# Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 1 (P1-9709/01) | Year 2003 | May-Jun | (P1-9709/01) | Q#11

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Question

The equation of a curve is .

i. Express  in the form , stating the numerical values of  and .

ii. Hence, or otherwise, find the coordinates of the stationary point of the curve.

iii. Find the set of values of  for which .

The function g is defined by  for .

iv. State the domain and range of .

v. Find an expression, in terms of , for

Solution

i.

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 ;

Hence;

ii.

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 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 downwards the vertex is the  maximum point on the graph.
Here y-coordinate of vertex represents maximum value of
and x-coordinate of vertex represents corresponding value of .

For the given case, vertex is . Therefore, least value of  is 16 and corresponding value of  is 4.

Vertex of a parabola is a stationary point on the curve.

Hence coordinates of the stationary point on the given curve  are;

iii.

To find the set of values of x for which ;

We solve the following equation to find critical values of ;

Now we have two options;

Hence the critical points on the curve for the given condition are -2 & 10.

Since, as demonstrated in (ii) it is an downwards opening parabola.

Therefore conditions for  are;

Therefore, the desired interval is;

iv.

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, if we have domain and range of , we can find domain and range of .

The given function is  for .

The set of numbers  for which function  is defined is called domain of the function.

We are given domain of  as;

The set of values a function  can take against its domain is called range of the function.

Finding range of a function:

· Substitute various values of  from given domain into the function to see what is happening to y.

· Make sure you look for minimum and maximum values of y by substituting extreme values of   from given domain.

We can find range of  for  as follows.

Hence range of  is;

Therefore;

Domain of  is;

Range of  is;

v.

We have;

We write it as;

To find the 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’;