A) \[2n\pi +\frac{\pi }{12},2n\pi -\frac{\pi }{3}\]
B) \[2n\pi +\frac{5\pi }{12},2n\pi -\frac{\pi }{12}\]
C) \[2n\pi -\frac{\pi }{12},2n\pi -\frac{\pi }{12}\]
D) None of these
Correct Answer: D
Solution :
Given, \[\sqrt{3}\cos x+\sin x=\sqrt{2}\] \[\Rightarrow \]\[\frac{\sqrt{3}}{2}\cos x+\frac{1}{2}\sin x=\frac{\sqrt{2}}{2}\] \[\Rightarrow \]\[\cos \left( x-\frac{\pi }{6} \right)=\cos \frac{\pi }{4}\] \[\Rightarrow \]\[x-\frac{\pi }{6}=2n\pi \pm \frac{\pi }{4}\] \[\Rightarrow \]\[x=2n\pi \pm \frac{\pi }{4}+\frac{\pi }{6}\] \[\Rightarrow \]\[x=2n\pi +\frac{5\pi }{12},2n\pi -\frac{\pi }{12},\]where\[n\in I\] Solutions for Q. No. 5 to 6. Given that, \[|\vec{a}|=|\vec{b}|=|\vec{c}|=4\] Now, \[\vec{a}.\vec{a}=|\vec{a}{{|}^{2}}=16\] \[\vec{a}.\vec{b}=\vec{b}.\vec{a}=|\vec{a}|\,\,|\,\vec{b}|\cos \pi /3\] \[=4.4.\frac{1}{2}=8\] \[\vec{a}.\vec{c}=\vec{c}.\vec{a}=|\vec{a}|\,|\vec{c}\cos \frac{\pi }{3}=4.4.\frac{1}{2}=8\] \[\vec{b}.\vec{b}=|\vec{b}{{|}^{2}}=16\] \[\vec{b}.\vec{c}=\vec{c}.\vec{b}=|\vec{b}|\,|\vec{c}|\cos \pi /3=4.4.\frac{1}{2}=8\] \[\vec{c}.\vec{c}=|\vec{c}{{|}^{2}}={{4}^{2}}=16\] Now, \[{{[\vec{a}\,\vec{b}\,\vec{c}]}^{2}}=[\vec{a}\,\vec{b}\,\vec{c}].[\vec{a}\,\vec{b}\,\vec{c}]\] \[=\left| \begin{matrix} \vec{a}.\vec{a} & \vec{b}.\vec{a} & \vec{c}.\vec{a} \\ \vec{a}.\vec{b} & \vec{b}.\vec{b} & \vec{c}.\vec{b} \\ \vec{a}.\vec{c} & \vec{b}.\vec{c} & \vec{c}.\vec{c} \\ \end{matrix} \right|\] \[=\left| \begin{matrix} 16 & 8 & 8 \\ 8 & 16 & 8 \\ 8 & 8 & 16 \\ \end{matrix} \right|\] \[={{8}^{3}}\left| \begin{matrix} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \\ \end{matrix} \right|\] \[={{8}^{3}}[2(4-1)-1(2-1)+1(1-2)]\] \[={{8}^{3}}\times 4=32\times 32\times 2\] \[\Rightarrow \]\[[\vec{a}\,\vec{b}\,\vec{c}]=32\sqrt{2}\]
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