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
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
If
We recognize that given curve
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
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
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
It can be seen that the two graphs of
ii.
We are required to verify by calculation that the only root of equation
To use the signchange method we need to write the given equation as
Therefore;
If the function
We can find the signs of
Since
iii.
If we can write the given equation
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
Therefore, if
We have already found in (ii) through signchange rule that root of the given equation lies between
Therefore, for iteration method we use;
We use



0 


1 


2 


3 


4 


5 


6 


7 


8 


9 


10 


11 


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