Past Papers’ Solutions  Cambridge International Examinations (CIE)  AS & A level  Mathematics 9709  Pure Mathematics 1 (P19709/01)  Year 2016  OctNov  (P19709/12)  Q#10
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Question
A function f is defined by for .
i. Find the range of f.
ii. Sketch the graph of
iii. Solve the equation , giving answers in terms of .
The function is defined by for , where k is a constant.
iv. State the largest value of k for which g has an inverse.
v. For this value of k, find an expression for .
Solution
i.
We are given the function with its domain.
The function is;
The domain of the given function is;
We are required to find the range of function.
The set of values a function can take against its domain is called range of the function.
Finding range of a function :
· Substitute various values of from given domain into the function to see what is happening to y.
· Make sure you look for minimum and maximum values of y by substituting extreme values of from given domain.
We know following properties of .
Properties of 

Domain 

Range 

We can deduce that for the given function;
We just utilize the extreme values of to find the extreme values of range.
For 
For 








Therefore range of can be expressed as;
ii.
We are required to sketch for .
We can find the points of the graph as follows.




























Now we can sketch required graph from these points as shown below.
iii.
We are required to solve .
We are given that for .
Therefore;
We are required to solve this for .
To solve this equation for ., we can substitute . Hence,
Since given interval is , for interval can be found as follows;
Multiplying the entire inequality with 2;
Since ;
Hence the given interval for is .
To solve equation for interval ,
Using calculator we can find the value of .
We have following properties of .
Properties of 

Domain 

Range 

Odd/Even 

Periodicity 



Translation/ Symmetry 






We utilize the symmetry property of to find other solutions (roots) of :
Translation/ Symmetry 


Therefore;
Hence;
Hence, we have following two solutions of ;


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


For the given case,




For 
For 


Now;
For 















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


Since ;


Therefore;


iv.
The function is defined by for , where k is a constant.
We are required to state the largest value of k for which g has an inverse.
A oneone function has only one value of against one value of . A function is oneone function if it passes horizontal line test i.e. horizontal line passes only through one point of function. However, if a function is not oneone, we can make it so by restricting its domain.
The given function is a sinusoidal as observed in (ii) and sinusoidal does not pass horizontal line test, so it is not a oneone function. However, we can make it so by restricting its domain.
We can see from the graph obtained in (ii) that starting from the graph remains oneone function till where .
After , the graph of function starts repeating itself and hence does not remain onone function.
Hence;
v.
As found in (iv), the function is defined by for .
We have found in (iv) that function is oneone for this domain and hence its inverse exists for this domain only.
We have;
We write it as;
To find the inverse of a given function we need to write it in terms of rather than in terms of .
Interchanging ‘x’ with ‘y’;
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