Past Papers’ Solutions | Assessment & Qualification Alliance (AQA) | AS & A level | Mathematics 6360 | Pure Core 1 (6360-MPC1) | Year 2014 | June | Q#4

 

Question

a.    

                           i.       Express  in the form  where p and q are integers.

                         ii.       Hence write down the maximum value of .

b.    

                           i.       Factorise  .

                           ii.       Sketch the curve with equation  , stating the values of x where the curve  crosses the x-axis and the value of the y-intercept.

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;

We are required to find the maximum value of .

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 downwards.

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 downwards 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 curve coordinates of vertex are .

Therefore, maximum value of   is 25.

 

b.
 

                           i.
 

                         ii.
 

We are required to sketch the given equation;

It is evident that it is quadratic equation which represents a parabola. 

First we find the coordinates of x-intercepts of the parabola.

The point  at which curve (or line) intercepts x-axis, the value of . So we can find the  value of  coordinate by substituting  in the equation of the curve (or line). 

Therefore;

As demonstrated in (b:i), we know that;

Therefore;

Now we have two options.

Two values of x indicate that there are two intersection points.

Next we find the y-intercept of the parabola.

The point  at which curve (or line) intercepts y-axis, the value of . So we can find the  value of  coordinate by substituting  in the equation of the curve (or line).

Therefore;

Now we can sketch the parabola represented by  as follows.

desmos-graph (15).png

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