Mathematics - Functions Question with Solution | TestHub

(i) for domain (i) $[x]>0$ and $[x]\ne 1$ so $[x]\ge 2$, so $x\in [2,\infty )$

for range if $x\in [2,\infty )$, then $\frac{|x|}{x}=1$ so $f(x)={\cos }^{-1}0=\frac{\pi }{2}$

Range of $\mathrm{f}(\mathrm{x})=\frac{\pi }{2}$

Ans. $\mathrm{D}:[2,\infty );\mathrm{R}:\{\pi /2\}\mathrm{c}$

(ii) $f(x)={\log }_{[x-1]}\sin x$

$\sin \mathrm{x}>0\Rightarrow \mathrm{x}\in (2\mathrm{n}\pi ,(2\mathrm{n}+1)\pi )$

here $[x-1]>0\&[x-1]\ne 1\Rightarrow x\in [3,\infty ]$

Domain $\mathrm{x}\in [3,\pi ]\cup [{\bigcup }_{\mathrm{n}=1}^{\infty }(2\mathrm{n}\pi ,(2\mathrm{n}+1)\pi )$.

For range $\sin \mathrm{x}\in (0,1]$ and $[\mathrm{x}-1]\in [2,\infty )$ so range $\in (-\infty ,0]$

$$\left\{\text{ Ans. }:D:[3,\pi )\cup \bigcup _{\mathrm{n}=1}^{\infty }(2\mathrm{n}\pi ,2\mathrm{n}\pi +\pi );\mathrm{R}:(-\infty ,0]\right\}$$