Submitted by maths123 on Fri, 05/01/2015 - 15:07
Forums:
for 7i, the mark scheme says that both turning points are below axis hence the curve crosses x axis once only.
What is the need to consider both turning points?
Surely proving that the turning point at $ x= - \sqrt q $ is less than zero would be sufficient in showing that the given cubic has only one real root?
Since it is a positive cubic, the maximum is at $ x= - \sqrt q $ and the minimum must be at $ x= \sqrt q $ . The minimum y value must be less than the maximum y value so if the maximum is less than zero so is the minimum.
Correct
Your reasoning is correct as $- \sqrt q$ is a local maxima of the function this does give you that there is only one intersection. They consider both points because they aren't assuming that either of the points are locally minima or maxima. It's worth noting you have implicitly considered the 2nd derivative in making your argument when asserting it's a local maxima, I think if you're going to do the question this way you should probably show the 2nd deriative is negative (only one line). Also when answering this I think a quick sketch is one of the best ways to be absolutely clear in what you mean.
Yes
I agree with your reasoning. There's often more than one way to answer a question; the mark scheme doesn't cover every route.