Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 1 (P1-9709/01) | Year 2015 | Oct-Nov | (P1-9709/13) | Q#10

Question

The function  is defined by  for .

i.       Find  and  and hence verify that the function  has a minimum value at .

The points  and  lie on the curve , as shown in the diagram.

i.       Find the distance AB.

ii.       Find, showing all necessary working, the area of the shaded region.

Solution

i.

We are given;

Rule for differentiation of  is:

Rule for differentiation of  is:

Rule for differentiation of  is:

Rule for differentiation of  is:

Rule for differentiation of  is:

Rule for differentiation of  is:

A stationary point  on the curve  is the point where gradient of the curve is equal to zero;

If the given function   has a minimum (stationary) value at  then  at  .

We have found above that for given function ;

At ;

Therefore, function  has a stationary value at .

Once we have the coordinates of the stationary point  of a curve, we can determine its  nature, whether minimum or maximum, by finding 2nd derivative of the curve.

We substitute  of the stationary point in the expression of 2nd derivative of the curve and  evaluate it;

If  or  then stationary point (or its value) is minimum.

If  or  then stationary point (or its value) is maximum.

We have found above that for given function ;

At ;

Since , function  has a minimum value at .

ii.

Expression to find distance between two given points  and is:

Therefore for points  and ;

iii.

It is evident from the diagram that;

To find the area of region under the curve , we need to integrate the curve from point  to  along x-axis.

We are given equation of the curve as follows;

However, we do not have equation of line but using points  and  we can write  equation of line AB.

Two-Point form of the equation of the line is;

Therefore;

Now we can find area of shaded region.

Rule for integration of  is:

Rule for integration of  is:

Rule for integration of  is:

Rule for integration of  is: