By the way, does anybody here want to see the answers to the problems posed in the OP?
I had a minimal interest in seeing if he could solve them himself, but I think that question has been adequately answered.
He can't.
Oh but
I can! Here are
the answers:
The point was to see your answers.
As in not the answers you're copying and pasting from someone else, but your answers — answers showing you can play the game you were asking others to join.
Doubling down on proof that you can't isn't helping you much.
Which makes the o/p itself essentially fraudulent.
I. Lets set A = {1, 2, 3, 4, 5, 6, 7} and set B = {4, 5, 6, 7, 8, 9, 10}. Find the symmetric difference A ∆ B.
Answer:
A ∆ B = {1, 2, 3, 8, 9, 10}
Answers, by themselves, don't show mastery. That's why I ask my students to submit their work even with online exams. it's why the work is mandatory, but the online submissions are not.
In this case, work could consist of calculating the differences ...
\( A \setminus B = \{1,2,3\}, B \setminus A = \{8, 9, 10\}\)
... and then joining them as a union to create the symmetric difference ...
\( A \oplus B = \{1,2,3\} \cup B \setminus \{8, 9, 10\}=\{1,2,3,8,9,10\}\)
... showing the student has mastered the concept of a symmetric difference.
Recall that the \(\Delta\) is no longer used for set difference conventionally as it conflicts with its usage for the Laplacian.
II. You poll ten people who drink cola asking each if they like Pepsi and dislike Coke, like Coke and dislike Pepsi, and possibly like both Pepsi and Coke. If 3 people like Pepsi and dislike Coke, and 4 people like Coke and dislike Pepsi, then how many of the ten people like Coke and Pepsi?
Answer: Three people like Coke and Pepsi.
This is incorrect as noted by myself, twice, and by SB as well ...
Sample space:
3 - Like only Pepsi
4 - Like only Coke
? - Like Coke and Pepsi
? - Like neither Coke nor Pepsi
(Like Coke And Pepsi) + (like neither Coke nor Pepsi) = 3
The universe (or sample space in SB's less precise terminology) is "cola drinkers," which obviously includes people who drink Pepsi Cola and Coca Cola, but also includes all of the other colas, like Royal Crown Cola, Jolt Cola, and some real oddballs like Mecca Cola I ran into once in west Africa.
For a mathematician, the first task when presented with any question is to check to see if it's well-defined. It was obvious from the way the question was stated that the question intended to restrict itself to just those two. But it didn't. So Steve's answer is correct and the one you provided — which is obviously not your own answer — is not correct.
III. If x is any real number and a, b, and c are real numbers where f(x) = ax2+ bx + c and f(1) = 1, f(2) = 1, and f(4) = 3, then find the values of a, b, and c.
Answer: a = 1/3, b = -1, and c = 5/3
The answers, in context, are clearly not your own. Everything else here is copied from someone else.
How did you solve it? Double-elimination with back-sub from the original equations you couldn't figure how to write? Do you even know what that means? Earlier, you showed you couldn't understand where the equations came from well enough to spot an obvious included error.
This answer could have come from anywhere.
IV. If x is any real number, and g(x) = 2sin(3x), then find the area under the curve of g(x) and above the x-axis over the interval 0 ≤ x ≤ π. Also, find the equation of the tangent line to the curve of g(x) when x = π/6.
Answer: The are under the curve and above the x-axis over the interval 0 ≤ x ≤ π when g(x) = 2sin(3x) =4/3,
Now this is just precious. I'd answered it previously, with an included deliberate error.
Area = G(pi)-G(0) with G(x)=-2 cos 3x / 3
\(\displaystyle G(\pi) = -\frac{2\cos(3\pi)}{3} = \frac{2}{3}, G(0) = -\frac{2\cos(0)}{3} = -\frac{2}{3} \to G(\pi)-G(0) = \frac{4}{3}\)
That being the answer you said — and I agreed — was incorrect.
And here you are, blithely unaware you're posting the exact same wrong answer. "Wrong" for me means incorrect. Wrong for you, as was obvious at the time, means "doesn't look like this answer I'm reading off but can't really understand."
