# Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 1 (P1-9709/01) | Year 2019 | Oct-Nov | (P1-9709/11) | Q#9

**Question**

A curve for which passes through the point (2,3).

** i. **Find the equation of the curve.

** ii. **Find .

** iii. **Find the coordinates of the stationary point on the curve and, showing all necessary working, determine the nature of this stationary point.

**Solution**

i.

We can find equation of the curve from its derivative through integration;

We are given that;

Therefore, to find the equation of the curve;

Rule for integration of is:

Rule for integration of is:

Rule for integration of is:

If a point lies on the curve , we can find out value of . We substitute values of and in the equation obtained from integration of the derivative of the curve i.e. .

We are given that curve passes through the point (2,3).

Therefore, substituting coordinates of the point in above found equation of the curve, we can find value of c.

Therefore, equation of the curve is;

ii.

We are required to find second derivative.

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 are given;

Therefore;

Rule for differentiation of is:

Rule for differentiation of is:

Rule for differentiation of is:

** iii.
**

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 are given derivative of the equation of the curve;

To find the coordinates of the stationary point on the curve;

One possible values of implies that there is only one stationary point on the curve at this 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.

We have found equation of the curve in (i);

Substituting ;

Hence coordinates of stationary points are .

Once we have the coordinates of the stationary point of a curve, we can determine its nature, whether minimum or maximum, by finding 2^{nd} derivative of the curve.

We have already found second derivative of the equation of the curve in (ii) as;

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.

Substituting ;

Since , the stationary point is minimum.

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