Mathematics - Sets and Relations Question with Solution | TestHub

$\begin{array}{rl}& \mathrm{A}:{\log }_{2\pi }|\sin \mathrm{x}|+{\log }_{2\pi }|\cos \mathrm{x}|=2\\ & \Rightarrow {\log }_{2\pi }(|\sin \mathrm{x}\cdot \cos \mathrm{x}|)=2\\ & \Rightarrow |\sin 2\mathrm{x}|=\frac{8}{{\pi }^2}\end{array}$
Number of solution 4 $\text{ B : let }\sqrt{\mathrm{x}}=\mathrm{t}<2$ Then $\sqrt{x}(\sqrt{x}-4)+3(\sqrt{x}-2)+6=0$ $\begin{array}{rl}& \Rightarrow {\mathrm{t}}^2-4\mathrm{t}+3\mathrm{t}-6+6=0\\ & \Rightarrow {\mathrm{t}}^2-\mathrm{t}=0,\mathrm{t}=0,\mathrm{t}=1\\ & \mathrm{x}=0,\mathrm{x}=1\end{array}$ again let $\sqrt{\mathrm{x}}=\mathrm{t}>2$ then ${\mathrm{t}}^2-4\mathrm{t}-3\mathrm{t}+6+6=0$ $\begin{array}{rl}& \Rightarrow {\mathrm{t}}^2-7\mathrm{t}+12=0\\ & \Rightarrow \mathrm{t}=3,4\\ & \mathrm{x}=9,16\end{array}$ Total number of solutions $\mathrm{n}(\mathrm{A}\cup \mathrm{B})=4+4=8$