Past Papers’ Solutions  Cambridge International Examinations (CIE)  AS & A level  Mathematics 9709  Pure Mathematics 1 (P19709/01)  Year 2016  OctNov  (P19709/11)  Q#10
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Question
A curve has equation and it is given that . The point A is the only point on the curve at which the gradient is −1.
i. Find the xcoordinate of A.
ii. Given that the curve also passes through the point (4,10), find the ycoordinate of A, giving your answer as a fraction.
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:
Therefore;
We are also given that point A is the only point on the curve at which the gradient is −1.
Gradient (slope) of the curve at the particular point is the derivative of equation of the curve at that particular point.
Gradient (slope) of the curve at a particular point can be found by substituting x coordinates of that point in the expression for gradient of the curve;
Therefore;
Hence;
Now we have two options.









Two possible values of imply that there are two points on the curve, one at each value of , where curve has gradient 1.
But we are given that point A is the only point on the curve at which the gradient is −1.
We can see that substitution of both values of gives;
For 
For 











Hence, is not valid option. 
Therefore;
ii.
We are required to find the ycoordinate of point A.
To find ycoordinate of a point on the curve, we substitute value of xcoordinate of the point on the curve in the equation of the curve.
We have already found xcoordinate of point A in (i).
Therefore, we need equation of the curve.
We are given that;
We can find equation of the curve from its derivative through integration;
Therefore;
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. .
We are given that the curve passes through the point (4,10).
Therefore, substituting and in the above derived equation of the curve;
Hence, equation of the curve is;
Now we can find ycoordinate of point A by substituting xcoordinate of point A in equation of the curve;
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