Past Papers’ Solutions  Cambridge International Examinations (CIE)  AS & A level  Mathematics 9709  Pure Mathematics 2 (P29709/02)  Year 2005  MayJun  (P29709/02)  Q#6
Hits: 12
Question
The diagram shows the part of the curve for . The curve cuts the xaxis at A and its maximum point is M.
i. Write down the coordinates of A.
ii. Show that the xcoordinate of M is e, and write down the ycoordinate of M in terms of e.
iii. Find the xcoordinate 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 underestimate or an overestimate of the true value of the integral in part (iii).
Solution
i.
We are required to find the coordinates of point A which is yintercept of the curve with equation;
The point at which curve (or line) intercepts yaxis, 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 antilogarithm 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 ilogarithm of both sides;
Since ;
Hence, xcoordinate of point M on the curve is e.
To find the ycoordinate of point M on the curve, we substitute value of xcoordinate in equation of the curve.
Since for any ;
Therefore;
Hence;
Hence, ycoordinate 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.
Comments