MCQ
If $\sin ^{-1} x=y,$ then
- A$-\frac{\pi}{2} < y < \frac{\pi}{2}$
- B$0 \leq \mathrm{y} \leq \pi$
- ✓$-\frac{\pi}{2} \leq y \leq \frac{\pi}{2}$
- D$0 < y < \pi$
We know that the range of the principal value branch of $\sin ^{-1}$ is $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$
Therefore, $-\frac{\pi}{2} \leq y \leq \frac{\pi}{2}$
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$(A)$ $\left|z-z_1\right|+\left|z-z_2\right|=\left|z_1-z_2\right|$
$(B)$ $\operatorname{Arg}\left(z-z_1\right)=\operatorname{Arg}\left(z-z_2\right)$
$(C)$ $\left|\begin{array}{cc}z-z_1 & \bar{z}-\bar{z}_1 \\ z_2-z_1 & \bar{z}_2-\bar{z}_1\end{array}\right|=0$
$(D)$ $\operatorname{Arg}\left(z-z_1\right)=\operatorname{Arg}\left(z_2-z_1\right)$