Past Papers’ Solutions  Cambridge International Examinations (CIE)  AS & A level  Mathematics 9709  Pure Mathematics 2 (P29709/02)  Year 2007  OctNov  (P29709/02)  Q#8
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Question
The diagram shows the curve and its maximum point M.
i. Find the xcoordinate of M.
ii. Show that the tangent to the curve at the point where x = 1 passes through the origin.
iii. Use the trapezium rule with two 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 maximum point of the curve;
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.
Hence, xcoordinate of point M on the curve is .
ii.
If coordinates of a point satisfies the equation of a curve (or line) then that particular point lies on that curve (or line).
Therefore, to show whether tangent to the given curve at x=1 passes through the origin we first need to find equation of the tangent to the curve at point where x=1.
Hence, we are required to find the equation of tangent to the curve at the point where .
To find the equation of the line either we need coordinates of the two points on the line (TwoPoint form of Equation of Line) or coordinates of one point on the line and slope of the line (PointSlope form of Equation of Line).
We need coordinates of the point on the curve (and tangent).
We are given equation of the curve;
We substitute desired value of xcoordinate in equation of curve to find the ycoordinates of this point on the curve (and tangent).
Therefore, coordinates of the point on the curve (and tangent) are .
Next we need to find slope of tangent at this point in order to write its equation.
The slope of a curve at a particular point is equal to the slope of the tangent to the curve at the same point;
Therefore, if we can find slope of the curve C at point then we can find slope of the tangent to the curve at this point.
Gradient (slope) of the curve at the particular point is the derivative of equation of the curve at that particular point.
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;
We have found in (i) that;
Therefore;
Hence;
With coordinates of a point on the tangent and its slope in hand, we can write equation of the tangent.
PointSlope form of the equation of the line is;
Now we can substitute coordinates of origin (0,0) in this equation of tangent to see whether it passes through origin or not.
Hence, equation of tangent is satisfied by coordinates of the origin, therefore, origin lies on the tangent.
iii.
We are required to apply Trapezium Rule to evaluate;
The trapezium rule with intervals states that;
We are given that there are two intervals, .
We are also given that and .
Hence;




1 



2 



3 


Therefore;
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