Submitted by Heirio on Tue, 08/08/2017 - 21:44
Okay, now I'm not sure if this is a mistake or if I'm doing something wrong. [less than or equal to is marked here as <~] I got the correct solution via ((pqrz)^0.25) <~ ((pq)^0.5 + (rz)^0.5)/2, after which I put this equal to (p + q + r + z)/4, in reference to part ii.
This doesn't seem right to me. Why? Well, in part ii, the three expressions are separated by a less than or equal to sign, including the ((pq)^0.5 + (rz)^0.5)/2 and the (p + q + r + z)/4. They are NOT separated by an equals sign. Why then does the solution require I use the equals sign, if it is not used prior? It seems inconsistent, hence I'm wondering if it was a mistake in the print.
However, a mistake also seems unlikely, since (p+q+r)/4 + (p + q + r)/12 does equal (p + q + r)/3. So an equivalence sign IS true, but in order for it to follow with part ii and thus give me a solution, it is only reasonable that it should be a less than or equal to sign.
Is this a mistake in the print of a mistake of my own?
Moreover, in assignment 9,
Moreover, in assignment 9, STEP question part 1, it states that when A < 0 and B > 0, (4/27)A^3 + B < 0. Why? We're never given any real definition of how large A and B could be. If A was 0.1 and B was 57, for example, the statement would not be true. Why do we assume that the expression is less than 0?
In part 2, the same thing happens again and my work s based off of assumptions on how large each value is. I can tell whether or not it is larger or smaller than the other values, but I cannot tell how much so and hence which quadrant the point is in and thus I cannot tell how many roots there are.
Assignment 9
In the solutions it says that when $A<0$ and $B>0$ then there are two different cases, one where $\frac 4 {27}A^3 + B <0$ (when the minimum point lies below the $x$-axis) and one where $\frac 4 {27}A^3 + B >0$ (when the minimum point lies above the $x$-axis). You need to take both cases into account separately (the solutions suggest that you sketch a graph in each case).
In part (ii) you need to consider the different cases.
Assignment 8 question 1iii
I don't think there is a mistake in 1iii (I assume you are talking about the assignment itself rather than the hints and solutions).
In part (ii) you are using the result from "the stem" twice (actually three times, but in two stages - once using $a=p$, $b=q$ and $a=r$, $b=a$ which gives the first inequality, and then from the middle bit using $a=\sqrt{pq}$, $b=\sqrt{rs}$. since you are using the result from the stem twice there are two inequality signs.
In part (iii) the "$\frac{p+q+r}3=\frac{p+q+r}4+\frac{p+q+r}{12}$" bit is two ways of writing the same thing, the second of which will be more useful to you in proving the result, but it's a bit of a "magic trick" coming up with it, so to be helpful we gave the equivalent expression to you. The next bit has an inequality sign as to get to this you need to use the AM-GM inequality.
Does this make sense?
While looking through my
While looking through my working to see where I went wrong, I think I MAY have stumbled upon some kind of answer? I'll do my best to relay it to you.
So, in part ii we learn that the quartic root is less than or equal to the sum of the two roots divided by two, which itself is less than the sum of all four variable divided by four. Hence, we can infer that the quartic root is less than or equal to the sum of the four variables divided by four.
With part iii, we can then use that newly formed inequality to prove that the new one is true. We call the (p + q + r)/3 part z and treat the entire equation the same as we did in part ii, coming up with (p + q + r + z)/3. Since z is as we said earlier, we can simplify the fraction which leads to (p + q + r)/3, thus proving the inequality. We can then split the fraction into the two parts with denominators 4 and 12 respectively, proving that equality.
I hope I made sense there. Was that method correct?
As to assignment 9...
I'll give you some feedback later, I can't seem to find my work on part three. It'll probably be tonight or tomorrow.
That sounds about right, yes!
That sounds about right, yes!
As what came as not so much
As what came as not so much of a surprise, my mistake lay in the fact that I thought i could only draw one graph for each (which sounds quite simple now that I look back at it), something you showed was incorrect. I've done the question, having drawn multiple graphs where needed and have proved what the question has asked of me
thanks!