Question

If $f\left(x\right)=sin\left\{\right[x+5]+\{x-\{x-\{x\left\}\right\}\left\}\right\}$ for $xϵ(0,\frac{\pi }{4})$ is invertible, where {.} and [.] represent fractional part and greatest integer functions respectively, then $f^{-1}\left(x\right)$ is

• A. $si{n}^{-1}x$
• B. $\frac{\pi }{2}-co{s}^{-1}x$
• C. $si{n}^{-1}\left\{x\right\}$
• D. $co{s}^{-1}\left\{x\right\}$

Answer 1

Answer: (A), (B), (C)

The correct options are
A

$si{n}^{-1}x$

B

$\frac{\pi }{2}-co{s}^{-1}x$

C

$si{n}^{-1}\left\{x\right\}$

$\because \left\{\right[x+5]+\{x-\{x-\{x\left\}\right\}\left\}\right\}$

$=\left\{\right[x+5]+\{x-\left\{\right[x\left]\right\}\}$

$=\left\{\right[x+5]+\{x\left\}\right\}=\left\{\right\{x\left\}\right\}=\left\{x\right\}$

so $f\left(x\right)=sin\left\{x\right\}$ for $xϵ$ $(0,\frac{\pi }{4})$

so $f\left(x\right)=sinx$

so $f^{-1}\left(x\right)=si{n}^{-1}\left(x\right)$

$=si{n}^{-1}\left\{x\right\}=\frac{\pi }{2}-co{s}^{-1}x$