# Past Papers’ Solutions | Assessment & Qualification Alliance (AQA) | AS & A level | Mathematics 6360 | Pure Core 1 (6360-MPC1) | Year 2013 | June | Q#4

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Question

a.   The polynomial f(x) is given by .

i.       Use the Factor Theorem to show that x+3 is a factor of f(x).

ii.       Express f(x) in the form  , where p and q are integers.

b.   A curve has equation .

i.       Find .

ii.       Show that the x-coordinates of any stationary points of the curve satisfy the equation

iii.       Use the results above to show that the only stationary point of the curve occurs when .

iv.       Find the value of  when .

v.       Hence determine, with a reason, whether the curve has a maximum point or a minimum point when .

Solution

a.

i.

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 .

ii.

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:i) 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:i) 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;

b.

i.

We are given equation of the curve as;

Rule for differentiation is of  is:

Rule for differentiation is of  is:

Rule for differentiation is of  is:

ii.

A stationary point  on the curve  is the point where gradient of the curve is equal to zero;

To find the x-coordinates of stationary points of the given curve we can equate;

iii.

A stationary point  on the curve  is the point where gradient of the curve is equal to zero;

We can utilize the equation obtained in (b:ii) to find the x-coordinate of  stationary point(s) of the  curve.

As demonstrated in (a:ii) this equation can be written as;

Now we have two options. First one is;

The second is;

Since we are required to show that the only stationary point of the curve occurs at x=-3, the above  quadratic factor of the equation of the curve must NOT yield any solution.

For a quadratic equation , the expression for solution is;

Where  is called discriminant.

If , the equation will have two distinct roots.

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

If , the equation will have no roots.

Therefore, for the quadratic factor of equation of the curve;

Since , there is no solution for the quadratic factor of the equation and hence, the only  stationary point of the curve occurs at x=-3.

iv.

Second derivative is the derivative of the derivative. If we have derivative of the curve   as  , then  expression for the second derivative of the curve  is;

We have found in (b:i) that;

Therefore;

Rule for differentiation is of  is:

Gradient (slope)  of the curve  at a particular point  can be found by substituting x- coordinates of that point in the expression for gradient of the curve;

Similar is the case of second derivative at a given point. Therefore;

v.

We substitute  of the stationary point in the expression of 2nd derivative of the curve and  evaluate it;

If  or  then stationary point (or its value) is minimum.

If  or  then stationary point (or its value) is maximum.

From (b:iv) we have found that;

Since , at the point  is a minimum point.