Past Papers’ Solutions  Cambridge International Examinations (CIE)  AS & A level  Mathematics 9709  Pure Mathematics 2 (P29709/02)  Year 2007  MayJun  (P29709/02)  Q#7
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Question
The diagram shows the part of the curve y=e^{x} cos x for . The curve meets the yaxis at the point A. The point M is a maximum point.
i. Write down the coordinates of A.
ii. Find the xcoordinate 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 underestimate or an overestimate of the true value of the integral in part (iii).
Solution
i.
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 yaxis, 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).
ii.
We are required to find the xcoordinate 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:
Therefore;
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 

Domain 

Range 

Periodicity 



Odd/Even 

Translation/ Symmetry 


We utilize the periodicity property to find other solutions of .
However, it will yield values outside the interval of interest .
Hence, only solution, and xcoordinate of 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.
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