# Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 1 (P1-9709/01) | Year 2018 | May-Jun | (P1-9709/13) | Q#11

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**Question**

The diagram shows part of the curve and the line x = 1. The point A is the minimum point on the curve.

** i. **Show that the x-coordinate of A satisfies the equation and find the exact value of at A.

** ii. **Find, showing all necessary working, the volume obtained when the shaded region is rotated through 36^{0} about the x-axis.

**Solution**

i.

We are given that the curve is;

The point A is the minimum point on the curve.

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

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.

Therefore, we need to find the derivative of equation of the curve.

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

Therefore;

Rule for differentiation of is:

Rule for differentiation of is:

To find the coordinates of the stationary (minimum) point of the curve;

We are required to find the exact value of at A.

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;

Therefore, first we need to 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;

We have found above that;

Therefore;

Rule for differentiation of is:

Rule for differentiation of is:

Rule for differentiation of is:

Rule for differentiation of is:

Rule for differentiation of is:

Rule for differentiation of is:

To find the exact value of at point A we need to substitute x-coordinate of A in above equation.

We have found above that x-coordinate of A satisfies the equation

Hence;

This is x-coordinate of point A.

Therefore;

ii.

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 the given case;

We are given that;

Hence, for x=0 and x=1;

Rule for integration of is:

Rule for integration of is:

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