# 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 2nd derivative of the curve.

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.

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;