yup
yup
Oh whoops... I read that as .
HSC 2018: English Adv. [80] • Maths Ext. 1 [98] • Maths Ext. 2 [95] • Chemistry [87] • Software Design [95]
ATAR: 97.40 | Uni Course: Advanced Mathematics (Hons) / Engineering (Hons) at UNSW
Question 7 Of Ekman's Compilation
for the working out of integrands clever solution if anyones interested
2017
4u99 - 3u98 - EngAdv88 - Phys94 - Chem94
Atar : 99.55
Course: Engineering(Hons)/Commerce @ UNSW
ATAR is just a number
I'm not sure what you did on this line, I'm guessing its in regard to the fact that Im(1) = 0
Also, did you mean:
HSC 2018
English Adv - MX1 - MX2 - Physics - Economics
If I am a conic section, then my e = ∞
Just so we don't have this discussion in the future, my definition of the natural numbers includes 0.
Sorry if I'm missing something obvious, but just to clarify
When you take the imaginary part of the sum of GP formula, does that instantly disregard the z^0 (=1) in the original series becuase Im(1) = 0? It seems weird because you use a = 1 for the formula, maybe I'm thinking about this too much haha
HSC 2018
English Adv - MX1 - MX2 - Physics - Economics
HSC 2018
English Adv - MX1 - MX2 - Physics - Economics
Is the answer ?
(I intend to post a writeup later)
HSC 2018: English Adv. [80] • Maths Ext. 1 [98] • Maths Ext. 2 [95] • Chemistry [87] • Software Design [95]
ATAR: 97.40 | Uni Course: Advanced Mathematics (Hons) / Engineering (Hons) at UNSW
Converting to polar form,
Since ,
The cosine function is even, so the first term on the LHS and RHS are equal. Cancelling,
Since the sine function is odd, for all .
And clearly this is only true when .
So therefore,
Last edited by fan96; 14 Feb 2018 at 3:24 PM.
HSC 2018: English Adv. [80] • Maths Ext. 1 [98] • Maths Ext. 2 [95] • Chemistry [87] • Software Design [95]
ATAR: 97.40 | Uni Course: Advanced Mathematics (Hons) / Engineering (Hons) at UNSW
2u integration+circle geo
i made this question myself in paint so dont mind the bad production quality.
edit: look for the question which has the circle with one edge on -5 instead of -3. ignore the question where the circle touches -3.
integral composite graph.png
Last edited by mrbunton; 17 Feb 2018 at 1:41 PM.
i) define the graph as
Let the lower bound be such that .
Now,
So for the integral to be zero,
Rearranging,
Using the quadratic formula,
Since ,
So
(One other solution is also )
_______________________________________________
ii) The area of the semicircle is , and
The graph doesn't seem to be defined for .
Clearly , so if one of the bounds is , the only bound that will make the integral zero is .
HSC 2018: English Adv. [80] • Maths Ext. 1 [98] • Maths Ext. 2 [95] • Chemistry [87] • Software Design [95]
ATAR: 97.40 | Uni Course: Advanced Mathematics (Hons) / Engineering (Hons) at UNSW
2^2*pi/2 = 2pi >6 therefore ii) is possible. remember it has radius 2 not 1
also u got the gradient of the second line for i) incorrect, getting the gradient wrong also affected the y-intercept.
noticing k=5 was impressive although.
Last edited by mrbunton; 17 Feb 2018 at 11:28 AM.
HSC 2018: English Adv. [80] • Maths Ext. 1 [98] • Maths Ext. 2 [95] • Chemistry [87] • Software Design [95]
ATAR: 97.40 | Uni Course: Advanced Mathematics (Hons) / Engineering (Hons) at UNSW
There are currently 1 users browsing this thread. (0 members and 1 guests)
Bookmarks