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.

Approved

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