# Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 2 (P2-9709/02) | Year 2005 | Oct-Nov | (P2-9709/02) | Q#7

Hits: 169

Question

The diagram shows the part of the curve y = sin2 x for  .

i.       Show that

ii.       Hence find the x-coordinates of the points on the curve at which the gradient of the curve is  0.5.

iii.       By expressing sin2 x in terms of cos 2x, find the area of the region bounded by the curve and the x-axis 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 double-angle formula;

Therefore;

ii.

We are required to find the x-coordinate 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 x-axis 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 x-axis.

Therefore;

Since we are required to find the area of the region bounded by the curve and the x-axis 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: