# 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

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**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 2^{nd }derivative of the curve.

We substitute of the stationary point in the expression of 2^{nd} 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:

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