Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 1 (P1-9709/01) | Year 2018 | May-Jun | (P1-9709/11) | Q#10

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Question

The curve with equation  passes through the origin.

    i.       Show that the curve has no stationary points.

   ii.       Denoting the gradient of the curve by m, find the stationary value of m and  determine its nature.

Solution


i.
 

We are required to show that curve has no stationary points.

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

Therefore, if curve has any stationary point(s) then;

Hence, first we find expression for gradient of the curve.

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

We are given that;

Therefore;

Rule for differentiation of  is:

Rule for differentiation of  is:

Rule for differentiation of  is:

To find the x-coordinate(s) of stationary point(s);

Hence;

It is evident that it is a quadratic equation.

Standard form of quadratic equation is;

Expression for discriminant of a quadratic equation is;

If  ; Quadratic equation has two real roots.

If  ; Quadratic equation has no real roots.

If  ; Quadratic equation has one real root/two equal roots.

For the given case;

Hence, discriminant of the equation is;

Since quadratic equation has no real roots and, therefore, there are no  stationary points on the curve. 


ii.
 

We are given that the gradient of the curve is m.

We have found in (i) that;

Hence;

We are required to find the stationary value of m and determine its nature.

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

Therefore, we need to find .

Rule for differentiation of  is:

Rule for differentiation of  is:

Rule for differentiation of  is:

Since we are looking for stationary value of .

To find the stationary value of  we substitute  in;

Next we are required to find the nature of this stationary value of .

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.

Therefore, we need second derivative of;

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;

Rule for differentiation of  is:

Rule for differentiation of  is:

Rule for differentiation of  is:

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 , the stationary value of  is minimum.

 


iii.
 

We are required to find the area of the region enclosed by the curve, the x-axis and  the line x = 6.

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;

Therefore;

Rule for integration of  is:

Rule for integration of  is:

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