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Bidisha Baruah Grade: 11
        
find the domain of definition of f(x)=(x+3)6^1/2/{(2-x)(x-5)}^1/2.i want the detailed answer of this question.THANK YOU
2 years ago

Answers : (12)

Nicho priyatham
626 Points
										since the denominator is not equal to 0
we get x is not=2,5
now since the function under root must be >0
we get
           (x+3)/(2-x)(x-5) >0
  so now
    (x+3)/(x-2)(x-5)  
their is something called descardes sign rule u need to apply that
if u dont no whats that rule  den
divide domain of x(wh u know now R-{2,5}) to different parts as shown
   x belongs to (-infinit,-3]U[-3,2)U(2,5)U(5,infinit)

    now consider x belongs to (-infinit,-3]

          so now if x belongs to (-infinit,-3]  
            then x+3 is -ve
                    x-2 is -ve
                    x-5 is -ve
            so now   (x+3)/(x-2)(x-5) = (-ve)/(-ve)(-ve)=-ve
            so wn x belongs to (-infinit,-3]   (x+3)/(x-2)(x-5) is -ve
               so (-inf,-3] is part of domain _____________________-(1)

    now take xE[-3,2)

?             so now if x belongs to [-3,2)  
            then x+3 is +ve
                    x-2 is -ve
                    x-5 is -ve
            so now   (x+3)/(x-2)(x-5) = (+ve)/(-ve)(-ve)=+ve
            so wn x belongs to [-3,2)   (x+3)/(x-2)(x-5) is +ve
               so [-3,2) is NOT part of domain
       

    now take xE(2,5)

?             so now if x belongs to (2,5)  
            then x+3 is +ve
                    x-2 is +ve
                    x-5 is -ve
            so now   (x+3)/(x-2)(x-5) = (+ve)/(+ve)(-ve)=-ve
            so wn x belongs to (2,5)   (x+3)/(x-2)(x-5) is -ve
               so (2,5) is  part of domain___________________(2)
         

    now take xE(5,inf)

?                          so now if x belongs to  (5,inf)
                                then x+3 is +ve
                                        x-2 is +ve
                                         x-5 is +ve
            so now   (x+3)/(x-2)(x-5) = (+ve)/(+ve)(+ve)=+ve
            so wn x belongs to (5,inf)   (x+3)/(x-2)(x-5) is +ve
               so (5,inf) is NOT part of domain
            
so now the domain can be either (1) or (2)
so domain can be (-inf ,-3]U(2,5)
PLZ APPROVE if use full and in case of any doubts ask in ans box
 
2 years ago
Bidisha Baruah
20 Points
										x=R-(2,5) i got that part but the rest i didnt get “x belongs to (-infinit,-3]U[-3,2)U(2,5)U(5,infinit) “where and how did u (-infinity,3] and how is it that if (x+3)/(x-2)(x-5)=+ ve it is not the part of the domain as the function is supposed to be greater than 0.sorry for bothering u but i need the answer.
										
2 years ago
Nicho priyatham
626 Points
										sorry it is a typing mistake it must be   (x+3)/(x-2)(x-5)=- ve (in the 7th line of solution)
c if  (x+3)/(2-x)(x-5)=+ve then (x+3)/(x-2)(x-5)=- ve ri8......?  
and its not R-(2,5) its R-{2,5}
the solution i gave u is correct expect these typing mistakes which made the solution rubbish
 
this problem wouldnt be this lengthy if u would know
Descartes rule of signs
 it is just 3 steps problem but u dont know whats dat so i took a risk  of typing such a lengthy solution it  took me 45 mins to do dat
i am sorry that their were some mistakes
n also sorry coz i confused u
better stop undusting the solution i gave u
if u r much worried abt the answer then  the ans is
 
(-inf,-3]U(2,5)
if u were much bothered abt the ans then why did u ask detailed explination
anyway this will be the last question i will ans u so sorry 1s again
all the best
if u feel solution is  WRONG PLZ DISAPPROVE it
 in case of any doubts put it in the ans box
2 years ago
Bidisha Baruah
20 Points
										
i am sorry!!u took it in d wrong way...i was sorry for u that i was bothering u by asking the same kind of questions again and again.i am really very thankful to u that u r answeing my questions.i just wanted to clear my doubt and i was not critisizing u.typing mistakes happen.its not a big deal.i just didnt get some steps coz i am not as intelligent as u r.i am extremely sorry but i didnt mean to disrespect u
2 years ago
Nicho priyatham
626 Points
										
what ever it is
if u want any futher explination plz ask 
anyway which school or acedamy u go ur quite laging
 
 
2 years ago
Bidisha Baruah
20 Points
										
my school will be starting from 19th.
2 years ago
Bidisha Baruah
20 Points
										
i have problem regarding this chapter and i am in search of someone who will clear my doubts.And before my school starts i wanted to complete some of the chapters but i am stuck!And i am worried.
2 years ago
Nicho priyatham
626 Points
										
if u have any doubts u can ask me if i wont respond u can send it to me on fb
2 years ago
Bidisha Baruah
20 Points
										
this whole expression is under the root so we need to set the numerator >=0 and denominator should not be zero.So if we set the numerator >=0 than we get x>=-3.Then in this case we write [-3,infi.).But the asnwer is starting from “-infinity”.How??
2 years ago
Nicho priyatham
626 Points
										
 
 
ur saying that WHOLE expression is under root 
so Numarator need b >zero but
(numator)/(denominator) must be > zero
 
see what happens wn x belongs to (-inf,-3]
numator
and denominator also
so as the whole expression becomes >0
 
incase of further doubts put it in ans box 
 
u no wt bidisha i can never explain u like this 
i huv sent u req on FB plz acept it if u dont mind
i  really wanna help u  i am sure i will clarify all ur doubts till ur final jee advanced
2 years ago
Bidisha Baruah
20 Points
										
i have sent u a request but u wont find me online much. 
2 years ago
Nicho priyatham
626 Points
										
i can send u things wn ur  offline ri8
2 years ago
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