Question

The domain of real valued function
f(x) =
75+X-V5–2
is​

Answer 1

Step-by-step explanation:

ANSWER

4

f(x)

+4

1−f(x)

=4

x

4

f(x)

+

4

f(x)

4

=4

x

4

2fx

+4=4

x+f(x)

1+4

2f(x)−1

=4

x+f(x)−1

-----{1}

1+4

1−2fx

=4

x−f(x)

-----{2}

both equation 1 & 2 are same.

So, 4

1f(x)−1

=4

1−2f(x)

and 4

x+f(x)−1

=4

x−f(x)

2f(x)−1=1−2f(f) & x+f(x)−1=x−f(x)

f(x)=

2

1

& f(x)=

2

1

i.e. f(x) is a constant function.

but from above equation 4

x

=f

f(x)

+4

1−f(x)

4

x

=4

2

1

+4

2

1

=4

x has to be 1 atleast

So, x∈[1,+∞)