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1. find domain of f(x)=logxcosx.here, xcosx inside the function log.
2. find domain of f(x)=arccos(x+[x]),here[.] representes greatest integer function.
For log(x*cos(x)) to exist x*cos(x) > 0.
For this purpose, either x and cos(x) must be simultaneously > 0 or they should be simultaneously less than 0.
Case I: For x>0 , x>0 and for Cos(x) > 0, x should belong to [0,pi/2) U (3pi/2,5pi/2) U (7pi/2, 9 pi/2) U ...
Case II: For x<0 , x<0 and for Cos(x) < 0, x should belong to (-pi/2,-3pi/2) U ( -5pi/2,-7pi/2) U (-9pi/2,-11pi/2) U ...
So Its defined for the following sets.
Pls Specify clearly what is arccos(x+[x]).
arccos means inverse of cos.and inside cos dere is x + [x] where [.] is greatest integer function.
Hi Manuj,
As we know cos-1(t) is defined for -1 to 1 and for t to be b/w -1 and 1 , -1< x + [x] <1
So domain of the following function is [0,1)
This is so as for x< 0 , [x] will take values = -1 and the sum will be < -1 which is not desirable
Similarly for x<1 , x is 0 so x + [x] will <1
Howeve if x= 1 then [x]=1 and in such a case the sum will again be >1 which is not desirable
As such domain is [0,1)
i understood ur solution......thanks for efforts.......but cant we solve mathematically tht equation (-1<=x+[x]<=1)
Hi, Manuj
No, this cant be done algebraically, however u can often use graphs for solving such problem. Graphing can be an important tool in ones hand for solving such problems. So better u start practicing graph sketching.
Eg. the graph of this problems ( x+[x] )appears something like this
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