Search results

ah yes they will let them know.... with what means of communication?

Hey you can't just commute the limit with the integral like that. Everyone knows limits don't commute in general. Please justify the commutation using only techniques from the syllabus. :)

In a square, the square of the diagonal length is twice the square of the side length. 2(a+b)·(a+b)=(3a-2b)·(3a-2b) 2|a|²+2|b|²+4(a·b)=9|a|²+4|b|²-12(a·b) ⇔7|a|²+2|b|²=16(a·b) I got PR is a-4b, so correct me if I am wrong. Diagonals of a square are perpendicular. Use the perpendicular...

For 11 c), a much more obvious (at least to me) solution is to do the following: α²(β + γ) = α(αβ + γα) = α(k - βγ) = -3α - αβγ where -3 is the coefficient corresponding to αβ + βγ + γα Repeating the same for the other terms results in α²(β + γ) + β²(γ + α) + γ²(α + β) = -3(α + β + γ) - 3αβγ...

It is perfectly legal to only check one case. However, this is only acceptable if the student wrote "Without Loss of Generality", or WLOG, or similar. This is just my opinion on the matter btw.

or let u = the integrand (in the few instances where it actually legitimately works)

the biggest meme in our school was saying "by inspection" whenever you didn't know the answer

The proof is extremely straightforward. You know n is a positive integer, so you can just expand it and delete all terms after 1+1. QED. In fact, there were many people who wrote the binomial expansion, and then suddenly gave up (???)

I am not joking. 5 people committed the exact same monstrosity.

except based on what Trebla said, that wasn't even need anyway, I'm just an idiot who hasn't bothered reviewing the syllabus outlines by the statement of the question, one can see that dy/dx (p,p) = dx/dy (p,p) = 1 and the conclusion follows

actually i think this is not the only way to do it using differentiation, but it is probably the most obvious now that i think about it from a student perspective

well fortunately, the other solution also works out (and a fair few students did go about that way solving it)

you don't remember learning slope of inverse function at inverse point = 1/(slope of function at function point) ? because i certainly do remember that class in 3U

apparently 5 people think that \frac{\text{d}}{\text{d}n} \left( \left(1+\frac{1}{n}\right)^n \right) = \left(-\frac{1}{n^2}\right) n \left( 1+\frac{1}{n} \right)^{n-1}

While this solution is correct, it is much more cumbersome to work with, especially since you are pulling it down to a specific case. If it was put in a more general form, then it would be slightly less cumbersome to write out. But then at that point it would just be easier to use the Inverse...

The intended solution: Inverse Function Theorem: Let b = f(a), f differentiable around a (f⁻¹)'(b) = 1/f'(a) Regarding the problem: By the Inverse Function Theorem, (f⁻¹)'(p) = 1/f'(p) But the slopes are equal, by the problem assumption. So let m = f'(p) = (f⁻¹)'(p), and we have m = 1/m ⇒...

trivial counterexample: x+y=0 and the inverse x+y=0 are tangent at (0, 0) with slope -1.

basically what i am saying is that almost nobody was able to justify m²=1 in the only way that is correct. they claimed (falsely, as well) that the slope must be 1. given the base of the exponential is less than 1, this is literally impossible.

apparently 90% of these students think that inverses that are tangent to each other are tangent to y=x. the syllabus teaches inverse functions so badly.

meme post don't take seriously >specifically says the base is less than 1 >mfw i see people drawing increasing exponentials instead of decreasing exponentials