# Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 1 (P1-9709/01) | Year 2017 | Oct-Nov | (P1-9709/12) | Q#8

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Question

A curve is such that .

i.       Find the x-coordinate 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 x-coordinate 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 x-coordinate of the stationary point on the curve.

Now we equate given derivative with zero to find the x-coordinate 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 2nd derivative of the curve.

We substitute of the stationary point in the expression of 2nd 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 x-coordinates as;  We find nature of both stationary points by substituting respective value of x-coordinates in .

 For For         Since , the stationary point at is a minimum. Since , the stationary point at is a maximum.

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; 