# Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 1 (P1-9709/01) | Year 2011 | Oct-Nov | (P1-9709/12) | Q#8

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**Question**

The equation of a curve is . Find

** i. **an expression for and the coordinates of the stationary point on the curve,

** ii. **the volume obtained when the region bounded by the curve and the x-axis is rotated through about the x-axis.

**Solution**

i.

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:

Rule for differentiation of is:

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

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, for the given case;

Single value of imply that there is only one stationary point on the curve.

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.

Equation of the curve is given as;

Substitution of in given equation of the curve yields;

Hence, coordinates of stationary point are .

ii.

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

For the limits we need to find the x-intercept of the curve.

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

Therefore,

Hence;

Rule for integration of is:

Rule for integration of is:

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