Past Papers’ Solutions  Cambridge International Examinations (CIE)  AS & A level  Mathematics 9709  Pure Mathematics 1 (P19709/01)  Year 2016  OctNov  (P19709/13)  Q#11
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Question
A curve has equation , where k is a nonzero constant.
i. Find the xcoordinates of the stationary points in terms of k, and determine the nature of each stationary point, justifying your answers.
ii.
The diagram shows part of the curve for the case when k = 1. Showing all necessary working, find the volume obtained when the region between the curve, the xaxis, the yaxis and the line x=2, shown shaded in the diagram, is rotated through 360^{o }about the xaxis.
Solution
i.
We are given that equation of the curve is;
First we are required to find the xcoordinates of the stationary points 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.
Therefore, first we need derivative of the equation of the curve.
Gradient (slope) of the curve is the derivative of equation of the curve. Hence gradient of curve with respect to is:
Therefore;
Rule for differentiation of is:
Rule for differentiation of is:
To find the xcoordinate of the stationary point;
Now we have two options.










Two possible values of imply that there are two stationary points on the curve one at each value of .
Next we are required to determine the nature of these stationary points.
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;
We already have found that;
Therefore;
Rule for differentiation of is:
Rule for differentiation of is:
Rule for differentiation of is:
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 given that k is a nonzero constant. Therefore, whether k is positive or negative, .
Therefore;


Minimum 
Maximum 
ii.
Expression for the volume of the solid formed when the shaded region under the curve is rotated completely about the xaxis is;
We have;
We are given that k=1;
Therefore;
Rule for integration of is:
Rule for integration of is:
Rule for integration of is:
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