# Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 1 (P1-9709/01) | Year 2012 | May-Jun | (P1-9709/13) | Q#11

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Question

The function  is such that , for .

i. Find the coordinates and the nature of the stationary point on the curve .

The function g is such that , , where  is a constant.

ii. State the smallest value of  for which  has an inverse.

For this value of k,

iii. Find an expression for .

iv. Sketch, on the same diagram, the graphs of  and .

Solution

i.

Vertex form of a quadratic equation is;

The given curve , is in vertex form.

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, a stationary point, is .

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.

To find the nature of stationary point we can expand the given function to write it in quadratic form;

Since  (‘a’ is negative), parabola opens downwards and its vertex is the maximum point on the graph.

Hence, vertex (stationary point)  of the function   is  maximum.

ii.

For a given function  the inverse function  exists only if  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.

Vertex of the given parabola, as demonstrated in (i) 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.

iii.

, ,

As demonstrated in (ii), function is one-one for the given domain and thus has inverse.

We can write 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’;

iv.

We have;

,

First we can sketch , as follows, but remember that  for .

 Point 2.0 2.5 3.0 3.5 4.0

It can be sketch as shown below.

To sketch  which is inverse of  we need not to do tedious calculations. We just need to know following;

A function  and its inverse  are reflections of each other in the line .

Therefore we just draw a line .

Now we sketch a reflection of  in this line. That will make sketch of .

Red line is , blue line is  and orange line is .