Past Papers’ Solutions | Assessment & Qualification Alliance (AQA) | AS & A level | Mathematics 6360 | Pure Core 1 (6360-MPC1) | Year 2014 | June | Q#3

 

Question

A curve has equation .

a.   Find:

                         i.       ;

                         ii.       

b.   The point on the curve where  is P.

                           i.       Determine whether y is increasing or decreasing at P, giving a reason for your answer.

                         ii.       Find an equation of the tangent to the curve at P.

c.   The point Q(-2,15) also lies on the curve. Verify that Q is a maximum point of the curve.

Solution

a.
 

                          i.
 

We are given;

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

Therefore;

Rule for differentiation is of  is:

Rule for differentiation is of  is:

Rule for differentiation is of  is:

                        ii.
 

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 have found in (a:i);

Therefore;

Rule for differentiation is of  is:

Rule for differentiation is of  is:

Rule for differentiation is of  is:

 

b.
 

                          i.
 

To test whether a function  is increasing or decreasing at a particular point , we  take derivative of a function at that point.

If  , the function  is increasing.

If  , the function  is decreasing.

If  , the test is inconclusive.

 

We have found in (a:i) that;

We are required to find whether y is increasing or decreasing at P where .

Therefore;

Since , therefore, y is decreasing at point P where .

 

                        ii.
 

We are required to find equation of the tangent to the curve at the point P(-1,y).

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).

We have coordinates of point on the tangent to the curve P(-1,y).

Therefore, first we need y-coordinate of point P.

We are given;

Hence;

Now we have coordinates of point P(-1,2) on the tangent to the curve at point P.

Next, we need slope of tangent to write its equation.

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

Since desired line is tangent to the curve at point P(-1,2), its slope (gradient) will be same as  gradient of the curve at this point. Let’s find gradient of the curve at point     A(-1,0).

Gradient (slope) of the curve at the particular point is the derivative of equation of the curve at that  particular point.

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 have found in (b:i) that;

Therefore;

Now we can write equation of tangent.

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

Therefore;

c.
 

We are required to verify that the point Q(-2,15) is a maximum point of the curve.

A stationary point is the maximum or minimum point of a curve.

Therefore, first we need to establish that point Q is a stationary point.

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

We have found in (a:i) that;

Let’s find gradient of the curve at point Q.

Gradient (slope) of the curve at the particular point is the derivative of equation of the curve at that  particular point.

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 have found in (a:i) that gradient of the curve is given by;

Hence, to find gradient of curve at point Q, we substitute x-coordinates of point Q, , in the  expression of gradient of the curve.

Hence, point Q is a stationary point.

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.

To determine the nature of a stationary point, we need second derivative of the equation of the curve which we have found in (a:ii).

Let’s substitute x-coordinate of stationary point Q in this equation.

Since , stationary point Q on the curve is a maximum point.

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