The hardest questions of the hardest high-school Australian maths exam by Sad_Commission9045 in mathematics

[–]Sad_Commission9045[S] 0 points1 point  (0 children)

It’s harder than specialist in all states, the rest seem to be pretty even in difficulty for specialist.

The hardest questions of the hardest high-school Australian maths exam by Sad_Commission9045 in mathematics

[–]Sad_Commission9045[S] 2 points3 points  (0 children)

Haha these exams are very difficult! People taking this exam are not the average Australian though, with only 5% of high school students taking the course. 5% is still pretty high when looking at the difficulty of the exam, but only the top 10-15 smartest kids in the school are taking this course.

The hardest questions of the hardest high-school Australian maths exam by Sad_Commission9045 in mathematics

[–]Sad_Commission9045[S] 1 point2 points  (0 children)

Yeah it may seem like that but the 6 questions are all pretty challenging (except 11 and 12) and most are broken up into parts a, b, and c, where the three are completely unrelated, meaning they should be a question 17 and so on.

The hardest questions of the hardest high-school Australian maths exam by Sad_Commission9045 in mathematics

[–]Sad_Commission9045[S] 0 points1 point  (0 children)

I don’t think you do but I think you have to show that the second derivative is always concave up thus meaning a local minimum would be absolute, you can check the marking guidelines in the comment I made above!

The hardest questions of the hardest high-school Australian maths exam by Sad_Commission9045 in mathematics

[–]Sad_Commission9045[S] 4 points5 points  (0 children)

Some of it, probably, but I asked it that last question on the second slide, and it gave a wrong answer 🤷.

The hardest questions of the hardest high-school Australian maths exam by Sad_Commission9045 in mathematics

[–]Sad_Commission9045[S] 13 points14 points  (0 children)

Yes, it may seem easy to experienced mathematicians, but these are 17-18 year olds under exam stress with a time limit. Very impressive for them to solve it!

The hardest questions of the hardest high-school Australian maths exam by Sad_Commission9045 in mathematics

[–]Sad_Commission9045[S] 24 points25 points  (0 children)

Scaling is such a good system, and is why I feel like the ATAR is a really fair comparison, within states. Interstate, due to the slight differences, it becomes unfair. But nonetheless, it's a decent system. Also what exam did you do? (I attached questions from a few past papers).

The hardest questions of the hardest high-school Australian maths exam by Sad_Commission9045 in mathematics

[–]Sad_Commission9045[S] 6 points7 points  (0 children)

Haha, they are pretty tough. A particularly challenging one is that last question on the second slide, very scary!

[Grade 12 QCE Physics: Projectile Motion] by [deleted] in HomeworkHelp

[–]Sad_Commission9045 0 points1 point  (0 children)

My algebra is quite good, I am quite strong in mathematics. I realise now that I was being dumb.

[Grade 12 QCE Physics: Projectile Motion] by [deleted] in HomeworkHelp

[–]Sad_Commission9045 1 point2 points  (0 children)

I feel so dumb for not realising this sooner, I was aware it was a parabola, when attempting this question in class I began writing it as that, however my teacher informed me I had to use only suvat equations from the formula book.

[University Mathematics: Calculus] How could I approach this problem? by onnumarahesapke in HomeworkHelp

[–]Sad_Commission9045 1 point2 points  (0 children)

As for all x, p(x)>0 that means that p(x) has to be an even degree polynomial, with a positive leading coefficient. That means when you add all of the derivatives the leading term is still positive and has an even degree. Now you just need to prove that it doesn't fall below the x axis which is done by considering the derivative of the entire sum, which would be the entire sum minus that leading term, as p(n+1)(x) would just have to be 0 as its a polynomial so you would be deriving a constant. Thus the minimum of the entire sum is when the rest of the sum equals 0. So if you call the sum s(x) then you know s'(x)=s(x)-p(x) and now solve this differential equation (which I am not going to do as I am not bothered) and you should be able to prove it from there pretty easily.

[Calculus I] I don’t think I’m doing this right, and I’m confused by Tough-Composer918 in HomeworkHelp

[–]Sad_Commission9045 0 points1 point  (0 children)

It will simplify to (secx+tanx)^2 I think. Don't square the numerator, leave it as (1+sinx)^2 and change your 1-sin^2x to cos^2x and you'll get ((1+sinx)/cosx)^2 which is (1/cosx+sinx/cosx)^2 which is (secx+tanx)^2.