Past Papers’ Solutions  Edexcel  AS & A level  Mathematics  Core Mathematics 1 (C16663/01)  Year 2011  January  Q#10
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Question
a. On the axes below, sketch the graphs of
i.
ii.
showing clearly the coordinates of all the points where the curves cross the coordinate axes.
b. Using your sketch state, giving a reason, the number of real solutions to the equation
Solution
a.
i.
We are required to sketch;
We need to expand it in order to sketch it.
We can now sketch the curve as follows.
ü Find the sign of the coefficient of . This gives the shape of the graph at the extremities.
It is evident that with negative coefficient of will shape the curve at extremities like decreasing from right to left.
ü Find the point where the graph crosses yaxis by finding the value of when .
We can find the coordinates of yintercept from the given equation of the curve.
ü Find the point(s) where the graph crosses the xaxis by finding the value of when . If there is repeated root the graph will touch the xaxis.
We can find the coordinates of xintercepts from the given equation of the curve.
Now we have three options.








Hence, the curve crosses xaxis at points , and .
ü Calculate the values of for some value of . This is particularly useful in determining the quadrant in which the graph might turn close to the yaxis.
ü Complete the sketch of the graph by joining the sections.
ü Sketch should show the main features of the graph and also, where possible, values where the graph intersects coordinate axes.
ii.
We are required to sketch;
If we first sketch , it would become easy to sketch .
A graph of the form is known as a reciprocal graph and looks like;
Next we sketch;
Transformation of the function into results from ‘stretch’ of in y direction by a scale factor of .
‘Stretch’ of the function in ydirection by a scale factor of transforms into .


Original 
Transformed 
Effect 

Function 


Expansion 
Coordinates 



Function 


Shrinking 
Coordinates 



Function 


Shrinking 
Coordinates 



Function 


Expansion 
Coordinates 


It is evident that is a case of vertical expansion by a scale factor of 2.
Now we are ready to sketch;
Transformation of the function into results from reflection of in xaxis.
Reflection of the function in xaxis transforms into .
Original 
Transformed 
Reflection 

Function 


xaxis 
Coordinates 


It is evident that we are required to transform the function into , therefore it is case of reflection of the given function in xaxis.
It is also evident from the above table that only ycoordinates of the graph change whereas x coordinates of the graph will remain unchanged.
Hence, the new function has all the xcoordinates same as that of original given function whereas all the ycoordinates are negative of original given function.
It is shown in the figure below.
b.
We are given that;
It is evident that this equation is obtained by equating the equations of the two given graphs in (a).
If two lines (or a line and a curve) intersect each other at a point then that point lies on both lines i.e. coordinates of that point have same values on both lines (or on the line and the curve). Therefore, we can equate coordinates of both lines i.e. equate equations of both the lines (or the line and the curve).
Therefore, the equation given in (b) gives the point(s) of intersection of the two given graphs in (a).
We can sketch the both graphs sketched in (a) on a single graph as shown below.
Since the two graphs intersect each other at two distinct points, therefore, there are will be only two solutions of the given equation.
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