Sol. We have,

f(x) + f(y) = f for all x, y ∈ R ... (i)

Putting x = y = 0, we get

f(0) + f(0) = f(0) ⇒ f(0) = 0

Again putting y = –x, we get

f(x) + f(–x) = f(0) for all x ∈ R

⇒ f(x) + f(–x) = 0 for all x ∈ R [  f(0) = 0]

⇒ f(–x) = –f(x) for all x ∈ R

Hence, f(x) is an odd function.

Now, f ′ (x) =

= [  –f(x) = f(–x)]

= [using (i)

= ×

=

⇒ f(x) = 2 tan –1 x + C [On integration]

⇒ f(0) = 2 tan –1 0 + C [Putting x = 0]

⇒ 0 = 0 + C ⇒ C = 0

∴ f(x) = 2 tan –1 x

∴ f = 2 tan –1 = 2 × =

Also, f ′ (x) =

⇒ f ′ (1) = = 1

Hence, f = and f ′ (1) = 1