Relations and Functions

f (f (x) = (a-f (x)/ (a+f (x) = x
Let f (x) = y. (a-y)/ (a+y) = x ⇒ a-y = ax + xy ⇒ a (1-x) = y (1+x) ⇒ y = a (1-x)/ (1+x)
⇒ f (x) = a (1-x)/ (1+x)
From the given options, we infer that comparing the derived f (x) leads to a=1.
⇒ a = 1
So f (x) = (1-x)/ (1+x)
f (-1/2) = (1 - (-1/2)/ (1 + (-1/2) = (3/2)/ (1/2) = 3

For R? let a = 1+√2, b=1-√2, c = 8¹/?
aR? b => a² + b² = (1+√2)² + (1-√2)² = 6 ∈ Q
aR? c => b² + c² = (1-√2)² + (8¹/? )² = 3 ∈ Q
aR? c => a² + c² = (1+√2)² + (8¹/? )² = 3 + 4√2 ∉ Q
∴ R? is not transitive.
For R? let a = 1+√2, b=√2, c=1-√2
aR? B => a² + b² = (1+√2)² + (√2)² = 5+2√2 ∉ Q
bR? b => b² + c² = (√2)² + (1-√2)² = 5-2√2 ∉ Q
aR? c => a² + c² = (1+√2)² + (1-√2)² = 6 ∈ Q
∴ R? is not transitive.

Kindly consider the following Image 

R { (x, y):x, y∈z, x²+3y²≤8}
For domain of R? ¹
Collection of all integral of 's
For x=0, 3y²≤8
⇒ y∈ {-1,0,1}

gogog (3n+1)=gog (3n+2)=g (3n+3)=3n+1. So gogog=I.
If fog=f, then f must map range of g to values consistent with f.
There exists a one-one function f: N→N such that fog=f. e.g. f (x)=x.

19.00
Only '2' in range → 1 function
one element out of 1, 3, 4 is in range with '2'
number of ways = ³C? * (3!/(2!1!)) * 2! = 18
(Select one from 1, 3, 4 and distribute among a, b, c)
Total function = 1 + 18 = 19