Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 1 (P1-9709/01) | Year 2011 | Oct-Nov | (P1-9709/13) | Q#9

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The functions  and  are defined by



     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 .



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 ;



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 x-coordinate of point of intersection of two lines (or line and the curve) at hand, we can find the y-coordinate of the point of intersection of two lines (or line and the curve) by substituting value of x-coordinate 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;

Slope-Intercept 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 x-axis, 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 y-axis, 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 x-axis.

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 .


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.
 (‘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;