A) \[\frac{7\pi }{24}\]
B) \[\frac{5\pi }{24}\]
C) \[\frac{11\pi }{2}\]
D) \[\frac{\pi }{24}\]
Correct Answer: A
Solution :
\[\left| \begin{matrix} 1+{{\sin }^{2}}\theta & {{\cos }^{2}}\theta & 4\sin 4\theta \\ {{\sin }^{2}}\theta & 1+{{\cos }^{2}}\theta & 4\sin 4\theta \\ {{\sin }^{2}}\theta & {{\cos }^{2}}\theta & 1+4\sin 4\theta \\ \end{matrix} \right|=0\] \[\Rightarrow \]\[\left| \begin{matrix} 2 & {{\cos }^{2}}\theta & 4\sin 4\theta \\ 2 & 1+{{\cos }^{2}}\theta & 4\sin 4\theta \\ 1 & {{\cos }^{2}}\theta & 1+4\sin 4\theta \\ \end{matrix} \right|=0\] \[({{C}_{1}}\to {{C}_{1}}+{{C}_{2}})\] \[\Rightarrow \] \[\left| \begin{matrix} 2 & {{\cos }^{2}}\theta & 4\sin 4\theta \\ 0 & 1 & 0 \\ 1 & {{\cos }^{2}}\theta & 1+4\sin 4\theta \\ \end{matrix} \right|=0\] \[({{R}_{2}}\to {{R}_{2}}-{{R}_{1}})\] \[\Rightarrow \] \[2(2+4\sin 4\theta )=0\] \[\Rightarrow \] \[1+2\sin 4\theta =0\] \[\Rightarrow \] \[\sin 4\theta =-\sin .\frac{\pi }{6}\] \[\Rightarrow \] \[4\theta =n\pi +{{(-1)}^{n}}\left( -\frac{\pi }{6} \right)\] \[\therefore \]The value of\[\theta \]between 0 and\[\frac{\pi }{2}\]will be \[\frac{7\pi }{24}\]and\[\frac{11\pi }{24}\]
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