# Past Papers’ Solutions | Assessment & Qualification Alliance (AQA) | AS & A level | Mathematics 6360 | Pure Core 1 (6360-MPC1) | Year 2009 | June | Q#3

**Question**

The curve with equation passes through the point P, where .

**a. **Find

** i. **

** ii. **

**b. **Verify that the point P is a stationary point of the curve.

**c. **

** i. **Find the value of at the point P.

** ii. **Hence, or otherwise, determine whether P is a maximum point or a minimum point.

**d. **Find an equation of the tangent to the curve at the point where .

**Solution**

**a.
**

We are given;

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

Rule for differentiation is of is:

Rule for differentiation is of is:

Rule for differentiation is of is:

**b. **** **

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

For the given curve, from (a) we have;

We are also given that x-coordinate of point P is .

Hence;

It shows that point P is a stationary point.

**c. **

** i.
**

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;

Rule for differentiation is of is:

Rule for differentiation is of is:

Rule for differentiation is of is:

To find the value of at point P we substitute x-coordinate of point P in this equation.

ii.

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 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.

As seen in (c:i), at point P, hence, point P is a maximum point.

**d.
**

We are required to find the equation of tangent to the curve.

To find the equation of the line either we need coordinates of the two points on the line (Two-Point form of Equation of Line) or coordinates of one point on the line and slope of the line (Point-Slope form of Equation of Line).

Let’s first find coordinates of point on the tangent line.

We already have x-coordinate of a point on the tangent to curve .

To find y-coordinate of the point on the curve, we substitute value of x-coordinate of the point on the curve in the equation of the curve.

We are given equation of the curve as;

Substituting ;

Hence a point (1,13) lies on the tangent to curve.

Next we need slope of the tangent to the curve.

The slope of a curve at a particular point is equal to the slope of the tangent to the curve at the same point;

Therefore, if we have gradient (slope) of the curve at point where x=1, we can find the slope of tangent to curve at this point.

For the given curve, from (a) we have;

Gradient (slope) of the curve at a particular point can be found by substituting x- coordinates of that point in the expression for gradient of the curve;

We are also given that x-coordinate of point is .

Hence;

Hence;

Now with coordinates of point on the tangent (1,13) and slope of the tangent , we can write equation of the tangent.

Point-Slope form of the equation of the line is;

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