Past Papers’ Solutions  Assessment & Qualification Alliance (AQA)  AS & A level  Mathematics 6360  Pure Core 1 (6360MPC1)  Year 2012  June  Q#5
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Question
a.
i. Express in the form .
ii. Hence write down the equation of the line of symmetry of the curve with equation .
b. The curve C with equation and the straight line intersect at the point A(0,5) and at the point B, as shown in the diagram below.
i. Find the coordinates of the point B.
ii. Find
iii. Find the area of the shaded region R bounded by the curve C and the line segment AB.
Solution
a.
i.
We have the expression;
We use method of “completing square” to obtain the desired form.
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 ;
ii.
We are given equation of the curve as;
We are required to write equation of line of symmetry of this curve.
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:i) can be written in vertex form as;
Coordinates of the vertex are .Since this is a parabola opening down the vertex is the maximum point on the graph. Here ycoordinate of vertex represents maximum value of and x coordinate of vertex represents corresponding value of .
For the given case, vertex is .
Line of symmetry of a parabola is the vertical line passing through its vertex.
Therefore, for the given case vertex is and equation of line of symmetry is;
b.
i.
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).
Equation of the line is;
Equation of the curve is;
Equating both equations;
Now we have two options.



Two values of x indicate that there are two intersection points.
It is evident from the diagram that is for point A on the curve, therefore, for B point .
With xcoordinate of point of intersection of two lines (or line and the curve) at hand, we can find the ycoordinate of the point of intersection of two lines (or line and the curve) by substituting value of xcoordinate of the point of intersection in any of the two equations.
We choose;
Hence, coordinates of point B are (4,9).
ii.
Rule for integration of is:
Rule for integration of is:
Rule for integration of is:
iii.
It is evident from the diagram that;
First we find area under line.
To find the area of region under the curve , we need to integrate the curve from point to along xaxis.
We are given equation of line as;
It is evident from the diagram that desired area under line extends from (Point A) to (point B). Hence;
Rule for integration of is:
Rule for integration of is:
Rule for integration of is:
Next we find area under curve.
To find the area of region under the curve , we need to integrate the curve from point to along xaxis.
We are given equation of curve as;
It is evident from the diagram that desired area under curve extends from (Point A) to (point B). Hence;
We have found in (c:ii) that;
Therefore;
Finally, we can find the desired area f shaded region;
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