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22 Jun 2008 21:46:10 IST
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find the domain of
sin inverse [sec x] where [.] represents the greatest integer functions..
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IMPOSSIBLES ARE OFTEN UNTRIED... |
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22 Jun 2008 21:51:06 IST
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it is -infinity to infinity
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22 Jun 2008 21:51:27 IST
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rate if right
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22 Jun 2008 21:53:04 IST
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nooo the ans. is [ (2n+1)pie) ] union [2npie , 2npie - pie /3 ]
where n belongs to integer.....
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22 Jun 2008 22:44:53 IST
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the range of secx is R-{-1,1}.but sin inverse x is defined for -1 to 1.so domain is null.
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22 Jun 2008 22:56:09 IST
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please rate me and tell whether my answer is true or false?
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22 Jun 2008 22:59:06 IST
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yup.........it is a null set i.e da function does nt exists as da value of [sec x ] never lies b/w [-1 , 1]
& sin inverse x is defined only fr x = [-1 ,1]
EDIT-lozzzz ......made a mistake secx =-1 so dat [sec x ] =[-1,1]
so domain aint null
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22 Jun 2008 23:03:46 IST
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the domain of sin -1 x is [-1,1]
sec x lies between [-infinity,-1] union [1,infinity]
sec x is equal to -1 at (2n+1)  ,so [sec x]=-1 at this value
[sec x] =1 means sec x lies between [1,2)
that happens when x lies between [0, /3) ,general eq. is [2n ,2n +- /3)
so the final answer is (2n+1) union [2n ,2n +- /3)
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22 Jun 2008 23:14:21 IST
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the inverse exists only if
[secx] = -1 & +1
so
secx belongs to [-1,0) U [1,2)
in [-1,0) only {-1} is possibl--------so x= (2n+1)pi
in [1,2)-----x takes values from [2n pi , 2n pi +- sec inverse2)
--------------[2n pi , 2n pi +- pi/3)
finally take the union....
hey y r u guyz missing the +- .........sec follows cos .......nd so wen u take gen solution u'll have( 2npi +- alpha)
Hope that Helps!
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2nd year
B.E.(Hons) Mechanical Engineering
BITS PILANI PILANI CAMPUS~! |
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22 Jun 2008 23:18:15 IST
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sorry,domain is all the values of -1&1.
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22 Jun 2008 23:24:29 IST
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but @ vasant , rohit
the secx is [1,2) not only in 0 to pie/3 it is also in -pie/3 to 0
so general solution can be ( 2npie - pie/3 , 2npie + pie/3)..
that is where my ans is not matching....
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22 Jun 2008 23:28:39 IST
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ya, sorry abt that .i forgot that in cos you have to take the +- part in the general eqn.
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22 Jun 2008 23:29:41 IST
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hi @akshaykhare
i have also got the same answer..........i used +- jus to make the sign omitters kno their mistakes.......
i stated [2n pi , 2npi + - pi/3)
unstead of my +- u've used the conventional method-------( 2n pi - pi/3, 2npi + pi/3)--------
we've got the same answer!
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2nd year
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BITS PILANI PILANI CAMPUS~! |
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22 Jun 2008 23:29:57 IST
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sooo ...how the ans .is going come ...????????
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22 Jun 2008 23:31:56 IST
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ya i also think the ans given in book is wrong...
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22 Jun 2008 23:35:29 IST
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but the ans is given only + pie/3 ...
i am also getting same ans a
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22 Jun 2008 23:43:35 IST
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domain -1to+1
sin inverse[sec]
d0main =[-1,1]
and it satifies it............!
rate if helped
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22 Jun 2008 23:43:57 IST
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domain -1to+1
sin inverse[sec]
d0main =[-1,1]
and it satifies it............!
rate if helped
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22 Jun 2008 23:45:28 IST
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sec never lies between -1 to +1 ...
that is not sec inverse that is sin inverse..
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22 Jun 2008 23:53:49 IST
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wll u hav 2 find domain of sininverse na sin ^-1[sec] sec= R-(2pie+1)pie/2 nd sin^-1 =[-1,1] .......................
in but there is not a common interval................!
so it is null...........
domain= 
not defined
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