Past Papers’ Solutions  Edexcel  AS & A level  Mathematics  Core Mathematics 1 (C16663/01)  Year 2010  January  Q#10
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Question
f(x) = x^{2} + 4kx + (3+11k), where k is a constant.
a) Express f(x) in the form (x + p)^{2} + q, where p and q are constants to be found in terms of k.
Given that the equation f(x) = 0 has no real roots,
b) find the set of possible values of k.
Given that k = 1,
c) sketch the graph of y = f(x), showing the coordinates of any point at which the graph crosses a coordinate axis.
Solution
a)
We have the expression;
We use method of “completing square” to obtain the desired form.
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 ;
b)
We are given that;
Therefore, for f(x)=0;
We are given that given equation has no real roots.
For a quadratic equation , the expression for solution is;
Where is called discriminant.
If , the equation will have two distinct roots.
If , the equation will have two identical/repeated roots.
If , the equation will have no roots.
Since given is a quadratic equation with no real roots, its discriminant must be;
We are required to solve the inequality;
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 & 3.
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 it is an upwards opening parabola.
Therefore conditions for are;
c)
We are given that;
Therefore, for k=1;
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 upwards.
Vertex form of a quadratic equation is;
The given curve , as demonstrated in (a) can be written in vertex form as;
We are given that k=1;
Coordinates of the vertex are .Since this is a parabola opening upwards the vertex is the minimum point on the graph.
Here ycoordinate of vertex represents least value of and xcoordinate of vertex represents corresponding value of .
For the given case, vertex is .
Next we need to find the x and y intercepts of the parabola.
The point at which curve (or line) intercepts xaxis, the value of . So we can find the value of coordinate by substituting in the equation of the curve (or line).
Therefore, we substitute y=0 in equation of parabola.
Therefore, there is no real solution and hence the parabola does not intersect the xaxis.
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).
Therefore, we substitute x=0 in equation of parabola.
Hence, parabola intersects yaxis at (0,14).
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