# Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 1 (P1-9709/01) | Year 2019 | Feb-Mar | (P1-9709/12) | Q#4

Question

A curve has equation .

i.       Find  and  .

ii.       Find the x-coordinates of the stationary points and, showing all necessary working, determine  the nature of each stationary point.

Solution

i.

We are required to find and  .

Therefore, we find the derivative of equation of the curve.

Gradient (slope) of the curve is the derivative of equation of the curve. Hence gradient of curve  with respect to is:

We are given equation of the curve as;

Therefore;

Rule for differentiation of  is:

Rule for differentiation of  is:

Rule for differentiation of  is:

Next, 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;

We have found in (i) that;

Therefore;

Rule for differentiation of  is:

Rule for differentiation of  is:

Rule for differentiation of  is:

ii.

We are required to find the coordinates of the stationary point of 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.

We have found in (i) that;

Therefore;

Now we have two options.

Two values of indicate two stationary points on the curve.

Next, we are required to find the nature of each 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.

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.

Above, we have found x-coordinates of stationary points on the curve as 0 & 1.

Therefore, substituting  and in expression of second derivative we can find nature of  both stationary points of the curve.

We have found in (i);

 For For Maximum Minimum