# Past Papers’ Solutions | Assessment & Qualification Alliance (AQA) | AS & A level | Mathematics 6360 | Pure Core 1 (6360-MPC1) | Year 2015 | June | Q#3

**Question**

The diagram shows a sketch of a curve and a line.

The curve has equation . The points A(-1,6) and B(2,30) lie on the curve.

**a. **Find an equation of the tangent to the curve at the point A.

**b. **

** i. **Find

** ii. **Calculate the area of the shaded region bounded by the curve and the line AB.

**Solution**

**a.
**

We are required to find the equation of the tangent to the curve at the point A.

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 one point on the line A(-1,6).

Therefore, we need slope of tangent to the curve at point A 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;

Therefore, if we have slope of the curve at point A(-1,6), we can find slope of tangent to the curve at that point.

Gradient (slope) of the curve is the derivative of equation of the curve. Hence gradient of curve with respect to is:

We have equation of the curve as;

Gradient (slope) of the curve is;

Rule for differentiation is of is:

Rule for differentiation is of is:

Rule for differentiation is of is:

We need gradient of the curve at point A(-1,6).

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;

Now we can find slope of tangent to the curve at point A(-1,6).

We have coordinates of a point on the tangent and also slope of the tangent, therefore, 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.

Hence;

Therefore, we need equation of the line AB.

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

We are given coordinates of both point A(-1,6) and B(2,30).

Therefore;

Rule for integration of is:

Rule for integration of is:

Rule for integration of is:

Next we need area under the curve.

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

Hence;

We are given equation of the curve as;

Therefore;

From (b:i) we have;

Therefore;

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