Past Papers’ Solutions  Cambridge International Examinations (CIE)  AS & A level  Mathematics 9709  Pure Mathematics 1 (P19709/01)  Year 2011  OctNov  (P19709/13)  Q#9
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Question
The functions and are defined by
for
for
i. Express in terms of and solve the equation .
ii. On the same diagram sketch the graphs of and , showing
the coordinates of their point of intersection and the relationship between the graphs.
iii. Find the set of values of which satisfy .
Solution
i.
We are given that;
We can write it as;
First we find .
We can rewrite the function 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’;
Now we solve the equation ;
Hence;
ii.
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 other line is;
Equating both equations;
Single value of x indicates that there is only one intersection point.
With xcoordinate of point of intersection of two lines (or line and the curve) at hand, we can find the ycoordinate of the point of intersection of two lines (or line and the curve) by substituting value of xcoordinate of the point of intersection in any of the two equations;
We choose;
Substituting ;
Hence, coordinates of point of intersection are;
We have;
We can write these as;
SlopeIntercept form of the equation of the line;
Where is the slope of the line.
It is evident both are the equations of line with positive slopes.
To sketch these two line graphs, we first find their and intercepts.
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).
To find intercepts we make coordinates ZERO. 













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).
To find intercepts we make coordinates ZERO. 









Now we have all the necessary data to sketch both lines. Remember that for
, hence do not plot positive xaxis.
But we need to define the domain i.e. values of .
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
Therefore, range of will be domain of . Now we find range of .
We are given that domain of is .
When ;
When ;
When ;
When ;
It is evident that range of can be expressed as;
Hence, domain of is .
However, we are also required to show the relationship between and .
A function and its inverse are reflections of each other in the line .
Therefore we also sketch the line on the same graph along with and as shown below.
Red line is , orange line is and green line is .
iii.
First we find .
We have;
For ;
To find the set of values of x for which ;
We solve the following equation to find critical values of ;
Hence the critical points on the curve for the given condition are & .
Standard form of quadratic equation is;
The graph of quadratic equation is a parabola. If (‘a’ is positive) then parabola opens upwards and its vertex is the minimum point on the graph.
If (‘a’ is negative) then parabola opens downwards and its vertex is the maximum point on the graph.
We recognize that given curve , is a parabola opening upwards.
Therefore conditions for are;
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