Past Papers’ Solutions  Cambridge International Examinations (CIE)  AS & A level  Mathematics 9709  Pure Mathematics 2 (P29709/02)  Year 2005  OctNov  (P29709/02)  Q#7
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Question
The diagram shows the part of the curve y = sin^{2} x for .
i. Show that
ii. Hence find the xcoordinates of the points on the curve at which the gradient of the curve is 0.5.
iii. By expressing sin^{2 }x in terms of cos 2x, find the area of the region bounded by the curve and the xaxis between 0 and π.
Solution
i.
We are given;
We are required to show that;
Therefore;
If we define , then derivative of is;
Therefore;
Rule for differentiation of is;
We have doubleangle formula;
Therefore;
ii.
We are required to find the xcoordinate of the point on the curve where gradient is 0.5.
Gradient (slope) of the curve at the particular point is the derivative of equation of the curve at that particular point.
Gradient (slope) of the curve at a particular point can be found by substituting x coordinates of that point in the expression for gradient of the curve;
We already have found in (i) that;
Therefore, we can equate expression of gradient of the curve with 0.5.
Using calculator, we can find;
Properties of 

Domain 

Range 

Odd/Even 

Periodicity 



Translation/ Symmetry 






We utilize the periodic property of to find another solution (root) of :
Hence;
Therefore;
Therefore, we have two solutions (roots) of the equation;
So we have two possible values of ,


We do not need to find any solutions beyond the desired range .
iii.
We are required to find the area of the region bounded by the curve and the xaxis between 0 and π.
We are given equation of the curve as;
To find the area of region under the curve , we need to integrate the curve from point to along xaxis.
Therefore;
Since we are required to find the area of the region bounded by the curve and the xaxis between 0 and π;
We have the trigonometric identity;
From this we can wirte , and hence;
Since , we can write;
Therefore;
Rule for integration of is:
Rule for integration of is:
Rule for integration of is:
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