# Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 1 (P1-9709/01) | Year 2017 | Oct-Nov | (P1-9709/13) | Q#10

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Question

A curve has equation y=f(x) and it is given that , where a and b are positive  constants.

i.
Find, in terms of a and b, the non-zero value of x for which the curve has a stationary point and  determine, showing all necessary working, the nature of the stationary point.

ii.       It is now given that the curve has a stationary point at (-2,-3) and that the gradient of the curve  at x=1 is 9. Find .

Solution

i.

We are given that curve y=f(x) has a stationary point.

A stationary point on the curve is the point where gradient of the curve is equal to zero; Gradient (slope) of the curve is the derivative of equation of the curve. Hence gradient of curve with respect to is: We are given that Therefore, at stationary point;   Now we have two options.    Since we are looking the non-zero value of x for which the curve has a stationary point, therefore; Next we are required to determine the nature of stationary point.

Once we have the x-coordinate of the stationary point of a curve, we can determine its  nature, whether minimum or maximum, by finding 2nd 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 are given; Therefore; Rule for differentiation of is:  Rule for differentiation of is:    We have found above that; Therefore;   If or then stationary point (or its value) is minimum.

If or then stationary point (or its value) is maximum.

Hence, stationary point in consideration is a maximum for .

ii.

We are given that the curve has a stationary point at (-2,-3) and that the gradient of the curve at x=1  is 9.

We are required to find .

We can find equation of the curve from its derivative through integration;  To find we first need to find .

We are given that the curve has a stationary point at (-2,-3). Therefore, substitution of x=-2 in must yield ZERO.      We are also given that the gradient of the curve at x=1 is 9.

Gradient (slope) of the curve is the derivative of equation of the curve. Hence gradient of curve with respect to is: Therefore, substitution of x=1 in must yield 9.     We substitute in .    Substitution of in gives us;   Hence; 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 has a stationary point at (-2,-3), therefore, this point lies on the curve.  Hence, we substitute coordinates of this point in the above equation.     Hence, equation of the curve can be written as; 