Past Papers’ Solutions | Assessment & Qualification Alliance (AQA) | AS & A level | Mathematics 6360 | Pure Core 1 (6360-MPC1) | Year 2015 | June | Q#7

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Question

a.   Sketch the curve with equation .

b.   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.       Hence show that the equation  has exactly one real root and state its value.

Solution

a.
 

We are required to sketch a cubical polynomial given as;

ü Find the sign of the coefficient of . This gives the shape of the graph at the extremities. If the  coefficient of  is such that  cubic graph is increasing from left to right at the extremities and  when  the cubic graph is inverted.

The coefficient of  in the given polynomial is positive. Therefore it will cause normal cubic graph  increasing from left to right at the extremities.

ü Find the point where the graph crosses y-axis by finding the value of  when

Therefore for ;

Hence, the cubic graph intercepts y-axis at .

ü Find the point(s) where the graph crosses the x-axis by finding the value of  when . If  there is repeated root the graph will touch the x-axis.

It is evident from factors that the function  will cross/intersect horizontal axis when;

It is evident cubic graph will intersect x-axis at . Since  is a repeated factor, therefore, the graph will touch the x-axis and return  at this point.

ü Calculate the values of  for some value of . The is particularly useful in determining the  quadrant in which the graph might turn close to the y-axis.

ü Complete the sketch of the graph by joining the sections. 

Sketch should show the main features of the graph and also, where possible, values where the  graph intersects coordinate axes.

desmos-graph (1).png

b.
 

                            i.
 

Remainder theorem states that if  is divided by  then;

For the given case   is divided by .

Here  and . Hence;

                          ii.
 

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 (b: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 (b:ii) is .

We may divide the given polynomial by factor .

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Therefore, we get the quadratic factor  for given polynomial.

Hence,  can be written as;

                         iv.

From (b:iii), we know that  can be written as;

It is evident from this that (x+2) is a factor of p(x).

Therefore, one of the roots of p(x) can be found from;

Remaining roots of p(x) can be found from;

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.

For the given equation;

Since , the equation will have no real roots.

Hence, p(x) has only one real root;

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