# Past Papers’ Solutions | Assessment & Qualification Alliance (AQA) | AS & A level | Mathematics 6360 | Pure Core 1 (6360-MPC1) | Year 2007 | January | Q#6

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Question

The curve with equation is sketched below. The curve cuts the x-axis at the point A (-1, 0) and cuts the y-axis at the point B.

a.

i.       State the coordinates of the point B and hence find the area of the triangle AOB, where  O is the origin.

ii.       Find iii.       Hence find the area of the shaded region bounded by the curve and the line AB.

b.

i.       Find the gradient of the curve with equation at the point A (-1, 0).

ii.       Hence find an equation of the tangent to the curve at the point A.

Solution

a.

i.

We are given that curve with equation intersects the y-axis at point B and we are  required to find the coordinates of point B which is y-intercept of the curve.

The point at which curve (or line) intercepts y-axis, the value of . So we can find the  value of coordinate by substituting in the equation of the curve (or line).

Therefore, substituting in given equation of the curve;    Hence coordinates of B(0,5).

Now we are required to find the area of triangle AOB.

Expression for the area of the triangle is; For the given case;  We need to find OA and OB to find area of the triangle.

Expression for distance between two given points and is: For OA & OB we have coordinates of all three points O(0,0), A(-1,0) and B(0,5). We find OA and OB  using expression for distance between two points.          Since distance cannot be negative;    ii. Rule for integration of is:  Rule for integration of is: Rule for integration of is:     iii.

It is evident from the diagram that; To find the area of region under the curve , we need to integrate the curve from point to along x-axis. It is evident from the diagram that shaded region lies under the curve and extends from to .

Therefore; Rule for integration of is: From (b:ii) we have Hence;    We have already calculated area of triangle in a(i) , so we need to find area under the curve.  b.

i.

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 that; To find the gradient of the curve at point A(-1,0), first we need derivative of the equation of the given  curve. Therefore; Rule for differentiation is of is:  Rule for differentiation is of is:     Now to find gradient of the curve at point A(-1,0), we use;    ii.

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).

Since tangent to the curve is at point A(-1,0), we have coordinates of one point on the tangent and  so we only need slope of the tangent to write equation of the tangent.

The slope of a curve at a particular point is equal to the slope of the tangent to the curve at the same point;  We have found gradient (slope) of the curve at point A in (b:i) as; Therefore;   Now we can write equation of the tangent;

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