Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 1 (P1-9709/01) | Year 2015 | Oct-Nov | (P1-9709/11) | Q#5

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Question

A curve has equation  .

     i.         Find  and

 

  ii.        Find the coordinates of the stationary points and state, with a reason, the nature of each                   stationary point.

Solution


i.
 

We are given that;

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

Rule for differentiation of  is:

Rule for differentiation of  is:

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;

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; 

From (i), we have;

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;

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.   

For

For

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;

From (i), we have;

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.

For

For

Maximum

Minimum

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