Here's yet another chance to redeem yourself.
Why is the actual answer 8/3 rather than 8 as I previously specified?
The best student will see at a glance that tripling the angular speed triples the unscaled domain yielding two full humps above the xxx-axis, recall that each half hump has area 1, take the area of the four half-humps and scale them by 2 to match g(x)g(x)g(x) and give the answer 8 "by inspection."
Other than the obvious reason, that is, because I included yet another error for you or anyone else to spot.
and the line tangent to the curve when x = π/6 is y = 2.
Again, that's not your answer. I'd answered it previously, including a minor issue by failing to substitute \(\pi/6\) into the right-hand side showing the derivative of \(2\sin(3x)\)
(y-2 sin 3pi/6)/(x-pi/6) = 6cos 3x
Plugging in the specified value ...
\(\displaystyle \frac{y-2\sin (\pi/2)}{x-\pi/6} = 6 (\cos \pi/2) = 0 \to y=2\)
... as desired.
V. Let A be the square matrix
y 2
3 x
What real values of x and y will result in matrix A having a determinant equal to 4?
Answer: yx = 10.
And this one, again, isn't your answer, because nobody answering this question themselves would fail to follow the convention of placing variables in their alphabetical order. Copying a doofus, as I mention as gently as possible for my students, makes you look like a doofus, too.
So you should make a habit of only copying from the best students!
VI. If a and b are both positive real numbers, then prove that a/b + b/a ≥ 2.
Answer:
Let a and b be positive real numbers. Then
(a – b)^2 ≥ 0 and so
(a – b)(a – b) ≥ 0 and so
a^2 – 2ab + b^2 ≥ 0 and so
a^2 + b^2 ≥ 2ab and so
(a^2 + b^2)/(ab) ≥ 2 and therefore
a/b + b/a ≥ 2. □
Just why? Everyone recognizes the sum of reciprocals here which leads immediately to a quadratic inequality. This is the hard way.
VII. Prove if arbitrary n ∈ N = {0, 1, 2, ... }, then 2 | n^2+ n.
Answer: Using mathematical induction, the proof is
Base Case: If n = 0, then 2 | 02 + 0 which is true.
Inductive Step:
Suppose it is true that 2 | n^2 + n. Then it follows that 2 | (n + 1)^2 + (n + 1).
To see why, observe that (n + 1)^2 + (n + 1) = n^2 + 2n + 1 + n + 1 = n^2 + 3n + 2 = (n^2 + n) + 2(n + 1). Because it is assumed that 2 | n^2 + n, then there is an integer k such that n^2 + n = 2k. Substituting 2k for n^2 + n in
(n^2 + n) + 2(n + 1), the result is 2k + 2n + 2 which is divisible by 2.
So whenever 2 | n^2 + n, then it follows that 2 | (n + 1)^2 + (n + 1).
Therefore, by mathematical induction it is proved that if arbitrary n ∈ N, then 2 | n^2 + n. □
And speaking of the hard way, dayum.
Induction is great when you can't solve a problem by inspection. As in, not for this problem. The dividend here factors as \(n(n+1)\), two sequential numbers, one of which must be even. That is ...
If \(n\) is even, you're done.
If \(n\) is not even, then it's odd, and \(n+1\) must be even.
And the only reason I'm expanding the proof out that far is because you've failed to recognize it even after two attempts.
Now it's possible to pose this question, or any question for that matter, by including a required method of solution to assess mastery of that method. But if the method is not specified, and if as usual, there are myriad ways to solve the problem, it's assessing how the answer was found.
\(9,999,999,999\cdot 10,000,000,001 - 9,999,999,997 \cdot 10,000,000,003\)
... can be answered by finding the two products and their difference — by hand, because it's deliberately set to be beyond the display abilities of any handheld calculator — but if the best student recognizes the pattern, the answer is immediate.
It's awesome that you're interested in math, but if you want to get good at it, you're going to have to stop pretending you're better than you are, because an inability to correct errors cements them in place, created a roadblock to learning.
