# 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 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 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;