# Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 2 (P2-9709/02) | Year 2012 | Oct-Nov | (P2-9709/22) | Q#2

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** Question**

The curve with equation has one stationary point in the interval . Find the exact x-coordinate of this point.

**Solution**

We are required to find the x-coordinate of stationary point of the curve with equation;

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

Since it is stationary point of the curve and, hence, gradient of the curve at this point must be ZERO.

We can find expression for gradient of the curve at this point 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 ;

We solve derivative of and one by one.

First we solve .

If we define , then derivative of is;

If we have and then derivative of is;

Let and , then;

Rule for differentiation of is;

Rule for differentiation of is:

Since ;

Next, we solve .

Let . Then;

Rule for differentiation natural exponential function , or ;

Hence;

Now we need expression for gradient of the curve at stationary 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;

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

Using calculator;

Hence x-coordinate of stationary point on the curve is .

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