Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 2 (P2-9709/02) | Year 2011 | Oct-Nov | (P2-9709/21) | Q#3

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The diagram shows the part of the curve  for . Find the x-coordinates of the  points on this part of the curve at which the gradient is 4.


We are required to find the x-coordinate of the points on the curve where gradient is 4.

Therefore first we need to find the expression for gradient of the given curve.

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;

We are given;


For  we use chain rule.

If we define , then derivative of  is;

If we have  and then derivative of  is;

Let , then;

Rule for differentiation of  is:

Rule for differentiation of  is;

Therefore, we can equate expression of gradient of the curve with 4.

  provided that

Now we have two options.

Using calculator, we can find;


Properties of







We utilize the odd/even property of   to find another solution (root) of :


Therefore, we have four solutions of of ;

However, only following values of are within the given interval .

Hence, these are the x-coordinates of the points on the curve where curve has gradient 4.