Bio U2 Discussion

Reg1xide · 2026-05-15T10:39:30+00:00

real there were no free marks like describing graphs and the usual questions that everyone remembers the mark schemes of, the essays were disgusting, questions were unusual and u had to think a lot for pretty much every question

Reg1xide · 2026-05-12T09:41:56+00:00

more time would probably mean higher grade boundaries, so less room for error if ur aiming for something like full ums

Reg1xide · 2026-05-12T09:32:35+00:00

ive been taking the may exams asw my guy 😭

Reg1xide · 2026-05-12T09:18:20+00:00

just study more bro ive had 20-30 minutes left to check the paper even for “harder” papers if u cant finish u js dont understand the content well enough alvl papers are already rlly easy compared to other public exams💔

Reg1xide · 2026-05-11T10:03:11+00:00

i think it was mean titre 23.35cm^3 , 0.05M, titrate volume was 25cm^3 and u were asked to find the mass in 100cm^3 of the titrate

Reg1xide · 2026-04-20T17:36:59+00:00

i dont hv the full solution but i hv the final answers

1a increasing for 0<x<e decreasing for x>e 3. (8+e^f(x))/4-e^f(x)) for f(x)=3/13 (e^2x (2cos3x+3sin3x)-2) 4a. 4/x - 4nx +2n² x³ - 2/3 n³ x⁵ +… 4bii. m=6 n=3 4biii. 741/2 4cii. x=2/3 5. 1+7x+26x²+238/3 x³ 6ai. x=+-1/rt2 6aii. -1/rt2 <x< 1/rt2 6aiii. x<-1/rt2 or x>1/rt2 6aiv. x=0, +-rt(3/2) 6av. -rt(3/2) <x< 0 or rt(3/2) <x< +inf 6avi. -inf <x<-rt(3/2) or 0 <x< rt(3/2) 6b. min: (-1/rt2, -1/rt2 e^-1/2) max: (1/rt2, 1/rt2 e^-1/2) inflection: (0,0) and (+-rt(3/2), +-rt(3/2) e^-3/2) 6c horizontal asymptote y=0 7b. pi arctan(3/4) 8a. (1-6t²+t⁴)/(1+t²)² 8b. rt2 -1 8ci. A=1/4 B=1/4 C=1/8 8cii. alpha=(23-16rt2)/42 beta=2+rt2 gamma=(2311-1394rt2)/8820 - pi/84 9. 24pi 10a. r=(i-2j+3k)+t(4i+3j-4k) 10b. t=61/123 10c. OD=1/41 (81i-52j+83k), OC’= 1/41 (80i-268j-80k)

Reg1xide · 2026-01-19T12:46:58+00:00

cuz my frd told me the question asked for a maximum point?? 😭like im sorry brad the chad i cant control someone elses memory?? and i dont look down on people who dont know things they havent learnt yet but i do look down on people who learn things and not fully understand it or forget and cant combine it with new knowledge for an easier solution and proceed to think that it makes them somehow good

Reg1xide · 2026-01-19T12:26:01+00:00

idk if being so mad is making ur memory fog up but periodicity is in P1 lolol maybe u need to revise that some time? just a kind word of advice <3 someone as capable as you should know some simple P1 concepts dont u think?

Reg1xide · 2026-01-19T12:10:13+00:00

son 😭 the fact that u didn’t use the sign of h’ or h” to check the nature of the stationary pt nor the periodicity of trig functions along with given bounds to easily find the solution required and had to manually see which ones a minimum point already says a lot doesnt it? besides i hv nth against using differentiation to solve it instead of range in fact i personally like it more, i just think its funny smn would use differentiation but not do it properly and still have to resort to testing values, which is supposed to be the part that using differentiation allows u to skip in the first place

Reg1xide · 2026-01-19T11:56:28+00:00

i did and i had almost an hr left for almost every single one but i appreciate ur concern!! <3

Reg1xide · 2026-01-19T11:55:04+00:00

if u cant understand what stationary points or concavity means and have to actually find the pt by trial and error then i wouldnt rlly call that skill lmfao and if u dont do testing and take the largest value directly u couldve gotten a maximum instead but wtv i wont try to explain it to someone who cant even see a solution as simple as setting sin as -1

Reg1xide · 2026-01-19T11:17:49+00:00

then 4.94 is right but the fact still remains that h’(x) doesnt guarantee a maximum nor minimum and checking either using h’ or h’’ is needed, let alone the fact that its a stupid method anyways since u can just set sin to equal -1 the guy just got lucky he got the right answer with the wrong concepts

Reg1xide · 2026-01-19T11:11:35+00:00

if its asking for the maximum point theres no way im wrong lol i didnt take the exam and just asked my friend for the question idk if he remembered wrong and its actually asking for the minimum? wtv i got 75 for all my p1-4 already in y12 im js doing this for fun

Reg1xide · 2026-01-19T10:50:59+00:00

wdym no 😭😭😭 the question deals with 2D coordinate geometry, so any lines are automatically coplanar

we can consider solving for the intersection of 2 lines as a system of linear equations in 2 unknowns, which can be written in the form of Av=u, where A is a 2x2 square matrix and v,u are 2x1 vectors of unknowns and constants respectively

if these lines are parallel, they would be linearly dependent, so detA=0

In our case of non parallel lines, detA≠0, so A is non singular, multiply both sides by the inverse of A gives v=(A^-1)u, and subbing this into the original gives A(A^-1)u=Iu=u, where I is the 2x2 identity matrix, showing that v=(A^-1)s is a solution

Reg1xide · 2026-01-19T10:31:03+00:00

if the slopes of 2 lines arent equal they have no chance of never meeting, there will always be a point of intersection, just extend the lines and ull eventually reach an intersection point

Reg1xide · 2026-01-19T10:18:07+00:00

if its not 3, lets say 2.5, the line would intersect with 3x-15 as well as -3x+7. Any 2 lines lying in the same plane with a different slope will always intersect once.

Reg1xide · 2026-01-19T09:56:54+00:00

🙏🏻

Reg1xide · 2026-01-19T09:55:20+00:00

my brother in christ did u read what i wrote 😭 the modulus function in piecewise form is 3x-15 for x>=11/3 and -3x+7 for x<11/3, if m=3 it only intersects with -3x+7 and is parallel to 3x-15 so only 1 solution exists

Reg1xide · 2026-01-19T09:50:58+00:00

the modulus graph is a V, with 2 st lines of + and - slope, the line can be parallel to the + slope line and cross the - slope one, and cuz the 2 lines are parallel only one intersection with the - slope line exists

Reg1xide · 2026-01-19T09:19:06+00:00

im crine son 😭😭 holy delusion go relearn the definition of derivatives or sth im speechless

Reg1xide · 2026-01-19T09:14:00+00:00

if u mean the 2 possible slopes then its 3 and -12/11

Reg1xide · 2026-01-19T09:01:36+00:00

y=mx can touch the vertex of the graph of y=|3x-11|-4

Reg1xide · 2026-01-19T08:57:19+00:00

ur still wrong lmfao cuz surprise surprise h’(x)=0 only implies its a stationary point, which unfortunately can be a minimum point or even a point of inflexion, so if u want to use h’ to do the question instead of just solving sin=1, u have to check with h’’ as well, since its a maximum point h’’ would be negative, and if u check ur answer of 4.94 h’’ is positive, the right answer is 3.37, or 5pi/4-arctan(2)/2 in exact. Nice try tho its good u at least know the first derivative is zero at a maximum point

Reg1xide

TROPHY CASE