Let be such that the functionfx=cos−11−x2sin−11−xx−x3,x≠0 α, x=0 is Continuous at x=0,where x=x−x,x is the greatest integer less than or equal to x. Then :

RHL=limx→0+cos-11-x2sin-1(1-x)x1-x2=π2limx→0+cos-11-x2x=π2limx→0+-11-1-x22(-2x) (L'Hospital Rule) =πlimx→0+x2x2-x4=πlimx→0+12-x2=π2LHL=limx→0-cos-11-(1+x)2sin-1(-x)(1+x)-(1+x)3=π2limx→0-sin-1x(1+x)(1+x)2-1=π2limx→0+sin-1xx2+2x=π212=π4 As LHL≠RHL so f(x) is not continuous at x=0.There is no such α such exists

Let be such that the functionfx=cos−11−x2sin−11−xx−x3,x≠0 α, x=0 is Continuous at x=0,where x=x−x,x is the greatest integer less than or equal to x. Then :

RHL=limx→0+cos-11-x2sin-1(1-x)x1-x2=π2limx→0+cos-11-x2x=π2limx→0+-11-1-x22(-2x) (L'Hospital Rule) =πlimx→0+x2x2-x4=πlimx→0+12-x2=π2LHL=limx→0-cos-11-(1+x)2sin-1(-x)(1+x)-(1+x)3=π2limx→0-sin-1x(1+x)(1+x)2-1=π2limx→0+sin-1xx2+2x=π212=π4 As LHL≠RHL so f(x) is not continuous at x=0.There is no such α such exists

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Let be such that the function $f (x) = \{\begin{matrix} \frac{\cos^{- 1} (1 - {\{x\}}^{2}) \sin^{- 1} (1 - \{x\})}{\{x\} - {\{x\}}^{3}}, x \neq 0 \\ α, & x = 0 \end{matrix}$
is Continuous at $x = 0,$ where $\{x\} = x - [x], [x]$ is the greatest integer less than or
equal to x. Then :

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$n o s u c h α e x i s t s$

$α = 0$

$α = \frac{π}{4}$

$α = \frac{π}{2}$

answer is B.

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Detailed Solution

$\begin{array}{l} RHL = \lim_{x \to 0^{+}} \frac{\cos^{- 1} (1 - x^{2}) \sin^{- 1} (1 - x)}{x (1 - x^{2})} = \frac{π}{2} \lim_{x \to 0^{+}} \frac{\cos^{- 1} (1 - x^{2})}{x} \\ = \frac{π}{2} \lim_{x \to 0^{+}} \frac{- 1}{\sqrt{1 - {(1 - x^{2})}^{2}}} (- 2 x) (L'Hospital Rule) \\ = π \lim_{x \to 0^{+}} \frac{x}{\sqrt{2 x^{2} - x^{4}}} = π \lim_{x \to 0^{+}} \frac{1}{\sqrt{2 - x^{2}}} = \frac{π}{\sqrt{2}} \end{array}$

$\begin{array}{l} LHL = \lim_{x \to 0^{-}} \frac{\cos^{- 1} (1 - (1 + x)^{2}) \sin^{- 1} (- x)}{(1 + x) - (1 + x)^{3}} = \frac{π}{2} \lim_{x \to 0^{-}} \frac{\sin^{- 1} x}{(1 + x) [(1 + x)^{2} - 1]} = \frac{π}{2} \lim_{x \to 0^{+}} \frac{\sin^{- 1} x}{x^{2} + 2 x} \\ = \frac{π}{2} (\frac{1}{2}) = \frac{π}{4} As LHL \neq RHL so f (x) is not continuous at x = 0 . \end{array}$