# Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 1 (P1-9709/01) | Year 2018 | May-Jun | (P1-9709/13) | Q#10

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Question

The one-one function f is defined by  for , where c is a  constant.

i.       State the smallest possible value of c.

In parts (ii) and (iii) the value of c is 4.

ii.       Find an expression for  and state the domain of .

iii.       Solve the equation , giving your answer in the form .

Solution

i.

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.

Vertex form of a quadratic equation is;

We have the function;

It is evident that given function is a quadratic equation in vertex form.

We need to find the smallest possible value of  when  so that is one-one  function.

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.

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

Therefore, we need coordinates of vertex of the given function.

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 . 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.

Therefore, smallest possible value of  is .

ii.

We are given that for ;

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

Since given function is defined for ,  that means   is not possible.

Hence;

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

Therefore, domain of can be found from range of ;

We are given that for ;

To find the range of , we substitute the least value of domain in the function;

Hence range of  is;

Range of a function  Domain of

Hence, domain of ;

iii.

We are required to solve the equation;

We are given that;

Since ;

We are given that for ;

Therefore, is not possible, hence, only possible value is;