Past Papers’ Solutions  Assessment & Qualification Alliance (AQA)  AS & A level  Mathematics 6360  Pure Core 1 (6360MPC1)  Year 2017  June  Q#2
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Question
A curve has the equation . The curve has a stationary point
at the point M where .
a. Find the coordinates of the other stationary point of the curve.
b. Find the value of at the point M, and hence determine, with a reason, whether M is a minimum point or a maximum point.
c. Sketch the curve.
Solution
a.
We are required to find the coordinates of stationary point on the curve with equation;
A stationary point on the curve is the point where gradient of the curve is equal to zero;
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 to find the expression for the derivative of the given 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 is of is:
Rule for differentiation is of is:
To find the coordinates of the stationary point on the curve, we equate this derivative equal to ZERO.
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 that point M, where x=2, is a stationary point on the curve and hence x coordinate of other stationary point on the curve is .
b.
We are required to find the second derivative.
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;
Therefore;
Rule for differentiation is of is:
Rule for differentiation is of is:
Rule for differentiation is 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 are given that point M, where x=2, is a stationary point on the curve. Therefore, to determine the nature of point M, we substitute xcoordinate of point M in the expression of second derivative of the curve obtained above;
Since , the point M on the curve is minimum.
c.
We are required to sketch the curve with equation given as;
We are required to sketch a cubical polynomial given as;
ü Find the sign of the coefficient of . This gives the shape of the graph at the extremities. If the coefficient of is such that cubic graph is increasing from left to right at the extremities and when the cubic graph is inverted.
The coefficient of in the given polynomial is negative. Therefore it will cause inverted cubic graph increasing from right to left at the extremities.
ü Find the point where the graph crosses yaxis by finding the value of when .
Therefore for ;
Hence, the cubic graph intercepts yaxis at , ie it passes through the origin (0,0).
ü Find the point(s) where the graph crosses the xaxis by finding the value of when . If there is repeated root the graph will touch the xaxis.
We already have xcoordinates of two stationary points on the curve at and .
We may like to find corresponding ycoordinate of stationary points on the curve for accurate sketching of the curve by substituting corresponding value of xcoordinate in equation of the
curve.
When ; 
When ; 







We can sketch the graph with this information. However, we may like to do the following as well.
ü Calculate the values of for some value of . The is particularly useful in determining the quadrant in which the graph might turn close to the yaxis.
ü Complete the sketch of the graph by joining the sections.
Sketch should show the main features of the graph and also, where possible, values where the graph intersects coordinate axes.
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