Past Papers’ Solutions  Cambridge International Examinations (CIE)  AS & A level  Mathematics 9709  Pure Mathematics 1 (P19709/01)  Year 2012  OctNov  (P19709/13)  Q#11
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Question
The diagram shows the curve with equation . The minimum point on the curve has coordinates and the xcoordinate of the maximum point is , here and are constants.
i. State the value of .
ii. Find the value of .
iii. Find the area of the shaded region.
iv. The gradient, , of the curve has a minimum value . Find the value of .
Solution
i.
It is evident from the diagram that point is the xintercept of the given curve.
The point at which curve (or line) intercepts xaxis, the value of . So we can find the value of coordinate by substituting in the equation of the curve (or line).
Therefore, to find the coordinates of xintercept , we can substitute in the equation of the curve.
We have two options.


From the diagram it 






Hence
ii.
It
is evident from the diagram that is the xcoordinate of the stationary point of the curve. Therefore, to find we need to find the coordinates of the stationary point of the curve.
A stationary point on the curve is the point where gradient of the curve is equal to zero;
We are given the equation of the curve as;
We can rewrite the given equation as follows;
Gradient (slope) of the curve is the derivative of equation of the curve. Hence gradient of curve with respect to is:
For the given case;
Rule for differentiation of is:
Rule for differentiation of is:
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;
We have two options.









We already know that is the xintercept of the curve (which is also a stationary point). Hence, xcoordinate of other stationary point of the curve is;
iii.
It is evident from the diagram that shaded region is the area under the curve from origin to xintercept of the curve.
To find the area of region under the curve , we need to integrate the curve from point to along xaxis.
For the given case point A is , point B is and as expanded in (ii);
Hence;
Rule for integration of is:
Rule for integration of is:
iv.
We are given that gradient of the curve has minimum value and we are required to find .
We have found gradient of the curve in (ii) as follows;
We are interested in minimum value of the gradient of the curve (not minimum value of the curve), and in this case gradient of the curve will be treated as a curve and its equation as equation of the curve.
Since minimum value of the this curve (actually gradient of the original curve) is also a stationary point, we need to find this stationary point.
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.
Hence, we find second derivative of the gradient 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 have found gradient of the curve in (ii) as follows;
Therefore;
Rule for differentiation of is:
Rule for differentiation of is:
Rule for differentiation of is:
Since this is stationary point
Hence, stationary value of the gradient of the curve will occur at .
We are given that it is a minimum point and hence minimum value of gradient of the curve will occur at this point.
To find this value we would substitute in the equation of the gradient of the curve;
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