mx1 and mx2 questions thread (PLEASE STAY ON TOPIC FOR JSUT THIS THREAD PELAS PELASE PEALSEPE) (1 Viewer)

[synthesisFR]

i post questions here for u to answer

 

[synthesisFR]

For this its its C because like initially u do the range of the x thing
so then (x-1) is between -pi/2 and pi/2 and for that cos is is positive so u just take the postive version?

 

[carrotsss]

look at the original eqns, y=1/sqrt(1-t^2) cant be negative

 

[synthesisFR]

look at the original eqns, y=1/sqrt(1-t^2) cant be negative

but what i did was also correct?

 

[carrotsss]

im gonna be honest i did not understand a word u said

 

[synthesisFR]

cus x = arcsint +1
so (x-1) is arcsint which is between -pi/2 to pi/2
so then cos (x-1) is postive in that range
so when u sub in sin(x-1) from parametric into y, and simplify, and u get an absolute value u just take the postive version?

 

[carrotsss]

yeah u right

 

[synthesisFR]

how would u go about solving this

 

[carrotsss]

no clue i got dx/dt=0.15x

 

[synthesisFR]

no clue i got dx/dt=0.15x

ok no this is impossible if u cant do it then imma move on

 

[carrotsss]

well its common sense like its removing 15% of x per one t i dont rlly see how itd be anything else

what paper is this from

 

[SadCeliac]

GUYS GUYS GUYS WHAT IS THE QUICK WAY TO SOLVE QUESTIONS?????

4u grammar 2021


answer is B

 

[synthesisFR]

GUYS GUYS GUYS WHAT IS THE QUICK WAY TO SOLVE QUESTIONS?????

4u grammar 2021
View attachment 39578

answer is B

u bastard this is my thread

well its common sense like its removing 15% of x per one t i dont rlly see how itd be anything else

what paper is this from

penrith 2022

 

[SadCeliac]

u bastard this is my thread

idrk, now it's become a general mx1 mx2 help thread <3

penrith 2022

bro tf I was born there

 

[nsw..wollongong]

bro tf I was born there

wow. that's awesome.

 

[SadCeliac]

GUYS GUYS GUYS WHAT IS THE QUICK WAY TO SOLVE QUESTIONS?????

4u grammar 2021
View attachment 39578

answer is B

^ ? $ ! ( ) ! zyx thanks

 

[synthesisFR]

wow. that's awesome.

go away math advanced student

 

[carrotsss]

GUYS GUYS GUYS WHAT IS THE QUICK WAY TO SOLVE QUESTIONS?????

4u grammar 2021
View attachment 39578

answer is B

for statement I and II, the limits have an equal length in the positive and negative quadrants, so if u draw it out, for them to be equal to 0 the integrals must be odd functions. I clearly isn't odd by virtue of having an x^2, in II the sec^3 and tan^2 are both even functions and only the remaining tan is an odd function, so an even*odd function = odd function therefore II is true. then its just logic for cos^3, cos has an x intercept at pi/2 so it makes sense that the positive and negative halves would cancel out. hence B

 

[nsw..wollongong]

go away math advanced student

i do 3u but okay... u piss off

 

[SadCeliac]

go away math advanced student

latin ext student*

but let's stay on topic

GUYS GUYS GUYS WHAT IS THE QUICK WAY TO SOLVE QUESTIONS?????

4u grammar 2021
View attachment 39578

answer is B

@carrotsss ^^ you have a shortcut for this?

EDIT THANKS CARROTS

for statement I and II, the limits have an equal length in the positive and negative quadrants, so if u draw it out, for them to be equal to 0 the integrals must be odd functions. I clearly isn't odd by virtue of having an x^2, in II the sec^3 and tan^2 are both even functions and only the remaining tan is an odd function, so an even*odd function = odd function therefore II is true. then its just logic for cos^3, cos has an x intercept at pi/2 so it makes sense that the positive and negative halves would cancel out. hence B