Past Papers’ Solutions | Assessment & Qualification Alliance (AQA) | AS & A level | Mathematics 6360 | Pure Core 1 (6360-MPC1) | Year 2016 | June | Q#8

Hits: 23



The gradient, , at the point (x,y) on a curve is given by


                    i.               Find

                  ii.               The curve passes through the point Verify that the curve has a minimum point at P.


                    i.               Show that at the points on the curve where y is decreasing

                  ii.               Solve the inequality



We are given;


We are required to find the 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; 


Rule for differentiation is of  is:

Rule for differentiation is of  is:

Rule for differentiation is of  is:


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.

From (a:i) we have found that;

Therefore, for  we substitute ;

Since , the point  on the curve is minimum.




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 are given that for the curve;

Therefore, if y is decreasing then;



We are required to solve the inequality;

We solve the following equation to find critical values of ;

Now we have two options;

Hence the critical points on the curve for the given condition are  & 6.

Standard form of quadratic equation is;

The graph of quadratic equation is a parabola. If  (‘a’ is positive) then parabola opens upwards  and its vertex is the minimum point on the graph.
 (‘a’ is negative) then parabola opens downwards and its vertex is the maximum point on the  graph.

We recognize that  it is an upwards opening parabola.

Therefore conditions for  are;

Please follow and like us: