# Past Papers’ Solutions | Assessment & Qualification Alliance (AQA) | AS & A level | Mathematics 6360 | Pure Core 1 (6360-MPC1) | Year 2009 | January | Q#6

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**Question**

**a. **The polynomial is given by .

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

** ii. **Express in the form , where a and b are constants.

**b. **The curve C with equation , sketched below, crosses the x-axis at the point .

** i. **Find the gradient of the curve C at the point Q.

** ii. **Hence find an equation of the tangent to the curve C at the point Q.

** iii. **Find

** iv. **Hence find the area of the shaded region bounded by the curve C and the coordinate axes.

**Solution**

**a.
**

** i.
**

Factor theorem states that if is a factor of then;

For the given case is factor of .

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 required to find the gradient of the curve at point .

Gradient (slope) of the curve at the particular point is the derivative of equation of the curve at that particular point.

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;

We are given equation of the curve as;

Therefore;

Rule for differentiation is of is:

Rule for differentiation is of is:

Rule for differentiation is of is:

For gradient of the curve at point Q(2,0), substitute in derivative of the equation of the curve.

** ii.
**

We are required to find the equation of tangent to the curve at point Q(2,0).

To find the equation of the line either we need coordinates of the two points on the line (Two-Point form of Equation of Line) or coordinates of one point on the line and slope of the line (Point-Slope form of Equation of Line).

We already have coordinates of a point on the tangent as Q(2,0).

We need to find slope of tangent at in order to write its equation.

The slope of a curve at a particular point is equal to the slope of the tangent to the curve at the same point;

From (b:i) we have slope (gradient) of the curve at point Q(2,0);

Therefore;

With coordinates of a point on the tangent Q(2,0) and its slope in hand, we can write equation of the tanget.

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

iii.

Rule for integration of is:

Rule for integration of is:

Rule for integration of is:

** iv.
**

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;

Rule for integration of is:

From (b:iii) we know that;

Therefore;

Hence;

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