# 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 2^{nd }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 2^{nd} 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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