Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 1 (P1-9709/01) | Year 2015 | Oct-Nov | (P1-9709/12) | Q#9

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Question

The curve y=f(x) has a stationary point at (2,10) and it is given that

     i.       Find f(x).

   ii.       Find the coordinates of the other stationary point.

  iii.       Find the nature of each of the stationary points.

Solution

     i.
 

We are given;

We are required to find .

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

But we are given second derivative of . Therefore, we need derivative of  to find .

To find the derivative of the curve from second derivative we can integrate the second derivative of the curve.

Therefore;

Rule for integration of  is:

If a stationary point   lies on the curve , we can find out value of . We equate the equation  of derivative obtained from integration of second derivative of the curve with zero i.e. . Then  we substitute value of  in the equation of derivative obtained from integration of second derivative  of the curve i.e. .

We are given that point (2,10) is a stationary point on the curve. Therefore

Hence;

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

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 are given that point (2,10) is a stationary point on the curve. Therefore

Hence;

   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.

We have found in (i);

Therefore;

Two possible values of  imply that there are two stationary points on the curve one at each value  of .

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.   

Now we have two options.

For

For

Therefore stationary points are   and .

  iii.
 

Once we have the coordinates of the stationary point  of a curve, we can determine its  nature, whether minimum or maximum, by finding 2nd derivative of the curve.

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 are already given;

We have found stationary points are   and .

Therefore;

Point

Point

Minimum Point

Maximum Point

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