If ⨍ (x)=1-1-x2, then ⨍ (x) is

fx=1-1-x2 Domain fx=-1, 1 f1x=11-1-x2×x1-x2x≠0, x±1 f1x is not defined if 1-1-x2=0 and 1-x2=0 ⇒f is not differentiable i.e. at x=0,1,-1 f is differentiable in -1,0∪0,1

If ⨍ (x)=1-1-x2, then ⨍ (x) is

fx=1-1-x2 Domain fx=-1, 1 f1x=11-1-x2×x1-x2x≠0, x±1 f1x is not defined if 1-1-x2=0 and 1-x2=0 ⇒f is not differentiable i.e. at x=0,1,-1 f is differentiable in -1,0∪0,1

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If $⨍ (x)= \sqrt{1 - \sqrt{1 - x^{2}},}$ then $⨍ (x)$ is

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continuous on $[- 1, 1]$ and differentiable on $(- 1, 1)$

continuous on $[- 1, 1]$ and differentiable on $(- 1, 0) \cup (0, 1)$

continuous and differentiable on $[- 1, 1]$

not continuous on $[- 1, 1]$ and differentiable on $(- 1, 1)$

answer is B.

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Detailed Solution

$f (x) = \sqrt{1 - \sqrt{1 - x^{2}}} Domain f (x) = [- 1, 1] f^{1} (x) = \frac{1}{\sqrt{1 - \sqrt{1 - x^{2}}}} \times \frac{x}{\sqrt{1 - x^{2}}} (x \neq 0, x \pm 1) f^{1} (x) is not defined if 1 - \sqrt{1 - x^{2}} = 0 and \sqrt{1 - x^{2}} = 0 \Rightarrow f is not differentiable i . e . at x = 0, 1, - 1 f is differentiable in (- 1, 0) \cup (0, 1)$