Past Papers’ Solutions  Cambridge International Examinations (CIE)  AS & A level  Mathematics 9709  Pure Mathematics 1 (P19709/01)  Year 2012  OctNov  (P19709/13)  Q#7
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Question
i. The diagram shows part of the curve and part of the straight line meeting at the point , where and are positive constants. Find the values of and .
ii. The function f is defined for the domain by
Express in a similar way.
Solution
i.
It is evident from the diagram that point is the point of intersection of the curve and the straight line . Therefore, we are required to find the coordinates of point of intersection a curve and a line.
If two lines (or a line and a curve) intersect each other at a point then that point lies on both lines i.e. coordinates of that point have same values on both lines (or on the line and the curve). Therefore, we can equate coordinates of both lines i.e. equate equations of both the lines (or the line and the curve).
Equation of the line is;
Equation of the curve is;
Equating both equations;
Now we have two options;






Two values of x indicate that there are two intersection points.
However, it is evident from the diagram that point is in first quadrant and hence its xcoordinate can be positive only. Therefore, we consider;
Corresponding value of y coordinate can be found by substituting value of x in any of the two equation i.e either equation of the line or equation of the curve.
We choose equation of line;
Hence, .
ii.
The function is given piecewise and hence we also find its inverse piecewise.
We take the first part;
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 .
We are given that and from (i) we can write as . Since is always positive, we only consider;
Interchanging ‘x’ with ‘y’;
Since, has its domain restricted to , the inverse must also have its domain restricted.
Domain and range of a function become range and domain, respectively, of its inverse function .
Domain of a function Range of
Range of a function Domain of
So to find domain of , it is easy to find range of .
Range of can be found by substituting extreme values of its domain in its equation.
When ;
When ;
Hence range of can be written as;
Therefore, domain of can be written as;
Now we consider the second part of given function which is;
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’;
Since, has its domain restricted to , the inverse must also have its
domain restricted.
Domain and range of a function become range and domain, respectively, of its inverse function .
Domain of a function Range of
Range of a function Domain of
So to find domain of , it is easy to find range of .
Range of can be found by substituting extreme values of its domain in its equation.
When ;
Hence range of can be written as;
Therefore, domain of can be written as;
Now we can write the inverse of actual function, by combing its pieces according to found domains of each piece.
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