IIT JEE , AIEEE Forum - Mathematics , Algebra

udbarguj

domain &range of

(a-x)₊

(x-b)

Wherea>b>o.

Plzzzzzzzz help

catch_arnnie

(a-x)₊

(x-b)

since a > b >0

so, for domain, x>a & x>b,

=> domain is [a,

)

shubham_sachdeva

domain is [b,a]

@
catch_arnie
put infinity in the qn. u will get -ve symbol in first root...

catch_arnnie

i'm sorry, my mistake, didn't see the question carefully....

smartguymihir

(a-x) +

(x-b)

the domain is

= a > b > 0

= -a > -b > 0 or = -a < -b < 0

=x-a > x-b > x or x-a < x-b < x

=

(x-a) >

(x-b) >

(x) or ------< ------< -------

=(a-x) + (x-a) >

(a-x) +

(x-b) > (a-x) + x

or

=(a-x) + (x-a) <

(a-x) +

(x-b) < (a-x) + x

udbarguj

But what about range?

neeraj_agarwal_1990

range is root(a-b) to infinity...

priyesh

f(x) =

a-x +

x-b

for domain

a > x & x>b

therefore domain is [b,a]

now for range

f ' (x) = 1/2 [ 1/

x-b - 1/

a-x ]

therefore for f ' (x) to be positive

x-b <

a-x

=) x < (a+b)/2

and for f ' (x) to be negative x > (a+b)/2

& f ' (x) is 0 at x = (a+b)/2

therefore maxima is attained at x = (a+b)/2

now minimum will be either x = b or x = a

but f(a) = f(b)

therefore range is [ f(b) , f((a+b)/2) ]

therefore range is [ a-b , 2(a-b)/2

sweta_04

a-x>o=xsince a>b,domain is from a to infinity
fo finding range fine the inverse of the given function and aplly necessary conditions on it
[sqroot(a-b),2 sqroot(a-b)\2]

priyesh

hey sweta u cannot find the inverse in such problems because
firstly because the function is not one-one
secondly u cannot find whether function is onto because for that the codomain should be given
NOTE: only a bijective function i.e one-one & onto can have an inverse.For other functions the inverse does not exist

neeraj_agarwal_1990

sorry for the wrong answer...

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