Past Papers’ Solutions  Cambridge International Examinations (CIE)  AS & A level  Mathematics 9709  Pure Mathematics 1 (P19709/01)  Year 2011  OctNov  (P19709/13)  Q#8
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Question
A curve has a stationary point at . It is given that, where k is a constant.
i. Show that and hence find the xcoordinate of the other stationary point, Q.
ii. Find and determine the nature of each of the stationary points P and Q.
iii. Find .
Solution
i.
A stationary point on the curve is the point where gradient of the curve is equal to zero;
Hence gradient of the given curve at given stationary point .
Since;
Therefore;
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.
Now we can write the derivative of the equation of given curve;
Now we have two options.









Two possible values of imply that there are two stationary points on the curve one at each value of .
We are already given one stationary point at . So the x coordinate of other stationary point is .
ii.
Second derivative is the derivative of the derivative. If we have derivative of the curve as , then expression for the second derivative of the curve is;
We are given that;
For second derivative;
Rule for differentiation of is:
Rule for differentiation of is:
Rule for differentiation of is:
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.
Since we have two stationary points namely P & Q with xcoordinates and , we check nature of both stationary points one by one.
For ;
Since
it is minimum point.
For ;
Since it is maximum point.
iii.
We can find equation of the curve from its derivative through integration;
For the given case;
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 stationary point at .
Therefore;
Hence;
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