MCQ
If ${\sin ^{ - 1}}\frac{1}{3} + {\sin ^{ - 1}}\frac{2}{3} = {\sin ^{ - 1}}x,$ then $ x$ is equal to
- A$0$
- B$\frac{{\sqrt 5 - 4\sqrt 2 }}{9}$
- ✓$\frac{{\sqrt 5 + 4\sqrt 2 }}{9}$
- D$\frac{\pi }{2}$
$ = {\sin ^{ - 1}}\left[ {\frac{1}{3}\sqrt {1 - \frac{4}{9}} + \frac{2}{3}\sqrt {1 - \frac{1}{9}} } \right] = {\sin ^{ - 1}}\,\left[ {\frac{{\sqrt 5 + 4\sqrt 2 }}{9}} \right]$
Therefore $x = \frac{{\sqrt 5 + 4\sqrt 2 }}{9}$.
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$(A)$ $\mathrm{S}_{\mathrm{n}}<\frac{\pi}{3 \sqrt{3}}$ $(B)$ $S_n>\frac{\pi}{3 \sqrt{3}}$
$(C)$ $T_n<\frac{\pi}{3 \sqrt{3}}$ $(D)$ $T_n>\frac{\pi}{3 \sqrt{3}}$
$I.$ There exists a real number $A$ such that $f(x) \leq A$ for all $x$.
$II.$ There exists a real number $B$ such that $f(x) \geq B$ for all $x$.