JEE Main 2025 — Mathematics Algebra

$\begin{aligned} & \mathrm{A}: \log _{2 \pi}|\sin \mathrm{x}|+\log _{2 \pi}|\cos \mathrm{x}|=2 \\ & \Rightarrow \log _{2 \pi}(|\sin \mathrm{x} \cdot \cos \mathrm{x}|)=2 \\ & \Rightarrow|\sin 2 \mathrm{x}|=\frac{8}{\pi^2}\end{aligned}$​A:log2π​∣sinx∣+log2π​∣cosx∣=2⇒log2π​(∣sinx⋅cosx∣)=2⇒∣sin2x∣=π28​​

Number of solution 4 $\text { B : let } \sqrt{\mathrm{x}}=\mathrm{t} < 2$ B : let x​=t<2 Then $\sqrt{x}(\sqrt{x}-4)+3(\sqrt{x}-2)+6=0$x​(x​−4)+3(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$againlet\sqrt{\mathrm{x}}=\mathrm{t}>2$then$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}$Total number of solutions$Totalnumberofsolutions\mathrm{n}(\mathrm{A} \cup \mathrm{B})=4+4=8$