# 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.