Let $\mathrm{A}=\left\{x \in(0, \pi)-\left\{\frac{\pi}{2}\right\}: \log _{(2 / \pi)}|\sin x|+\log _{(2 / \pi)}|\cos x|=2\right\}$ and $\mathrm{B}=\{x \geqslant 0: \sqrt{x}(\sqrt{x}-4)-3|\sqrt{x}-2|+6=0\}$. Then $\mathrm{n}(\mathrm{A} \cup \mathrm{B})$ is equal to :

• A

  4
• Correct answer
  B

  8
• C

  6
• D

  2

Solution

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

Hence, the correct answer is option: 'B'.