# Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 2 (P2-9709/02) | Year 2012 | Oct-Nov | (P2-9709/23) | Q#5

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Question The diagram shows the curve , for . A rectangle OABC is drawn, where B is the  point on the curve with x-coordinate , and A and C are on the axes, as shown. The shaded region  R is bounded by the curve and by the lines x = and y = 0.

The shaded region R is bounded by the curve and the lines y = 1 and x = p, where p is a constant.

i.       Find the area of R in terms of .

ii.       The area of the rectangle OABC is equal to the area of R. Show that iii.       Use the iterative formula , with initial value , to determine the value of correct to 2 decimal places. Give the result  of each iteration to 4 decimal places.

Solution

i.

We are required to find area of shaded region R.

It is evident that area of shaded region R is area under the curve as x varies from to .

To find the area of region under the curve , we need to integrate the curve from point to along x-axis. For the given case; Therefore; Rule for integration of is:    Hence, area of shaded region R is; ii.

We are given that;

R = Area of Rectangle OABC

We have found in (i) that; Now we find area of rectangle.

Expression for the area of the rectangle is; Therefore; Therefore;  iii.

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 .

Therefore, iterative formula we use; We use as initial value.   1  2  3  4  5  6  7  8  9  10  11  12  13  14  15  16  17  18  19  20  21  22  23  It is evident that .

Hence, is a root of .

The root given correct to 2 decimal places is .