# Past Papers’ Solutions | Assessment & Qualification Alliance (AQA) | AS & A level | Mathematics 6360 | Pure Core 1 (6360-MPC1) | Year 2009 | June | Q#4

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Question

a.  The polynomial  is given by

i.       Find the remainder when  is divided by .

ii.       Use the Factor Theorem to show that  is a factor of .

iii.       Express  in the form  , where b and c are integers.

iv.       The equation  has one root equal to -2. Show that equation has no other real roots.

b. The curve C with equation  , sketched below.

The curve cuts the x-axis at the point  and the y-axis at point B.

i.       State the y-coordinate of the point B.

ii.       Find

iii.       Hence find the area of the shaded region bounded by the curve  and the line AB.

Solution

a.

i.

We are required to find the remainder when  is divided by .

Remainder theorem states that if  is divided by  then;

For the given case  is divided by .

Here  and . Hence;

ii.

We are required to to show that  is a factor of  .

Factor theorem states that if  is a factor of   then;

For the given case  is factor of  .

We can write the factor in standard form as;

Here  and . Hence;

Hence,  is factor of .

iii.

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 01 factor  of .

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 factor from (a:ii) is .

We may divide the given polynomial by this factor .

Therefore, we get the quadratic factor  for given polynomial.

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

iv.

We are required to show that equation  has only one real root and that is the given one -2.

We are given that;

From (iii) we have also found that  can be expresses as;

Since ;

From (ii) we have found that is factor of  and therefore  is a root of  .

It is evident that one root of  is obtained from

For other roots we use;

It is quadratic factor of  and solution of this quadratic equations would yield other roots of

For a quadratic equation , the expression for solution is;

Where  is called discriminant.

If , the equation will have two roots.

If , the equation will have two identical/repeated roots.

If , the equation will have no roots.

As we are given that  has no real roots other than -2; in this case;

Therefore,  has no real roots.

Hence,  has only one real root -2.

b.

i.

We are required to find y-coordinate of point B.

It is evident that point B is the y-intercept of the curve with equation .

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;

Hence, y-coordinate of point B is 6.

ii.

Rule for integration of  is:

Rule for integration of  is:

Rule for integration of  is:

iii.

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

For the given case;

First we find area under curve ;

We have found from (b:ii) that;

Now we find area under the line AB.

We need equation of line AB with points A(-2,0) and B(0,6).

Two-Point form of the equation of the line is;

Therefore;

Therefore;

Rule for integration of  is:

Rule for integration of  is:

Rule for integration of  is:

Finally, we can calculate area of shaded region;