Post by: c23 on October 07, 2009, 01:11:09 pm
Q. If ∫apie/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
Post by: Flaming_Arrow on October 07, 2009, 01:46:42 pm
2a = pi/4, 3pi/4
a = pi/8, 3pi/8
but hence the lower integral = pi/8 a= 3pi/8
Post by: c23 on October 07, 2009, 01:51:06 pm
thanks so much:)
Post by: c23 on October 07, 2009, 02:10:21 pm
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
Post by: Flaming_Arrow on October 07, 2009, 02:51:49 pm
Post by: c23 on October 07, 2009, 03:12:09 pm
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
Post by: Flaming_Arrow on October 07, 2009, 03:25:11 pm
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
Post by: Flaming_Arrow on October 07, 2009, 03:25:56 pm
- (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?
Post by: c23 on October 07, 2009, 03:34:28 pm
the answer says 1/4 is the minimum but 7/4 is the maximum
Post by: Flaming_Arrow on October 07, 2009, 03:37:58 pm