# Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 1 (P1-9709/01) | Year 2016 | May-Jun | (P1-9709/12) | Q#10

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**Question**

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

** i. **Find expressions for , and .

** ii. **Find the coordinates of M and determine the coordinates and nature of the stationary point on the part of the curve for which .

** iii. **Find the volume obtained when the region bounded by the curve, the x-axis and the lines x = 1 and x = 2 is rotated through 360^{O} about the x-axis.

**Solution**

i.

We are given equation of the curve as;

Let’s find .

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

Rule for differentiation of is:

Rule for differentiation of is:

Let’s now find .

Second derivative is the derivative of the derivative. If we have derivative of the curve as , then expression for the second derivative of the curve is;

Therefore, for the given case;

Rule for differentiation of is:

Rule for differentiation of is:

Finally we find .

Rule for integration of is:

Rule for integration of is:

ii.

We are given that minimum point on the curve is M.

A stationary value is the maximum or minimum value of a function.

There point M is a stationary point on the curve.

A stationary point on the curve is the point where gradient of the curve is equal to zero;

Now we find coordinates of stationary point M.

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 in (i) that;

Two possible values of imply that there are two stationary points on the curve one at each value of .

But we are given that for , therefore, only possibility is .

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.

Therefore, we substitute in equation of the curve;

Hence, coordinates of stationary (minimum) point M are (2,4).

As seen above, for coordinates of stationary point there is also possibility of which is evidently on part of the curve where .

We can find coordinates of this point as well.

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.

Therefore, we substitute in equation of the curve;

Hence, coordinates of stationary (minimum) point on other part of the given curve are (-2,-8).

Now we determine the nature of stationary point (-2,-8).

Once we have the coordinates of the stationary point of a curve, we can determine its nature, whether minimum or maximum, by finding 2^{nd }derivative of the curve.

We substitute of the stationary point in the expression of 2^{nd} derivative of the curve and evaluate it;

If or then stationary point (or its value) is minimum.

If or then stationary point (or its value) is maximum.

We have found in (i) that;

Therefore, to find the nature of stationary point (-2,-8), we substitute in expression of second derivative of the curve;

Since, , the point (-2,-8) is a maximum point.

iii.

Expression for the volume of the solid formed when the shaded region under the curve is rotated completely about the x-axis is;

Therefore, for ;

Rule for integration of is:

Since region is bounded by the curve, the x-axis and the lines x = 1 and x = 2;

We have found from (i) that;

Therefore;

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