) Putting x=0,y=0, we get f(0)=0‌‌............(i) and putting y=−x, we get f(x)+f(−x)=f(0)=0 ∴‌‌f(x)=−f(−x) .........(ii) ‌ Now, ‌f′(x)=
lim
h→0
‌
f(x+h)−f(x)
h
‌‌=
lim
h→0
‌
f(x+h)+f(−x)
h
[fromEq.(ii)] =
lim
h→0
‌
f(
x+h−x
1+(x+h)x
)
h
=
lim
h→0
‌
f(
h
1+x2+xh
)
(‌
h
1+x2+xh
)â‹…(1+x2+xh)
⇒f′(x)=‌
2
1+x2
‌‌(∵
lim
x→0
‌
f(x)
x
=2) ∴f′(−2)=‌
2
1+4
=‌
2
5
‌ and ‌‌‌f(x)=2tan−1x+c ‌‌f(0)=0+c=0 ∴‌‌c=0 ∴‌‌f(x)=2tan−1x ⇒‌‌f(√3)=2tan−1√3=‌
