maths-functions

[ritu]

find the domain of following function..........
f(x)=√1-|x|/√(2-|x|

note:under root is over whole of numerator and denominator

[anon_econ]

Is the answer (-oo, -2) union [-1,1] union (2, oo)?

[Bellatrix]

@Vasudha- anything greater than 2 or lesser than -2 will give a negative number inside the root of the denominator. So try again :)

[anon_econ]

But i thot since there is a square root sign over the entire thing, the numerator & denominator could both be either (-) at the same time or (+) at the same time (to make the ratio +, rather than each + separately)..

[anon_econ]

I think i did not interpret the note properly..i thot the root is over the entire expression. like only 1 square root sign.

[ritu]

hey vasudha...bang on...ur ans is correct..........
see firstly...explain me one thing..since whole expression is under the root sign so it has to be >=0......so as u said either both numerator and denominator should be >0 or both <0....this means we have following cases.............
1. if both positive then........mod x <=1 and mod x<2...COMMON AREA ON NUMBER LINE IS...mod x <1
2.if both negative.......mod x >=1 and mod x >2......COMMON AREA ON NUMBER LINE IS  mod x >2......


NOW....MOD X <=1 MEANS X<=1 OR X>= (-1)......so common area is [-1,1]

MOD X>2..........MEANS  X>2....OR  X<(-2).....BUT IN THIS CASE THERE'S NO COMMON AREA ON REAL LINE.....X>2....AND   X<(-2). BEING DISJOINT SETS......SO TATS WHY U HAVE TAKEN THE UNION OF THESE 3 AREAS---ie.....[-1,1] U  (-infinity,-2)   U   (2,infinity)??????????is this is the whole logic behind ur answer....??????m sry for this long query....:)))

[ritu]

hey bellatrix hi:))))ya  <(-2).....or >2....would make not only denominator negative but also the numerator.....so it will cancel out....and we still have a positive expression:))))))))

[Bellatrix]

Aare haan! Thanks :)

[anon_econ]

Hi ritu,
For the case when both are negative, mod x>2 gives us the set (-oo, -2) U (2, oo) [v r not concerned abt it being disjoint], and mod x>1 gives us the set (-oo, -1) U (1, oo).
The intersection of the 2 is (-oo, -2) U (2, oo).
So the final answer is its union with [-1,1].

[ritu]

@vasudha....thanx dear now my concept is clear....:)))))