Past Papers’ Solutions  Cambridge International Examinations (CIE)  AS & A level  Mathematics 9709  Pure Mathematics 1 (P19709/01)  Year 2015  OctNov  (P19709/12)  Q#9
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Question
The curve y=f(x) has a stationary point at (2,10) and it is given that
i. Find f(x).
ii. Find the coordinates of the other stationary point.
iii. Find the nature of each of the stationary points.
Solution
i.
We are given;
We are required to find .
We can find equation of the curve from its derivative through integration;
But we are given second derivative of . Therefore, we need derivative of to find .
To find the derivative of the curve from second derivative we can integrate the second derivative of the curve.
Therefore;
Rule for integration of is:
If a stationary point lies on the curve , we can find out value of . We equate the equation of derivative obtained from integration of second derivative of the curve with zero i.e. . Then we substitute value of in the equation of derivative obtained from integration of second derivative of the curve i.e. .
We are given that point (2,10) is a stationary point on the curve. Therefore
Hence;
We can find equation of the curve from its derivative through integration;
Therefore;
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 point (2,10) is a stationary point on the curve. Therefore
Hence;
ii.
Coordinates of stationary point on the curve can be found from the derivative of equation of the curve by equating it with ZERO. This results in value of xcoordinate of the stationary point on the curve.
We have found in (i);
Therefore;
Two possible values of imply that there are two stationary points on the curve one at each value of .
To find ycoordinate of the stationary point on the curve, we substitute value of xcoordinate of the stationary point on the curve (found by equating derivative of equation of the curve with ZERO) in the equation of the curve.
Now we have two options.
For 
For 






Therefore stationary points are and .
iii.
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 already given;
We have found stationary points are and .
Therefore;
Point 
Point 










Minimum Point 
Maximum Point 
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