Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 1 (P1-9709/01) | Year 2015 | Oct-Nov | (P1-9709/11) | Q#9

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     i.       Express  in the form , where a, b and c are constants.

The function  is defined for , where  is a constant.

   ii.       State the smallest value of  for which  is one-one.

  iii.       For the case where , find n expression for and state the domain of .



We have the expression;

We use method of “completing square” to obtain the desired form. We take out factor ‘-1’ from the  terms which involve ;

Next we complete the square for the terms which involve .

We have the algebraic formula;

For the given case we can compare the given terms with the formula as below;

Therefore we can deduce that;

Hence we can write;

To complete the square we can add and subtract the deduced value of ;


We are given function  for .

We can write the given function as;

It is evident that we have a quadratic equation.

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.

It is evident that given function is a parabola opening downwards. Let’s check for what values of  given function is one-one.

A one-one function has only one value of  against one value of . A function is one-one function if  it passes horizontal line test i.e. horizontal line passes only through one point of function. However,  if a function is not one-one, we can make it so by restricting its domain.

The given function is a parabola and parabola does not pass horizontal line test, so it is not a one- one function. However, we can make it so by restricting its domain around its line of symmetry.

We already have recognized that given curve , is a parabola opening downwards.

Vertex form of a quadratic equation is;

The given curve , as demonstrated in (i) can be written in vertex form as;

Coordinates of the vertex are .Since this is a parabola opening upwards the vertex is the  minimum point on the graph. 

Here y-coordinate of vertex represents least value of  and x-coordinate of vertex represents  corresponding value of .

For the given case, vertex is .

The line of symmetry of a parabola is a vertical line passing through its vertex.

Vertex of the given parabola is . Hence the vertical line passing through the vertex is .  By restricting the domain of the given function to , we can make it one-one function. On both  sides of the vertical line  given function is a one-one function.

Therefore, for , given function is a one-one function.


We are given that  is defined for , when  is a constant.


Since , domain of  is .

To find he inverse of a given function  we need to write it in terms of  rather than in terms of .

As demonstrated in (i), we can write the given function as;

Interchanging ‘x’ with ‘y’;

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 

The given function is is defined for .

Domain of  is;

For range of ;

As demonstrated in (i), we can write it as;

Hence range of  is;

We know that;

Range of a function  Domain of


Domain of

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