Identify which of the following pair(s) of functions is(are) identical?
[Note : [k], {k} and sgn(k) denote greatest integer function, fractional part function and signum function of k respectively.]
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We analyze each pair for identical functions (same domain and mapping).
First pair: f(x) = sin-1([{x}]), g(x) = 1 - cos-1({[x]}). Since {x} ∈ [0,1), [{x}] is always 0. Thus f(x)=sin-1(0)=0. Similarly, [x] is integer, so {[x]}=0, and g(x)=1-cos-1(0)=1-π/2, a constant. f and g are different constant functions. Not identical.
Second pair: f(x) = (sin x + cos x)/(sin x - cos x), g(x) = (cot x + 1)/(1 - cot x). Simplify f(x) by dividing numerator and denominator by cos x: f(x) = (tan x + 1)/(tan x - 1). Since cot x = 1/tan x, g(x) = (1/tan x + 1)/(1 - 1/tan x) = ( (1+tan x)/tan x ) / ( (tan x - 1)/tan x ) = (1+tan x)/(tan x - 1) = -f(x). So f(x) ≠ g(x). Not identical.
Third pair: f(x) = |x|/(1+|x|), g(x) = (x sgn(x))/(1+√(x²)). Note sgn(x) = |x|/x for x≠0, so x sgn(x)=|x|. Also √(x²)=|x|. Thus g(x)=|x|/(1+|x|) = f(x). Domain is all real numbers for both. Identical.
Fourth pair: f: R→R, f(x)=(16-x³)1/3, g(x)=f-1(x). f is its own inverse? Check: f(f(x)) = (16 - (16-x³))1/3 = (x³)1/3=x. So f is an involution, meaning f = f-1. Thus g(x)=f(x). Identical.
Final Answer: The third and fourth pairs are identical.