Let A=x (0, )-( )/(2): _(2/ )| x|+ _(2/ )| x|=2 and B=x 0:x(x-4)-3|x-2|+6=0.

Question

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

Accepted Answer

$$\begin{aligned}
& \mathrm{A}: \log _{2 /\pi}|\sin \mathrm{x}|+\log _{2 /\pi}|\cos \mathrm{x}|=2 \
& \Rightarrow \log _{2 /\pi}(|\sin \mathrm{x} \cdot \cos \mathrm{x}|)=2 \
& \Rightarrow|\sin 2 \mathrm{x}|=\frac{8}{\pi^2}
\end{aligned}$$

Number of solution 4

B : let $\sqrt{\mathrm{x}}=\mathrm{t}<2$

Then $\sqrt{\mathrm{x}}(\sqrt{\mathrm{x}}-4)+3(\sqrt{\mathrm{x}}-2)+6=0$

$$\begin{aligned}
& \Rightarrow \mathrm{t}^2-4 \mathrm{t}+3 \mathrm{t}-6+6=0 \
& \Rightarrow \mathrm{t}^2-\mathrm{t}=0, \mathrm{t}=0, \mathrm{t}=1 \
& \mathrm{x}=0, \mathrm{x}=1
\end{aligned}$$

again let $\sqrt{x}=t>2$

then $\mathrm{t}^2-4 \mathrm{t}-3 \mathrm{t}+6+6=0$

$$\begin{aligned}
& \Rightarrow t^2-7 \mathrm{t}+12=0 \
& \Rightarrow \mathrm{t}=3,4 \
& \mathrm{x}=9,16
\end{aligned}$$

Total number of solutions

$$\mathrm{n}(\mathrm{~A} \cup \mathrm{~B})=4+4=8$$

Solutions · 1