Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 1 (P1-9709/01) | Year 2010 | May-Jun | (P1-9709/11) | Q#6

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Question

A curve is such that    and the point (9, 2) lies on the curve.

    i.       Find the equation of the curve.

   ii.       Find the x-coordinate of the stationary point on the curve and determine the nature of the stationary point.

Solution


i.
 

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

For the given case;

Therefore;

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 know that curve passes through the point (9, 2). Therefore, we substitute  and  in the equation above.

Hence, equation of the curve is;


ii.

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 derivative (gradient) of the given curve.

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

For the given case, the equation of the curve is;

Therefore;

Rule for differentiation of  is:

Therefore;

Rule for differentiation of  is:

Rule for differentiation of  is:

To find the coordinates of the stationary point we equate the derivative with ZERO.

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. 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;

For the given case;

Therefore;

Rule for differentiation of  is:

Rule for differentiation of  is:

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.

We have x-coordinate of the stationary point 4. Therefore, we substitute  in the expression of
second derivative.

Since, , stationary point is minimum.

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