Past Papers’ Solutions | Assessment & Qualification Alliance (AQA) | AS & A level | Mathematics 6360 | Pure Core 1 (6360-MPC1) | Year 2005 | January | Q#4

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a.   The function f is defined for all values of  by  .

                           i.       Find the remainder when  is divided by .

                         ii.       Given that  and  , write down two linear factors of .

                       iii.       Hence express  as the product of three linear factors.

b.   The curve with equation  is sketched below.

                           i.       The curve intersects the y-axis at the point . Find the y-coordinate of .

                         ii.       The curve crosses the x-axis when  , when  and also at the point .

                       iii.       Use the results from part (a) to find the x-coordinate of .


                           i.       Find

                         ii.       Hence find the area of the shaded region bounded by the curve and the x-axis.




Remainder theorem states that if  is divided by  then;

For the given case  is divided by .

We can write the divisor in standard form as;

Here  and . Hence;


Factor theorem states that if  is a factor of   then; 

For the given case;






If  is a polynomial of degree  then  will have exactly  factors, some of which may repeat.

We are given a polynomial of degree ,

Therefore, it will have 03 factors some of which may repeat.

From (a:ii) we already have 02 factors  and .

We may divide the given polynomial with any of the already known linear factor(s) to get a quadratic  factor and then factorize the obtained quadratic factor to find the 2nd and 3rd linear  factors.

For the given case;

Already known factors from (a:ii) are  and .

We may divide the given polynomial by any of these two factors. We choose .


Therefore, we get the quadratic factor  for given polynomial. 

Now we factorize this quadratic factor.

Hence,  can be written as product of three linear factors as follows;



Equation of the given curve is;

We are given that point  is the y-intercept of the given curve.

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, at ;


 From (a:iii) we have;

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).

Since for each root/factor , that means curve intersects x-axis at each factor.

Therefore, x-intercepts of the given curve can be found from its factors;



Rule for integration of  is:

Rule for integration of  is:

Rule for integration of  is:


To find the area of region under the curve , we need to integrate the curve from point  to   along x-axis.

It is evident from the diagram that shaded region lies under the curve and extends from  to  .


Rule for integration of  is:

From (c:i) we have