Past Papers’ Solutions  Cambridge International Examinations (CIE)  AS & A level  Mathematics 9709  Pure Mathematics 2 (P29709/02)  Year 2003  OctNov  (P29709/02)  Q#5
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Question
i. By sketching a suitable pair of graphs, for x < 0, show that exactly one root of the equation is negative.
ii. Verify by calculation that this root lies between 1.0 and 0.5.
iii. Use the iterative formula
to determine the root correct to 2 decimal places, showing the result of each iteration.
Solution
i.
We are required to show that there is only one root of the following equation by sketching.
Root of an equation is the xcoordinate of a point of intersection of the graphs of and .
Therefore, first we sketch .
It can be seen that it is a quadratic equation;
To sketch a quadratic equation, a parabola, we need the coordinates of its vertex and x and y intercepts, if any.
First we find the coordinates of vertex of this parabola.
Standard form of quadratic equation is;
The graph of quadratic equation is a parabola. If (‘a’ is positive) then parabola opens upwards and its vertex is the minimum point on the graph.
If (‘a’ is negative) then parabola opens downwards and its vertex is the maximum point on the graph.
We recognize that given curve , is a parabola opening downwards.
Vertex form of a quadratic equation is;
We first write given quadratic equation in vertex form.
We have the equation;
For the given case, vertex is .
Next, we need x and yintercepts of the parabola.
Since vertex of parabola is on there would be no x and y intercepts as is evident from the sketch of parabola as shown below.
For the case x < 0 , we can restrict the sketch as shown below.
Next we sketch .
It is evident that it is an exponential function.
For an exponential function
Where
Some properties which help to plot/sketch this graph are as follows.
· The graphs of all exponential functions contain the point .
· The domain is all real numbers .
· The range is only the positive .
· The graph is increasing.
· The graph is asymptotic to the xaxis as x approaches negative infinity
· The graph increases without bound as x approaches positive infinity
· The graph is continuous and smooth.
Therefore the graph of will look like as shown below.
To be more accurate we may like to calculate coordinates of a couple of points of the graph and then sketch it.
We can calculate a couple of values to be accurate in sketching the graph.

3 
2 
1 
0 
1 
2 
3 
4 
5 

0.125 
0.25 
0.5 
1 
2 
4 
8 
16 
32 
We can plot these points to sketch the following graph of
However, as per given condition, we restrict the graph to desired interval .
Sketching both graphs on the same axes to desired interval , we get following.
It can be seen that the two graphs of and intersect each other at only a single point and that too on the negative xaxis side, therefore, the equation has a single roots and at a single value of which is negative.
ii.
We are required to verify by calculation that the only root of equation lies between 1.0 and 0.5. 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 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 required to find a root of the equation
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.



0 


1 


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3 


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5 


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11 


It is evident that .
Hence, is a root of .
The root given correct to 2 significant figures is 0.77.
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