Past Papers’ Solutions  Cambridge International Examinations (CIE)  AS & A level  Mathematics 9709  Pure Mathematics 2 (P29709/02)  Year 2009  OctNov  (P29709/22)  Q#7
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Question
The diagram shows the curve y = x^{2} cos x, for , and its maximum point M.
i. Show by differentiation that the xcoordinate of M satisfies the equation
ii. Verify by calculation that this equation has a root (in radians) between 1 and 1.2.
iii. Use the iterative formula to determine this root correct to 2 decimal places. Give the result of each iteration to 4 decimal places.
Solution
i.
We are required to find the equation for xcoordinates 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 of 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.
Since ;
Hence, xcoordinate of point M is found from .
ii.
We are required to verify by calculation that the root of equation lies between 1 and 1.2 radians. We need to use signchange rule.
To use the signchange method we need to write the given equation as .
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.
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 are given the iterative formula as;
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 2 decimal places is 1.08.
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