A) \[\Rightarrow \]
B) \[t=7.5(\sqrt{3}-1)s\]
C) \[I<II<III<IV\]
D) \[NO_{3}^{-}\]
Correct Answer: C
Solution :
We have,\[\sin A+\cos A=m\]and\[{{\sin }^{3}}A+{{\cos }^{3}}A=n\] Now,\[\sin A+\cos A=m\] \[\Rightarrow \]\[{{(\sin A+\cos A)}^{3}}={{m}^{3}}\] \[\Rightarrow \]\[{{\sin }^{3}}A+{{\cos }^{3}}A+3\sin A\cos A\] \[(\sin A+\cos A)={{m}^{3}}\] \[\Rightarrow \]\[n+3\sin A\cos Am={{m}^{3}}\] ?(i) Again,\[\sin A+\cos A=m\] \[\Rightarrow \]\[\sin A+{{\cos }^{2}}A+2\sin A\cos A={{m}^{2}}\] \[\Rightarrow \]\[\sin A\cos A=\frac{{{m}^{2}}-1}{2}\] ?(ii) From Eqs. (i) and (ii), we get \[n+3m\frac{({{m}^{2}}-1)}{2}={{m}^{3}}\] \[\Rightarrow \]\[2n+3{{m}^{3}}-3m=2{{m}^{3}}\] \[\Rightarrow \]\[{{m}^{3}}-3m+2n=0\]
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