Let f: (-1 ,1) → B , be a function defined by f(x) = tan-1 (2x/1 - x2) , then f is both one - one and onto when B is the interval

Question

Let f: (-1 ,1) → B , be a function defined by f(x) = tan-1 (2x/1 - x2) , then f is both one - one and onto when B is the interval (A) (0, (π/2))  (B) [0, (π/2))  (C) [- (π/2) , (π/2)]  (D) (- (π/2) , (π/2))

Tardigrade Admin · Accepted Answer

Given f(x) = tan-1 ((2x/1-x2)) = 2 tan-1x for x ∈(-1,1) If x ∈(-1,1) ⇒ tan-1 x ∈((- π/4), (π/4)) ⇒ 2 tan-1 x ∈((-π/2), (π/2)) Clearly, range of f(x) = (- (π/2) , (π/2)) For f to be onto, codomain = range ∴ Co-domain of function = B = (- (π /2) , (π /2))

Q. Let $f : (- 1, 1) \to B$ , be a function defined by $f (x) = tan^{- 1} \frac{2 x}{1 - x ^{2}}$ , then f is both one - one and onto when B is the interval

$(0, \frac{π}{2})$

$[0, \frac{π}{2})$

$[- \frac{π}{2}, \frac{π}{2}]$

$(- \frac{π}{2}, \frac{π}{2})$

Q. Let f:(−1,1)→B , be a function defined by f(x)=tan−11−x22x​ , then f is both one - one and onto when B is the interval

(0,2π​)

[0,2π​)

[−2π​,2π​]

(−2π​,2π​)

Given f(x)=tan−1(1−x22x​)=2tan−1x for x∈(−1,1) If x∈(−1,1)⇒tan−1x∈(4−π​,4π​) ⇒2tan−1x∈(2−π​,2π​) Clearly, range of f(x)=(−2π​,2π​) For f to be onto, codomain = range ∴ Co-domain of function =B=(−2π​,2π​)

Q. Let $f : (- 1, 1) \to B$ , be a function defined by $f (x) = tan^{- 1} \frac{2 x}{1 - x ^{2}}$ , then f is both one - one and onto when B is the interval

$(0, \frac{π}{2})$

$[0, \frac{π}{2})$

$[- \frac{π}{2}, \frac{π}{2}]$

$(- \frac{π}{2}, \frac{π}{2})$