# Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 1 (P1-9709/01) | Year 2019 | May-Jun | (P1-9709/12) | Q#11

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Question

The diagram shows part of the curve and the minimum point M.

i.Find the expressions for and

ii.Find the coordinates of M.

The shaded region is bounded by the curve, the y-axis and the line through M parallel to the x-axis.

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:

ii.

We are required to find the coordinates of point M which is minimum 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 ;

However, it is evident from the diagram that point M lies on positive x and y axes.

Therefore;

It is also evident from the diagram that point M has not y-coordinate as ZERO.

Therefore;

Hence, coordinates of the minimum 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;

We need equation of the line through point M and parallel to x-axis.

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 line .

We need slope of the line.

If two lines are parallel to each other, then their slopes and are equal;

Since line is parallel to the x-axis and slope of x-axis is ZERO, therefore;

Hence, now we can write equation of the line.

Point-Slope 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;

We have found in (i) that;

Therefore;

Next, we find area under line.

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

Hence;

Rule for integration of  is:

Finally,