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;

Area of Shaded Region
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.

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It is evident that .

Hence, is a root of .

The root given correct to 2 decimal places is .

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