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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
4 years ago

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

4 years ago
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.

4 years ago
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
4 years ago

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
4 years ago

what ever it is
if u want any futher explination plz ask
anyway which school or acedamy u go ur quite laging

4 years ago

my school will be starting from 19th.
4 years ago

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.
4 years ago

if u have any doubts u can ask me if i wont respond u can send it to me on fb
4 years ago

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??
4 years ago

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
4 years ago

i have sent u a request but u wont find me online much.
4 years ago

i can send u things wn ur  offline ri8
4 years ago
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