Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 1 (P1-9709/01) | Year 2016 | May-Jun | (P1-9709/12) | Q#10

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Question

The diagram shows the part of the curve  for , and the minimum point M.

     i.         Find expressions for ,  and .

   ii.       Find the coordinates of M and determine the coordinates and nature of the stationary point on          the part of the curve for which .

  iii.       Find the volume obtained when the region bounded by the curve, the x-axis and the lines                   x = 1  and x = 2 is rotated through 360O about the x-axis.

Solution


i.
 

We are given equation of the curve as;

Let’s find .

 

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:

Let’s now find .

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;

Therefore, for the given case;

Rule for differentiation of  is:

Rule for differentiation of  is:

Finally we find .

Rule for integration of  is:

Rule for integration of  is:


ii.
 

We are given that minimum point on the curve is M. 

A stationary value is the maximum or minimum value of a function.

There point M is a stationary point on the curve.

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

Now we find coordinates of stationary point M.

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) that;

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

But we are given that  for , therefore, only possibility is .

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.  

Therefore, we substitute  in equation of the curve;

Hence, coordinates of stationary (minimum) point M are (2,4).

As seen above, for coordinates of stationary point there is also possibility of  which is evidently on part of the curve where .

We can find coordinates of this point as well.

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.  

Therefore, we substitute  in equation of the curve; 

Hence, coordinates of stationary (minimum) point on other part of the given curve are (-2,-8).

Now we determine the nature of stationary point (-2,-8).

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 have found in (i) that;

Therefore, to find the nature of stationary point (-2,-8), we substitute  in expression of second  derivative of the curve;

Since, , the point (-2,-8) is a maximum point.


iii.
 

Expression for the volume of the solid formed when the shaded region under the curve  is rotated completely about the x-axis is;

Therefore, for ;

Rule for integration of  is:

Since region is bounded by the curve, the x-axis and the lines x = 1 and x = 2;

We have found from (i) that;

Therefore;

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