Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 2 (P2-9709/02) | Year 2007 | May-Jun | (P2-9709/02) | Q#7



The diagram shows the part of the curve y=ex cos x for . The curve meets the y-axis at the  point A. The point M is a maximum point.

Write down the coordinates of A.

ii. Find the x-coordinate of M.

iii. Use the trapezium rule with three intervals to estimate the value of

giving your answer correct to 2 decimal places.

  iv. State, with a reason, whether the trapezium rule gives an under-estimate or an over-estimate of  the true value of the integral in part (iii).



We are required to find coordinates of point A.

It is evident that point A is y intercepts of the curve with equation;

The point at which curve (or line) intercepts y-axis, the value of . So we can find the  value of coordinate by substituting in the equation of the curve (or line). 

Therefore we substitute x=0 in equation of the curve;

Hence, coordinates of point A are (0,1).



We are required to find the x-coordinate of point M which is given as maximum point of the curve  with equation;

The maximum or minimum point of a curve is a stationary point.

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

Since point M is maximum 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:


If  and  are functions of , and if , then;

If , then;

Let  and ;

Rule for differentiation natural exponential function , or ;

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.

Since ;

Using calculator;

Properties of







We utilize the periodicity property to find other solutions of

However, it will yield values outside the interval of interest .

Hence, only solution, and x-coordinate of point M, is .



We are required to apply Trapezium Rule to evaluate;

The trapezium rule with  intervals states that;

We are given that there are three intervals, .

We are also given that and .









If the graph is bending downwards over the whole interval from  to , then trapezium rule will give  an underestimate of the true area (as shown in the diagram below).

It is evident that for the given graph trapezium rule will give an underestimate.