# Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 1 (P1-9709/01) | Year 2018 | Oct-Nov | (P1-9709/12) | Q#11

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**Question**

The diagram shows part of the curve . The curve crosses the y-axis at A and the stationary point on the curve is M.

**
i. **Obtain expressions for and

**
ii. **Find the coordinates of M.

**
iii. **Find, showing all necessary working, the area of the shaded region.

**Solution**

i.

We are given;

First, we find .

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

Hence;

Rule for differentiation of is:

Rule for differentiation of is:

Next, we find .

Rule for integration of is:

Rule for integration of is:

Rule for integration of is:

** ii.
**

We are required to find the coordinates of point M which is stationary point on the curve.

Coordinates of stationary point on the curve can be found from the derivative of equation of the curve by equating it with ZERO. This results in value of x-coordinate of the stationary point on the curve.

We have found derivative of equation of the curve in (i);

We equate it with ZERO;

One possible value of implies that there is only one stationary point on the curve at this value of .

To find y-coordinate of the stationary point on the curve, we substitute value of x-coordinate of the stationary point on the curve (found by equating derivative of equation of the curve with ZERO) in the equation of the curve.

We have equation of the curve;

Substitute ;

Hence, coordinates of the stationary point are .

** iii.
**

We are required to find the area of the shaded region.

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.

We are given equation of the curve as;

The equation of line AM can be found from coordinates of points A and M.

We have found in (ii) that coordinates of point M are .

We can find coordinates of point A from equation of the curve.

It is evident from the diagram that point A 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;

Hence coordinates of point A are (0, 3).

Now we can find equation of line AM.

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

Hence;

First, we find area under the curve.

It is evident from the diagram that for area under the curve the varies from to .

Hence;

Rule for integration of is:

From (i), we have;

Therefore;

Next, we find area under line AM.

It is evident from the diagram that for area under the curve the varies from to .

Hence;

Rule for integration of is:

Rule for integration of is:

Rule for integration of is:

Finally,

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