MCQ
વિધેય $\sin([x]\pi)$ નો વિસ્તાર
- A$(0,1)$
- B$[-1,1]$
- ✓$\left\{0\right\}$
- D$0$
$\therefore \sin([x]\pi)=0$ ( $\because [x] $ પૂર્ણાંક છે.)
$R_f=\left\{0\right\}$
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