# 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 2^{nd} and 3^{rd} 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 2^{nd} 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.

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