Practice exam question

[c23]

hey, i just have a quick question from one of the practice exam papers (neap 2006 exam 2)

Q. If ∫^a_pie/8 cos(2x)dx=0 and 0≤x≤pie then a equals?
i understand that i should integrate it as i would normally and so on
but i got stuck at this step:
sin2a=√(2)/2
what should i do after that step to obtain the answer? or is it even correct :/

thanks

[Flaming_Arrow]

sin 2a = √(2)/2
2a = pi/4, 3pi/4
a = pi/8, 3pi/8

but hence the lower integral = pi/8  a= 3pi/8

[c23]

ah i get you. man i suck at trig

thanks so much:)

[c23]

another trig question i don't know how to work out is :

Q. Let f:R--->R, f(x)=sin(2piex/3)
a. solve the eqn sin(2piex/3)=-_/(3)/3 for [0,3]

i was able to solve this part of the question but the second part is the one i can't seem to figure out

b. Let g:R--->R, g(x)=3f(x-1)+2, find the smallest value of x for which g(x) is a minimum

it's from vcaa exam 1 2007

[Flaming_Arrow]

uhh whats -_/(3)/3?

[c23]

- (square root) 3 / 3

sorry, i don't know how to type those symbols on computer

i want to give you karma but i dont know how do that either lol

[Flaming_Arrow]

b. Let g:R--->R, g(x)=3f(x-1)+2, find the smallest value of x for which g(x) is a minimum



Dilation of factor 3 from the x axis



translation of 1 unit to the right



translation of 2 units up



minimum value of sin is -1

[Flaming_Arrow]

- (square root) 3 / 3

sorry, i don't know how to type those symbols on computer

i want to give you karma but i dont know how do that either lol

are you sure its rt[3]/3? not rt[3]/2?

[c23]

sorry, yeah it's -(square root)[3]/2

the answer says 1/4 is the minimum but 7/4 is the maximum

[Flaming_Arrow]