C3 January 2013 25/01/2013

about the domain for part b

ok look at the restriction of f(x) u have from 0 to pi/2 inclusive and for g(x) you have greater than or equal to 0.
now when you shift the graph by pi/2 to the left the domain is restricted by that of g(x) coz it is always greater than or equal to 0. so the domain has to be greater than or equal to 0 and where will you stop then you stop at pi/2 coz u shifted that pi by pi/2 to the left.
Its just like seeing what will be common for both of the functions.

so do u mean the answer is x>equal to -pie/2 ?

no its 0<=fg(x)<=pi/2

but tell will that give maths error and why 0

its cos sin was negative! and also obtuse means betwn 90 and 180 and tan is positive in 3rd and first quadrant so it was negative.
if cos was negative then that means only sin is positive coz sin is positive in second quadrant which is between 90 and 180

when u keep 0 you get cospi/2 right? which is 0
why do u get maths error?

I still have no idea where you got obtuse and all that stuff from when it wasn't mentioned in the question.

I've always thought that if cos was positive... that means its either in A or C quadrant.... if its negative then that means its positive in S and T quadrant

Am I missing something here... sorry this is going on a bit now

cause we have been taught by the teacher that the domain is usually the value when you sub a value into the equation and that gives u the maths error and that is the domain.

no! its not always true! if u get maths error first thing is it shouldn't be domain! coz a domain is one which has element in range

yes if you get maths error you put the value which you got but saying > for example

so what method would u chose to sure u are correct? is it the values they provided like the restricted domain and sub those values into the equation

domain is x-axis so most of the time I try drawing graph to be sure. but u can't always work out by graph but there will be some more thing that they will give of those type.

do u think i should draw graphs? is there any way u can check ur answer

Hi can somebody please help me with JAN 2007 Q6a, I did the inverse function but am confused for the domain of the inverse.

The domain of the function is x<2 so why is the domain of the inverse all real values...

Thanks

LOL! I don't know how can I make you understand.
obtuse- betwn 90 and 180.
relfex-greater than 180 but less than 360
now tell me the angle at different quadrant?

ya graph is always good to visualise anything you are confused with. don't think there is.

Sorry

given that x=cos^-1(-5/13) find exam value of:

a) tan(x)
b) cosec(2x)

^^^ okay now I don't understand how you figured out it was obtuse Yes because if cos is negative then it must be in sin quadrant.. but what about the tan quadrant ? that is in reflex territory.




Last time If I still don't get it , then god help me

will do that!

one thing you know sometimes we are given like 0-270 for example and we need to find sin will sin be positive in this cause as it does pass through the range of sin?

Sorry

given that x=cos^-1(-5/13) find exam value of:

a) tan(x)
b) cosec(2x)

^^^ okay now I don't understand how you figured out it was obtuse Yes because if cos is negative then it must be in sin quadrant.. but what about the tan quadrant ? that is in reflex territory.




Last time If I still don't get it , then god help me