The range of the function f ( x )= sec -1 x + tan -1 x + operatornamecosec-1 x, is

Question

The range of the function f ( x )= sec -1 x + tan -1 x + operatornamecosec-1 x, is (A) (0, (π/2)) (B) (0, π) (C) (0, (π/4)) ∪[(3 π/4), π) (D) (0, (π/4)] ∪[(3 π/4), π)

Tardigrade Admin · Accepted Answer

The domain of the function f ( x ) is (-∞,-1] ∪[1, ∞) f ( x )=(π/2)+ tan -1 x tan -1 x ∈((-π/2), (-π/4)] ∪[(π/4), (π/2)) Hence, range of f(x) is (0, (π/4)] ∪[(3 π/4), π).

Q. The range of the function $f (x) = sec^{- 1} x + tan^{- 1} x + cosec^{- 1} x$ , is

$(0, \frac{π}{2})$

$(0, π)$

$(0, \frac{π}{4}) \cup [\frac{3 π}{4}, π)$

$(0, \frac{π}{4}] \cup [\frac{3 π}{4}, π)$

The domain of the function $f (x)$ is $(- \infty, - 1] \cup [1, \infty)$
$f (x) = \frac{π}{2} + tan^{- 1} x$
$tan^{- 1} x \in (\frac{- π}{2}, \frac{- π}{4}] \cup [\frac{π}{4}, \frac{π}{2})$
Hence, range of $f (x)$ is $(0, \frac{π}{4}] \cup [\frac{3 π}{4}, π)$ .

Q. The range of the function f(x)=sec−1x+tan−1x+cosec−1x, is

(0,2π​)

(0,π)

(0,4π​)∪[43π​,π)

(0,4π​]∪[43π​,π)