# Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 2 (P2-9709/02) | Year 2006 | Oct-Nov | (P2-9709/02) | Q#5

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Question

The diagram shows a chord joining two points, A and B, on the circumference of a circle with centre  O and radius r. The angle AOB is radians, where . The area of the shaded  segment is one sixth of the area of the circle.

i.       Show that  satisfies the equation

ii.       Verify by calculation that lies between  and .

iii.       Use the iterative formula

with initial value x1 = 2, to determine correct to 2 decimal places. Give the result of each iteration  to 4 decimal places.

Solution

i.

We are given that area of the shaded segment is one sixth of the area of the circle.

Expression for area of a circle with radius  is;

From the given diagram it is evident that;

Expression for area of a circular sector with radius and angle  rad is;

Expression for the area of a triangle for which two sides (a and b) and the included angle (C ) is  given;

According to given condition;

Therefore;

Hence if ;

ii.

We are required to verify by calculation that that  lies between  and .

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 given the 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 (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

It is evident that .

Hence, is a root of .

The root given correct to 2 decimal places is 1.97.