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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

6 years ago

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

6 years ago

6 years ago

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

6 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

6 years ago

what ever it is

if u want any futher explination plz ask

anyway which school or acedamy u go ur quite laging

6 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.

6 years ago

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

6 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??

6 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

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