Let f:(-1, 1) to B be a function defined by f | vedclass

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(B) For $-1 < x < 1$,we use the substitution $x = 	an 	heta$,where $	heta \in (-\frac{\pi}{4}, \frac{\pi}{4})$.Then,$\frac{2x}{1-x^2} = \frac{2	an

Vedclass · Accepted Answer

$(-\frac{\pi}{2}, \frac{\pi}{2})$

Vedclass · Answer

(B) For $-1 Then,$\frac{2x}{1-x^2} = \frac{2	an 	heta}{1-	an^2 	heta} = 	an(2	heta)$.Thus,$f(x) = 	an^{-1}(	an(2	heta)) = 2	heta = 2	an^{-1}x$.Since $x \in (-1, 1)$,$	an^{-1}x \in (-\frac{\pi}{4}, \frac{\pi}{4})$.Therefore,$f(x) \in (2 	imes -\frac{\pi}{4}, 2 	imes \frac{\pi}{4}) = (-\frac{\pi}{2}, \frac{\pi}{2})$.For the function to be onto,the codomain $B$ must be equal to the range of the function.Hence,$B = (-\frac{\pi}{2}, \frac{\pi}{2})$.

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

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