Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 2 (P2-9709/02) | Year 2005 | May-Jun | (P2-9709/02) | Q#6

Hits: 24

 

  Question

The diagram shows the part of the curve  for . The curve cuts the x-axis at A and  its maximum point is M.

     i.       Write down the coordinates of A.

   ii.       Show that the x-coordinate of M is e, and write down the y-coordinate of M in terms of e.

  iii.       Find the x-coordinate of M.

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

correct to 2 decimal places.

   v.       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).

Solution

     i.
 

We are required to find the coordinates of point A which is y-intercept 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).

Let us substitute in equation of the curve;

Taking anti-logarithm of both sides;

Since ;

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

 

   ii.
 

We are required to find the coordinates of point M which is given as maximum 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 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:

Therefore;

If  and  are functions of , and if , then;

If , then;

Let  and ;

Rule for differentiation natural logarithmic function , for  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.

Taking anti-logarithm of both sides;

Since ;

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

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

Since for any ;

Therefore;

Hence;

Hence, y-coordinate of the point M is .

 

  iii.
 

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 .

Hence;

1

2

3

4

Therefore;

 

  iv.
 

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.

Please follow and like us:
0

Comments