ;

Therefore;

We are given that;

Therefore;

ii.

We are required to verify by calculation that has a root that lies between a=0.2 and  a=0.3.

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.

Iteration method can be used to find the root of the given equation using iterative formula;

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 (i) 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

It is evident that .

Hence, is a root of .

The root given correct to 2 decimal places is 0.29.