# Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 1 (P1-9709/01) | Year 2014 | Oct-Nov | (P1-9709/13) | Q#8

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**Question**

A curve has a stationary point at and is such that .

** i. **State, with a reason, whether this stationary point is a maximum or a minimum.

** ii. **Find and .

**Solution**

i.

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 are given second derivative of the curve ;

We are also given coordinates of stationary point on the curve . Therefore, we can substitute x-coordinate of the stationary point in given second derivative of the curve.

Since , stationary point is minimum.

ii.

We are given second derivative of the curve and are required to find first derivative and equation of the curve .

To find the derivative of the curve from second derivative we can integrate the second derivative of the curve.

Rule for integration of is:

If a point lies on the curve , we can find out value of . We substitute values of and in the equation obtained from integration of the derivative of the curve i.e. .

We know that a stationary point at is on the curve.

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

Since derivative is equal to zero at stationary point , we can substitute x-coordinate in this equation.

Hence;

We can find equation of the curve from its derivative through integration;

For the given case, we have found above;

Rule for integration of is:

Rule for integration of is:

Rule for integration of is:

If a point lies on the curve , we can find out value of . We substitute values of and in the equation obtained from integration of the derivative of the curve i.e. .

We are given that curve has a stationary point , therefore, we can substitute x and y coordinates in this equation.

Hence;

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