Past Papers’ Solutions  Cambridge International Examinations (CIE)  AS & A level  Mathematics 9709  Pure Mathematics 1 (P19709/01)  Year 2012  MayJun  (P19709/13)  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 this value of k,
iii. Find an expression for
iv. Sketch, on the same diagram, the graphs of
Solution
i.
Vertex form of a quadratic equation is;
The given curve
Coordinates of the vertex are
Here ycoordinate of vertex represents least 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
If
To find the nature of stationary point we can expand the given function to write it in quadratic form;
Since
Hence, vertex (stationary point)
ii.
For a given function
function.
A oneone function has only one value of
The given function is a parabola and parabola does not pass horizontal line test, so it is not a oneone 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
Therefore, for
iii.
As demonstrated in (ii), function is oneone for the given domain and thus has inverse.
We can write the function as;
To find the inverse of a given function
Interchanging ‘x’ with ‘y’;
iv.
We have;
First we can sketch


Point 

2.0 





2.5 





3.0 





3.5 





4.0 





It can be sketch as shown below.
To sketch
A function
Therefore we just draw a line
Now we sketch a reflection of
Red line is