Past Papers’ Solutions  Cambridge International Examinations (CIE)  AS & A level  Mathematics 9709  Pure Mathematics 2 (P29709/02)  Year 2018  MayJun  (P29709/21)  Q#4
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Question
The diagram shows the curve .
The curve crosses the xaxis at the point P and has a maximum point M.
i. Find the gradient of the curve at the point P.
ii. Show that the xcoordinate of the point M satisfies the equation
iii. Use an iterative formula based on the equation in part (ii) to find the xcoordinate of M correct to 4 significant figures. Show the result of each iteration to 6 significant figures.
Solution
i.
We are required to the gradient of the curve at P which is xintercept of the curve.
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 are given equation of the curve;
Therefore first we find from given equation of the curve.
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 ;
First we differentiate .
Rule for differentiation natural logarithmic function , for is;
Next we differentiate .
Rule for differentiation of is:
Rule for differentiation of is:
Rule for differentiation of is:
Hence;
Now we need expression for gradient of the curve at point P.
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 know that point P is the xintercept of the curve.
The point at which curve (or line) intercepts xaxis, the value of . So we can find the value of coordinate by substituting in the equation of the curve (or line).
Hence, to find the xcoordinate of point, we substitute in given equation of the curve.
and are inverse functions. The composite function is an identity function, with domain the positive real numbers. Therefore;
Taking antilogarithm of both sides;
Therefore, coordinates of point P are (1, 0).
Hence;
ii.
We are required to find the xcoordinates 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, 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.
From (i) we have;
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 stationary point, the gradient of the curve at this point must be equal to ZERO.
Hence, xcoordinate of point M on the curve is .
iii.
We are required to apply an iterative formula based on the equation in part (ii) to find the value of m correct to 4 significant figures.
We have found in (i) that;
We can write an iterative formula based on this equation.
If the sequence given by the inductive definition , with some initial value , converges to a limit , then is the root of the equation .
Therefore, if , then is a root of .
It is evident from the diagram that root of the curve lies where .
We use , as initial value since will yield , whereas will yield zero in denominator.



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It is evident that .
Hence, is a root of .
The root given correct to 4 significant figures is 3.181.
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