# Past Papers’ Solutions | Assessment & Qualification Alliance (AQA) | AS & A level | Mathematics 6360 | Pure Core 1 (6360-MPC1) | 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 y-coordinate 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 x-coordinate of point of intersection of two lines (or line and the curve) at hand, we can find the  y-coordinate of the point of intersection of two lines (or line and the curve) by substituting value  of x-coordinate 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 x-axis.

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 x-axis.

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;