# Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 2 (P2-9709/02) | Year 2013 | Oct-Nov | (P2-9709/22) | Q#2

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Question

The curve  has one stationary point. Find the coordinates of this stationary point.

Solution

We are required to find the coordinates of point M which is minimum point of the curve;

A stationary point on the curve is the point where gradient of the curve is equal to zero;

Since point M is minimum point, therefore, it is stationary point of the curve and, hence, gradient of  the curve at point M must ZERO.

We can find expression for gradient of the curve at point M and equate it with ZERO to find the x- coordinate of point M.

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

Therefore;

If  and  are functions of , and if , then;

If , then;

Let  and ;

Rule for differentiation of natural exponential function , or ;

Rule for differentiation of  is:

Rule for differentiation of  is:

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

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;

Since point M is a maximum point, the gradient of the curve at this point must be equal to ZERO.

Hence, x-coordinate of point M on the curve is .

To find the y-coordinate of point M on the curve, we substitute value of x-coordinate in equation of  the curve.

Hence;

Hence, y-coordinate of the point M is .

Therefore, coordinates of point M