Past Papers’ Solutions  Cambridge International Examinations (CIE)  AS & A level  Mathematics 9709  Pure Mathematics 2 (P29709/02)  Year 2010  OctNov  (P29709/23)  Q#7
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Question
The diagram shows the curve and its maximum point M.
i. Find the exact coordinates of M.
ii. Use the trapezium rule with three intervals to estimate the value of
giving your answer correct to 2 decimal places.
Solution
i.
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.
Power Rule;
Taking antilogarithm of both sides;
Since ;
Hence, xcoordinate of point M on the curve is .
To find the ycoordinate of point M on the curve, we substitute value of xcoordinate in equation of the curve.
Since ;
Hence;
Hence, ycoordinate of the point M is .
ii.
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;
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