# Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 2 (P2-9709/02) | Year 2010 | Oct-Nov | (P2-9709/23) | 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 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 anti-logarithm of both sides; Since ;  Hence, x-coordinate of point M on the curve is .

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