Past Papers’ Solutions  Cambridge International Examinations (CIE)  AS & A level  Mathematics 9709  Pure Mathematics 2 (P29709/02)  Year 2016  MayJun  (P29709/23)  Q#5
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Question
The equation of a curve is . At the point on the curve with xcoordinate p, the gradient of the curve is .
i. Show that .
ii. Show by calculation that 3.3 < p < 3.5.
iii. Use an iterative formula based on the equation in part (i) to find the value of p correct to 3
decimal places. Give the result of each iteration to 5 decimal places.
Solution
i.
We are given that;
First we are required to find the xcoordinate of the point where gradient of the curve is .
Therefore, first find the expression for gradient of this curve.
Gradient (slope) of the curve is the derivative of equation of the curve. Hence gradient of curve with respect to is:
Therefore;
We utilize the Product Rule to differentiate.
If and are functions of , and if , then;
If , then;
Let and , then;
Rule for differentiation of natural exponential function , or ;
Rule for differentiation of is:
Since at this point the gradient of the curve is 40.
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;
Hence;
Since ;
Taking logarithm of both sides;
for any
ii.
We are required to show by calculation that the xcoordinate of the point i.e., p lies between 3.3 and 3.5.
We need to use signchange rule.
To use the signchange method we need to write the given equation as .
From (i) we have;
Therefore;
Therefore;
If the function is continuous in an interval of its domain, and if and have opposite signs, then has at least one root between and .
We can find the signs of at and as follows;
Since and have opposite signs for function , the function has root between and .
iii.
Iteration method can be used to find the root of the given equation using sequence defined by;
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 .
We have already found in (ii) through signchange rule that root of the given equation lies between
and .
Therefore, for iteration method we use;
We use as initial value.



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It is evident that .
Hence, is a root of .
The root given correct to 3 decimal places is 3.412.
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