# 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 2nd derivative of the curve.

We have just verified that curve has a stationary point at .

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

We substitute  of the stationary point in the expression of 2nd 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 2nd 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;