# Past Papers’ Solutions | Assessment & Qualification Alliance (AQA) | AS & A level | Mathematics 6360 | Pure Core 1 (6360-MPC1) | Year 2017 | June | Q#5

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**Question**

A curve and the line AB are sketched below.

The curves has the equation and the points A(-1,9) and B(2,12) lie on the curve.

**a. **Find the equation of the normal to the curve at the point A, giving your answer in the form .

**b. **

** i. **Find .

** ii. **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 equation of the normal to the curve at the point A(-1,9).

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 normal A(-1,9). Therefore, we need slope of the normal to write its equation.

If two lines are perpendicular (normal) to each other, then product of their slopes and is;

Therefore;

We can find slope of normal to the curve at point A if we have slope of tangent to the curve at the same point.

First we find slope of the tangent to the curve at point A.

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

Therefore, if we have gradient for the curve at point A, we can find the slope of tangent to the curve at point A.

First we find slope of the curve at point A.

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 the equation of the curve as;

Therefore;

Rule for differentiation is of is:

Rule for differentiation is of is:

Rule for differentiation is of is:

Now we need gradient of the curve at point A(-1,9). Therefore, we substitute ;

Hence,

Now we can find slope of tangent to the curve at point A(-1,9).

Now we can find slope of the normal to the curve at point A.

Now we can write equation of the tangent.

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

**b.
**

** i.
**

Rule for integration of is:

Rule for integration of is:

Rule for integration of is:

** ii.
**

It is evident from the diagram that;

First we find area under line.

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

We do not have equation of the line AB but we have coordinates of points A(-1,9) and B(2,12) from which we can write equation of the line AB.

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

Therefore;

It is evident from the diagram that shaded region and area under line extend from point A to points B ie from x-coordinate of A (-1) to x-coordinate of B (2) along x-axis. Therefore;

Rule for integration of is:

Rule for integration of is:

Rule for integration of is:

Next we find area under curve.

First we find area under line.

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

We are given equation of the curve as;

It is evident that shaded region extends from x=-1 to x=2 along x-axis.

Therefore;

We have found in (b:i) that;

Therefore;

Hence;

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