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

Hits: 11

Question

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

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

b.

i.               Show that at the points on the curve where y is decreasing ii.               Solve the inequality Solution

a.

We are given; i.

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; Therefore; Rule for differentiation is of is:  Rule for differentiation is of is: Rule for differentiation is of is:     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.

From (a:i) we have found that; Therefore, for we substitute ;    Since , the point on the curve is minimum.

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 are given that for the curve; Therefore, if y is decreasing then;      ii.

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.
If (‘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;  