Past Papers’ Solutions  Cambridge International Examinations (CIE)  AS & A level  Mathematics 9709  Pure Mathematics 1 (P19709/01)  Year 2003  MayJun  (P19709/01)  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 ycoordinate of vertex represents maximum value of and xcoordinate 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’;
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