Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 1 (P1-9709/01) | Year 2011 | Oct-Nov | (P1-9709/13) | Q#8

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Question

A curve  has a stationary point at . It is given that, where k is a constant.


i.       
Show that  and hence find the x-coordinate of the other stationary point, Q.

   ii.       Find  and determine the nature of each of the stationary points P and Q.

  iii.       Find .

Solution


i.
 

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

Hence gradient of the given curve at given stationary point .

Since;

Therefore;

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.

Now we can write the derivative of the equation of given curve;

Now we have two options.

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

We are already given one stationary point at . So the x coordinate of other stationary point is .


ii.
 

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;

For second derivative;

Rule for differentiation of  is:

Rule for differentiation of  is:

Rule for differentiation of  is:

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.

Since we have two stationary points namely P & Q with x-coordinates  and  , we check nature of both stationary  points one by one.

For  ;

Since
 it is minimum point.

For  ;

Since  it is maximum point.


iii.
 

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

For the given case;

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 at .

Therefore;

Hence;

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