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

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  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 an i-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;

If the graph is bending downwards over the whole interval  from  to , then trapezium rule will give  an underestimate of the true area.

We are given that there are two intervals, .

We are also given that and .

Hence;

1

2

3

4

Therefore;

  iv.
 

The trapezium rule with  intervals states that;

If the graph is bending downwards over the whole interval  from  to , then trapezium rule will give  an underestimate of the true area.

The given figure shows that the graph is bending downwards over the whole interval  ,  therefore, area found by trapezium rule will give an underestimate of the true area. 

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