393) gof =ln(√1-x)
now √1-x is always greater than 0 when x <1
sop domain is (-∞,1)
range is (-∞,∞)
1) its pretty obvious
f(x0=√x-2
f(f(x))=√√(x-2) -2
so for this to be valid √x-2 >2
so x>6
and continue so on
I can find the domain and range when there is a case of composition of general function. But,I face difficulty in finding out the range and domain of real functions.
Am posting some doubts, please help me out.
1)Let f be a real function given by f(x) = √(x-2)
find each of the following
a)fof
b)fofof
c) (fofof)38
2)If f : (-π/2, π/2)→R and g :[-1,1]→R be defined as
f(x) = tan x and g(x) = √(1-x2)
Describe fog and gof.
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8 Answers
113) If f(x) = √(1 - x) and g(x) = logex are two real functions, then describe functions
fog and gof.
Please teach me to find out the range and domain, achchhe se [1]
Thanks in advance [1]
393)
fog=√(1-lnx)
now domain of f is (-∞,1)
domain of g is (0,∞)_
so domain of fog is intersection of these 2 which is (0,1)
now range is as x approaches 0 , ln x tends to -∞ and as x approaches 1 , ln x tends to 0
so range will be (0,∞)
39392)
left it so that u could give a try
fog=√(1-tan2x)
for this to be valid tan2x<1 implies from the given domain of f , for this inequality to be satisfied, we have
x belongs to (-Î /4,Î /4)U(-3Î /4,3Î /4)
and range as we can see is (0,√2)
similarly try for gof
11@ Shanks, I tried but nahi kar payi , so was asking, in trigo functions some times I get stuck in range some times in domain, algebraic ones toh got, thanks a lot, [2]
@Qwerty, am using R D Sharma for basics and Amit M Aggarwal, agar koi dusri standard book hai toh bata do
Anyways Thanks for the help
1well, these seek basic common sense and if u arent able to do
i suggest u once again go through ncert text bo