Past Papers’ Solutions  Cambridge International Examinations (CIE)  AS & A level  Mathematics 9709  Pure Mathematics 1 (P19709/01)  Year 2017  OctNov  (P19709/12)  Q#8
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Question
A curve is such that .
i. Find the xcoordinate of each of the stationary point on the curve.
ii. Obtain an expression for and hence or otherwise find the nature of each of the stationary points.
iii. Given that the curve passes through the point (6,2), find the equation of the curve.
Solution
i.
We are given that;
We are required to find the xcoordinate of each of the stationary point on the curve.
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 equate given derivative with zero to find the xcoordinate of stationary point(s).
Now we have two options.




Two possible values of imply that there are two stationary points on the curve one at each value of .
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;
Therefore;
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.
We have found in (i) that stationary points have xcoordinates as;


We find nature of both stationary points by substituting respective value of xcoordinates in .
For 
For 








Since , the stationary point 
Since , the stationary point 
iii.
We are required to find the equation of the curve if it passes through the point (6,2).
We can find equation of the curve from its derivative through integration;
We are given that;
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. .
Therefore, we substitute coordinates (6,2) in above equation.
Hence, equation of the curve is;
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