Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 1 (P1-9709/01) | Year 2012 | Oct-Nov | (P1-9709/13) | Q#11

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Question

The diagram shows the curve with equation . The minimum point on the curve has coordinates  and the x-coordinate of the maximum point is , here  and  are constants.

     i.       State the value of .

   ii.       Find the value of .

  iii.       Find the area of the shaded region.

  iv.       The gradient, , of the curve has a minimum value . Find the value of .

Solution


i.
 

It is evident from the diagram that point  is the x-intercept of the given curve.

The point  at which curve (or line) intercepts x-axis, the value of . So we can find the value of  coordinate by substituting  in the equation of the curve (or line).

Therefore, to find the coordinates of x-intercept , we can substitute  in the equation of the curve.

We have two options.

From the diagram it
is evident that
 because x-intercept
is on the right side of the origin.

Hence


ii.
 

It
is evident from the diagram that
 is the x-coordinate of the stationary point of the curve. Therefore, to find  we need to find the coordinates of the stationary point of the curve.

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

We are given the equation of the curve as;

We can rewrite the given equation as follows;

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;

Rule for differentiation of  is:

Rule for differentiation of  is:

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 have two options.

We already know that  is the x-intercept of the curve (which is also a stationary point). Hence, x-coordinate of other stationary point of the curve is;

 

  iii.
 

It is evident from the diagram that shaded region is the area under the curve from origin to x-intercept of the curve.

To find the area of region under the curve , we need to integrate the curve from point  to  along x-axis.

For the given case point A is , point B is   and  as expanded in (ii);

Hence;

Rule for integration of  is:

Rule for integration of  is:

  iv.
 

We are given that gradient of the curve has minimum value  and we are required to find .

We have found gradient of the curve in (ii) as follows;

We are interested in minimum value of the gradient of the curve (not minimum value of the curve), and in this case gradient of the curve will be treated as a curve and its equation as equation of the curve. 

Since minimum value of the this curve (actually gradient of the original curve) is also a stationary point, we need to find this stationary point.

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.

Hence, we find second derivative of the gradient 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;

We have found gradient of the curve in (ii) as follows;

Therefore;

Rule for differentiation of  is:

Rule for differentiation of  is:

Rule for differentiation of  is:

Since this is stationary point 

Hence, stationary value of the gradient of the curve will occur at .

We are given that it is a minimum point and hence minimum value of gradient of the curve will occur at this point.

To find this value we would substitute   in the equation of the gradient of the curve;

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