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

JEE Main 2025 (Online) 24th January Evening Shift Mathematics - Sets and Relations Question 26 English Explanation

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

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

19:T5ed,

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

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