Past Papers’ Solutions  Cambridge International Examinations (CIE)  AS & A level  Mathematics 9709  Pure Mathematics 2 (P29709/02)  Year 2013  MayJun  (P29709/23)  Q#7
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Question
i. Express in the form , where and , giving the exact value of R and the value of a correct to 2 decimal places.
ii. Hence solve the equation
Giving all solutions in the interval correct to 1 decimal place.
iii. Determine the least value of as varies.
Solution
i.
We are given the expression;
We are required to write it in the form;
If and are positive, then;
can be written in the form
can be written in the form
where,
and , , with
Considering the given equation, we have following case at hand;
can be written in the for
Comparing it with given equation Therefore


Therefore;
Finally, we can find , utilizing the equation;
Using calculator we can find that;
Therefore;
ii.
We are required to solve the equation;
As demonstrated in (i), we can write;
Therefore, we need to solve;
Let , then:
Using calculator we can find that;
We utilize the symmetry property of to find another solution (root) of :
Properties of 

Domain 

Range 

Odd/Even 

Periodicity 



Translation/ Symmetry 






Hence;
Therefore;
We have found two solutions of equation .


Since given interval is , for interval can be found as follows;
Multiplying the entire inequality with 2;
Adding in the entire inequality;
Since ;
Hence the given interval for is .
To find all other solutions of the equation in the given interval we utilize the periodicity property of :
For the given case,




For 
For 


Now;
For 















Only following solutions (roots) of the equation are within interval;




Since ;












iii.
We are required the find the least value of
As demonstrated in (i), we can write;
Hence;
To find the least value of we need to find the greatest value of because it is in the denominator.
We know that , hence the greatest value of will occur when and therefore;
Hence the least value of will be ;
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