Past Papers’ Solutions | Edexcel | AS & A level | Mathematics | Core Mathematics 1 (C1-6663/01R) | Year 2013 | June | Q#8

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Question

A rectangular room has a width of x m.

The length of the room is 4 m longer than its width.

Given that the perimeter of the room is greater than 19.2 m,

a.   Show that

Given also that the area of the room is less than 21 m2,

b.    

           i. Write down an inequality, in terms of x, for the area of the room.

          ii. Solve this inequality.

c.   Hence find the range of possible values of x.

Solution

a.
 

We can collect following data from the given equation.

Expression for perimeter of a rectangle with width  and length   is:

Therefore;

b.
         

            i.
 

We are also given that the area of the room is less than 21 m2.

Expression for area of a rectangle with width  and length   is:

Therefore;


              ii.
 

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;

Since x represents width of the rectangle, it cannot be negative, therefore;

c.    

As demonstrated in (a), due to perimeter greater than 19.2 we need .

We have also seen in (b:ii) that  for area to be less than 21.

Hence, only possible range of values of x is;

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