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

Question

The range of the function f(x)= sec -1(x)+ tan -1(x), is (A) (0, π) (B) ((-π/2), (3 π/2)) (C) (0, (3 π/4)] (D) none

Tardigrade Admin · Accepted Answer

D f =(-∞,-1] ∪[1, ∞) Also, f is an increasing function. For, x ∈(-∞,-1], f(x) ∈(0, (3 π/4)]......(1) and for x ∈[1, ∞), f(x) ∈[(π/4), π).....(2) ∴ For range of f ( x ),(1) ∪(2) ⇒(0, π).

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

$(0, π)$

$(\frac{- π}{2}, \frac{3 π}{2})$

$(0, \frac{3 π}{4}]$

none

$D_{f} = (- \infty, - 1] \cup [1, \infty)$
Also, $f$ is an increasing function.
For, $x \in (- \infty, - 1], f (x) \in (0, \frac{3 π}{4}]$ ......(1)
and for $x \in [1, \infty), f (x) \in [\frac{π}{4}, π)$ .....(2)
$∴$ For range of $f (x), (1) \cup (2) \Rightarrow (0, π)$ .

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

(0,π)

(2−π​,23π​)

(0,43π​]