MCQ
Find the values of $\sin \left(\frac{\pi}{3}-\sin ^{-1}\left(-\frac{1}{2}\right)\right)$ is equal to
- A$\frac{1}{2}$
- B$\frac{1}{3}$
- ✓$1$
- D$\frac{1}{4}$
Then $, \sin x=\frac{-1}{2}=-\sin \frac{\pi}{6}=\sin \left(\frac{-\pi}{6}\right)$
We know that the range of the principal value branch of $\sin ^{-1}$ is $\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]$ $\sin ^{-1}\left(\frac{-1}{2}\right)=\frac{\pi}{6}$
$\therefore \sin \left(\frac{\pi}{3}-\sin ^{-1}\left(\frac{-1}{2}\right)\right)=\sin \left(\frac{\pi}{3}+\frac{\pi}{6}\right)$$=\sin \left(\frac{3 \pi}{6}\right)=\sin \left(\frac{\pi}{2}\right)=1$
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($A$) There exists a function $f \in S$ such that $X_f=0$
($B$) For every function $f \in S$, we have $X_f \leq 2$
($C$) There exists a function $f \in S$ such that $X_f=2$
($D$) There does $NOT$ exist any function $f$ in $\mathrm{S}$ such that $\mathrm{X}_f=1$