Past Papers’ Solutions  Cambridge International Examinations (CIE)  AS & A level  Mathematics 9709  Pure Mathematics 1 (P19709/01)  Year 2015  OctNov  (P19709/13)  Q#7
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Question
a. Show that the equation can be expressed as
and hence solve the equation for .
b.
The diagram shows part of the graph of , where a and b are constants. The graph crosses the xaxis at the point and the yaxis at the point ,. Find c and d in terms of a and b.
Solution
a.
We are given the equation;
We know that
Therefore;
We have the trigonometric identity;
It can be rearranged as;
Therefore;
Now we are required to solve for .
We have seen above that can be expressed as;
Let;
Therefore can be written as,
Now we have two options;









Since;
We have two options;


We know that;
Therefore is not possible. Hence;
Using calculator we can find the value of .
We utilize the periodic property of to find other solutions (roots) of :


Symmetry 

Hence;
Therefore, another possible solution is;
Therefore, we have two solutions (roots) of the equation;


To find all the solutions (roots) over the interval , we utilize the periodic property of for both these values of .


Periodic 
or

Therefore;


For 
For 




For















Hence all the solutions (roots) of the equation for are;


b.
We are given coordinates of x and y intercepts of the graph with equation;
The coordinates of xintercept are while that of yintercept are .
First we consider xintercept of the graph.
The point at which curve (or line) intercepts xaxis, the value of . So we can find the value of coordinate by substituting in the equation of the curve (or line).
Therefore substitution of and in the equation of the graph;
First we consider yintercept of the graph.
The point at which curve (or line) intercepts yaxis, the value of . So we can find the value of coordinate by substituting in the equation of the curve (or line).
Therefore substitution of and in the equation of the graph;
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