Past Papers’ Solutions  Cambridge International Examinations (CIE)  AS & A level  Mathematics 9709  Pure Mathematics 1 (P19709/01)  Year 2015  OctNov  (P19709/11)  Q#5
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Question
A curve has equation .
i. Find and
ii. Find the coordinates of the stationary points and state, with a reason, the nature of each stationary point.
Solution
i.
We are given that;
Gradient (slope) of the curve is the derivative of equation of the curve. Hence gradient of curve with respect to is:
Rule for differentiation of is:
Rule for differentiation of is:
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;
Rule for differentiation of is:
Rule for differentiation of is:
Rule for differentiation of is:
ii.
A stationary point on the curve is the point where gradient of the curve is equal to zero;
From (i), we have;
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.
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.
For 
For 






Once we have the xcoordinate of the stationary point of a curve, we can determine its nature, whether minimum or maximum, by finding 2^{nd }derivative of the curve.
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;
From (i), we have;
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.
For 
For 












Maximum 
Minimum 
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