If f(x)= operatornamecosec-1( operatornamecosec x) and operatornamecosec( operatornamecosec-1 x) are equal functions then maximum range of values of x is

Question

If f(x)= operatornamecosec-1( operatornamecosec x) and operatornamecosec( operatornamecosec-1 x) are equal functions then maximum range of values of x is (A) [-(π/2),-1] ∪[1, (π/2)] (B) [-(π/2), 0] ∪[0, (π/2)] (C) (-∞,-1] ∪[1, ∞) (D) [-1,0) ∪[0,1)

Tardigrade Admin · Accepted Answer

y = operatornamecosec-1( operatornamecosec x ) ; x ∈ R - n π, n ∈ I ; y ∈[-π / 2,0) ∪(0, π / 2] and y = operatornamecosec( operatornamecosec-1 x ) ;| x | ≥ 1 ;| y | ≥ 1 ∴ range of value of x y ∈[-(π/2),-1] ∪[1, (π/2)]

Q. If $f (x) = cosec^{- 1} (cosec x)$ and $cosec (cosec^{- 1} x)$ are equal functions then maximum range of values of $x$ is

$[- \frac{π}{2}, - 1] \cup [1, \frac{π}{2}]$

$[- \frac{π}{2}, 0] \cup [0, \frac{π}{2}]$

$(- \infty, - 1] \cup [1, \infty)$

$[- 1, 0) \cup [0, 1)$

$y = cosec^{- 1} (cosec x); x \in R - {nπ, n \in I}; y \in [- π /2, 0) \cup (0, π /2]$
and $y = cosec (cosec^{- 1} x); ∣ x ∣ \geq 1; ∣ y ∣ \geq 1$
$∴$ range of value of $x y \in [- \frac{π}{2}, - 1] \cup [1, \frac{π}{2}]$

Q. If f(x)=cosec−1(cosecx) and cosec(cosec−1x) are equal functions then maximum range of values of x is

[−2π​,−1]∪[1,2π​]

[−2π​,0]∪[0,2π​]

(−∞,−1]∪[1,∞)

[−1,0)∪[0,1)