Past Papers’ Solutions  Cambridge International Examinations (CIE)  AS & A level  Mathematics 9709  Pure Mathematics 2 (P29709/02)  Year 2012  MayJun  (P29709/22)  Q#6
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Question
The diagram shows the curve , for . The xcoordinate of the maximum point M is denoted by .
i. Find and show that satisfies the equation tan 2x = 2x + 4.
ii. Show by calculation that lies between 0.6 and 0.7.
iii. Use the iterative formula to find the value of a correct to 3 decimal places.
Solution
i.
We are required to find the equation for xcoordinate 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 ;
We differentiate and one by one.
First we differentiate ;
If we define , then derivative of is;
If we have and then derivative of is;
Let , then , therefore;
Rule for differentiation of is;
Rule for differentiation of is:
First we differentiate ;
Rule for differentiation of is:
Rule for differentiation of is:
Rule for differentiation of is:
Finally;
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.
Since ;
ii.
We are required to verify by calculation that lies between 0.6 and 0.7. We need to use sign change rule.
To use the signchange method we need to write the given equation as .
From (i) we have found;
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.
If we can write the given equation and transform it to , then we can find the root of the equation by iteration method using sequence defined as.
We are also given the iterative formula as;
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 .
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