If ⨍ (x)=sin-12x1+x2, then ⨍ (x) is differentiable on

fx=sin-2x1+x2=2 tan-1x ; -1≤x≤1π-2tan-1x ; x>1-π-2tan-1x ; x 1-21+x2 ; x<-1f1-1-=-1, f1-1+=1 f11-=1, f11-=-1 ∴fx=sin-12x1+x2 is not differentiable at x=±1

If ⨍ (x)=sin-12x1+x2, then ⨍ (x) is differentiable on

fx=sin-2x1+x2=2 tan-1x ; -1≤x≤1π-2tan-1x ; x>1-π-2tan-1x ; x 1-21+x2 ; x<-1f1-1-=-1, f1-1+=1 f11-=1, f11-=-1 ∴fx=sin-12x1+x2 is not differentiable at x=±1

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If $⨍ (x)= \sin^{- 1} (\frac{2 x}{1 + x^{2}}),$ then $⨍ (x)$ is differentiable on

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$[- 1, 1]$

$R - \{- 1, 1\}$

$R - (- 1, 1)$

$R - \{0, 1\}$

answer is B.

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Detailed Solution

$f (x) = \sin^{-} \frac{2 x}{1 + x^{2}} = \{\begin{cases} 2 \tan^{- 1} x; - 1 \leq x \leq 1 \\ π - 2 \tan^{- 1} x; x > 1 \\ - π - 2 \tan^{- 1} x; x < - 1 \end{cases} f^{1} (x) = \{\begin{cases} \frac{2}{1 + x^{2}}; - 1 < x < 1 \\ \frac{- 2}{1 + x^{2}}; x > 1 \\ \frac{- 2}{1 + x^{2}}; x < - 1 \end{cases}$

$f^{1} (- 1^{-}) = - 1, f^{1} (- 1^{+}) = 1 f^{1} (1^{-}) = 1, f^{1} (1^{-}) = - 1 ∴ f (x) = \sin^{- 1} (\frac{2 x}{1 + x^{2}}) is not differentiable at x = \pm 1$