Past Papers’ Solutions  Edexcel  AS & A level  Mathematics  Core Mathematics 1 (C16663/01)  Year 2012  January  Q#8
Question
The curve has equation
a. Find .
b. Sketch , showing the coordinates of the points where C1 meets the xaxis.
c. Find the gradient of at each point where C1 meets the xaxis.
The curve has equation
where k is a constant and .
d. Sketch , showing the coordinates of the points where meets the x and y axes.
Solution
a.
We are given;
We can rewrite it as;
We are required to find .
Gradient (slope) of the curve is the derivative of equation of the curve. Hence gradient of curve with respect to is:
Therefore;
Rule for differentiation is of is:
Rule for differentiation is of is:
b.
We are required to sketch;
As demonstrated in (a), we can write it as;
It is evident that it is a cubic equation.
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 positive coefficient of will shape the curve at extremities like increasing from left to right.
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.
Hence, the curve crosses yaxis at point .
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 two 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.
c.
We are required o find the gradient of the curve at each point it meets the xaxis.
The curve meets xaxis at two points as demonstrated in (b); and .
First we find the gradient of the curve at point (0,0).
Gradient (slope) of the curve at the particular point is the derivative of equation of the curve at that particular point.
Gradient (slope) of the curve at a particular point can be found by substituting x coordinates of that point in the expression for gradient of the curve;
Therefore, we need .
From (a) we already have found that;
For gradient of the curve at points (0,0), substitute in derivative of the equation of the curve.
Hence, gradient of the curve at points (0,0) is;
Next we find the gradient of the curve at point (2,0).
Gradient (slope) of the curve at the particular point is the derivative of equation of the curve at that particular point.
Gradient (slope) of the curve at a particular point can be found by substituting x coordinates of that point in the expression for gradient of the curve;
Therefore, we need .
From (a) we already have found that;
For gradient of the curve at points (0,0), substitute in derivative of the equation of the curve.
Hence, gradient of the curve at points (2,0) is;
d.
We are required to sketch;
It is evident that we have a translated version of the equation of curve .
Translation through vector transforms the graph of into the graph of .
Transformation of the function into results from translation through vector .
Translation through vector represents the move, units in the positive xdirection and units in the positive ydirection.
Original 
Transformed 
Translation Vector 
Movement 

Function 



units in units in 
Coordinates 


However, for the given case we consider following.
Translation through vector represents the move, units in the positive xdirection and units in the ydirection.
Translation through vector transforms the function into .
Transformation of the function into results from translation through vector .
Translation through vector transforms the function into which means shift towards right along xaxis.
Original 
Transformed 
Translation Vector 
Movement 

Function 



units in 
Coordinates 


It is evident that we are required to transform the function into , therefore it is case of translation of along negative xaxis by k units.
It is also evident from the above table that only xcoordinates of the graph change whereas y coordinates of the graph will remain unchanged.
Hence, the new function has all the ycoordinates same as that of original given function whereas all the xcoordinates are shifted towards positive xaxis of original given function.
Since k>2, we shift graph of original curve as sketched in (b), towards positive xaxis at least 3 units.
However, we need to mark the yintercept of the curve as well.
The point at which curve (or line) intercepts yaxis, the value of . So we can find the value of coordinate by substituting in the equation of the curve (or line).
We have equation of the curve as;
We substitute x=0 to find the coordinates of yintercept.
Hence, coordinates of the yintercept of the curve are .
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