# Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 1 (P1-9709/01) | Year 2012 | Oct-Nov | (P1-9709/13) | Q#8

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**Question**

A curve is such that

** i. **Find

** ii. **Verify that the curve has a stationary point when and determine its nature.

**iii. **It is now given that the stationary point on the curve has coordinates (−1, 5). Find the equation of the curve.

**Solution**

i.

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 are given that

Therefore;

Rule for differentiation of is:

Rule for differentiation of is:

Rule for differentiation of is:

Rule for differentiation of is:

** ii.
**

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

We are given gradient of the curve;

Therefore, if curve has stationary point at then gradient of the curve must be equal to zero at this point. Substituting in the equation of gradient of the curve;

Hence curve has a stationary point at .

Once we have the x-coordinate 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 have just verified that curve has a stationary point at .

We have also found 2^{nd} derivative in (i);

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.

Substituting in 2^{nd} derivative;

Since , stationary point of the curve at is minimum.

** iii.
**

We can find equation of the curve from its derivative through integration;

We are given gradient of the curve;

Therefore, equation of the curve can be found as follows;

Rule for integration of is:

Rule for integration of is:

Rule for integration of is:

Rule for integration of is:

If a point lies on the curve , we can find out value of . We substitute values of and in the equation obtained from integration of the derivative of the curve i.e. .

We are given that stationary point lies on the curve. Therefore, we can substitute and in above equation of the curve obtained through integration.

Hence, equation of the curve can be written as;

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