MCQ
જો${\sin ^{ - 1}}\frac{x}{5} + {\rm{cose}}{{\rm{c}}^{ - 1}}\left( {\frac{5}{4}} \right) = \frac{\pi }{2},$તો $x = $
- A$4$
- B$5$
- C$1$
- ✓$3$
$\Rightarrow \sin ^{-1}\left(\frac{x}{5}\right)+\sin ^{-1}\left(\frac{4}{5}\right)=\frac{\pi}{2}$
$\Rightarrow \sin ^{-1}\left(\frac{x}{5}\right)=\frac{\pi}{2}-\sin ^{-1}\left(\frac{4}{5}\right)$
$\Rightarrow \sin ^{-1}\left(\frac{x}{5}\right)=\cos ^{-1}\left(\frac{4}{5}\right)$
$\Rightarrow \frac{x}{5}=\sin \left(\cos ^{-1} \frac{4}{5}\right)$
$\Rightarrow \frac{x}{5}=\sin \left(\sin ^{-1} \frac{3}{5}\right)$
$\Rightarrow \frac{x}{5}=\frac{3}{5}$
$\Rightarrow x=3$
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$(P)$ જો $A \neq I_{2},$ હોય તો $|A|=-1$:
$(Q)$ જો $|\mathrm{A}|=1,$ હોય તો $\operatorname{tr}(\mathrm{A})=2$
જ્યાં $I_{2}$ એ $2 \times 2$ નો એકમ શ્રેણિક અને $\operatorname{tr}(A)$ એ શ્રેણિક $A$ ના અગ્ર વિકર્ણના ઘટકોનો સરવાળો દર્શાવે તો