Test: Composition of Functions - JEE MCQ

We have f(x) = ax+b, g(x) = cx+d
Therefore, f{g(x)} = g{f(x)} 
⇔ f(cx+d) = g(ax+b)
⇔ a(cx+d)+b = c(ax+b)+d
⇔ ad+b = cb+d 
⇔ f(d) - g(b)

let y = (4x + 3)/(6x - 4)
⇒ y * (6x - 4) = 4x + 3
⇒ 6xy - 4y = 4x + 3
⇒ 6xy - 4y - 4x = 3
⇒ 6xy - 4x = 3 + 4y   
⇒ x(6y - 4) = 4y + 3
⇒ x = (4y + 3)/(6y - 4)
So, f^-1 (x) = (4x + 3)/(6x - 4)

Given f:R→R and g: R→R 
=> fog : R→R and gof :  R→R 
We know that I(R) : R→R 
So the domains of gof, fog I(R) are the same.
fog(x) = f(g(x)) = f(x-1) = x-1+1 = I(R)x…..(1)
gof(x) = g(f(x)) = g(x-1) = x-1+1 = I(R)x……(2)
From (1) and (2), we get
fog(x) = gof(x) = I(R) for all x belongs to R
Hence fog = gof

f:A→B and g:B→C are both one-to-one functions.
Suppose a₁,a₂∈A such that (gof)(a₁)=(gof)(a₂)
⇒ g(f(a₁)) = g(f(a₂)) (definition of composition) 
Since gg is one-to-one, therefore,
f(a₁) = f(a₂)
And since ff is one-to-one, therefore,
a₁ = a₂
Thus, we have shown that if (gof)(a₁)=(gof)(a₂ then a₁=a₂
Hence, gof is one-to-one function.

since g : B → C is onto
Suppose z implies C, there exist a pre image in B
Let the pre image be y
Hence y implies B such that g(y) = z
Similarly  f : A → B is onto
Suppose y implies B, there exist a pre image in A
Let the pre image be x
Hence y implies A such that f(x) = y
Now, gof : A → C gof = g(f(x))
= g(y) 
= z
So, for every x in A, there is an image z in C
Thus, gof is onto.

f(x) = x² + 5 = y(let)
⇒ y = x²+5
⇒ x² = y-5 
⇒ x = (y-5)^1/2
⇒ f-1(x) = (x-5)^1/2

R₁ is a reflexive on A , because (a,a) ∈ R₁  for each  a ∈ A