# Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 2 (P2-9709/02) | Year 2019 | Oct-Nov | (P2-9709/22) | Q#5

Question

Find the exact coordinates of the stationary point of the curve with equation

Solution

We are required to find the exact 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;

Therefore, we find the expression for gradient of the curve and equate it to ZERO.

We are given equation of the curve;

Therefore first we find from given 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:

Therefore;

We utilize Product Rule to differentiate .

If  and  are functions of , and if , then;

If , then;

Let  and ;

First we differentiate .

Rule for differentiation of natural exponential function is;

Next we differentiate .

Rule for differentiation of  is:

Rule for differentiation of  is:

Rule for differentiation of  is:

Hence;

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;

Now we need expression for gradient of the curve at point P.

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;

We know that point P is the stationary curve of the curve.

Therefore;

Single value of x indicates that there is only one stationary point.

Corresponding values of y coordinate can be found by substituting value of x in  equation of the curve.

Hence, coordinates of the stationary point of the curve are .