Test: Composition Of Functions - JEE MCQ

We are given that functions 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 g is one-to-one, therefore,
f(a¹) = f(a²)
And since f 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.

A composite function is denoted by (g o f) (x) = g (f(x)). The notation g o f is read as “g of f”.
Example : Consider the functions f: A → B and g: B → C. f = {1, 2, 3, 4, 5}→ {1, 4, 9, 16, 25} and g = {1, 4, 9, 16, 25} → {2, 8, 18, 32, 50}. A = {1, 2, 3, 4, 5}, B = {16, 4, 25, 1, 9}, C = {32, 18, 8, 50, 2}.Here, g o f = {(1, 2), (2, 8), (3, 18), (4, 32), (5, 50)}.

If f : A → B such that y implies B then f^-1(y) = {x implies A : f(x) = y}
Let f^-1(9) = x
f(x) = 9
x² = 9
x = +-3
Thus, f^-1(9) = {-3,3}

g(x) = 1 + x - [x]
f(x) = {-1, x<0   0, x=0    1, x>0}
f(g(x)) = f[1 + x - [x]]
= f[1 + x - 1]
= f(x)
f(g(x)) is equal to f(x) 

f(0)=0×sec0
f(0)=0×1
f(0)=0

gof(2)=g[f(2)]
=g[1]=2

 f(x) = |x|   g(x) = |5x - 2|
fog = f(g(x) = g(x) = |5x - 2|

f : R → R     g : R → R
f(g) : R → R
f(x) = 2x + 3      g(x²) = x² + 7
f(g(x)) = 2g(x) + 3
f(g(x)) = 2(x² + 7) + 3
= 2x² + 17
As it is given f(g(x)) = 25
Comparing both the eq, we get
2x² + 17 = 25
2x² = 8
x = +2