# 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; 