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JEE Main 202524 JanuaryEvening Shift
Exam
JEE Main
Year
2025
Date
24 January
Shift
Evening Shift
Subject
Mathematics
Chapter
Sets and Relations

JEE Main 2025 · 24 January · Evening Shift · Mathematics · Sets and Relations

Let A=\ x ∈(0, π)-\ (π)/(2)\ : _ (2 / π) | x|+ _ (2 / π) | x|=2\...

Let A={x∈(0,π)−{π2}:log⁡(2/π)∣sin⁡x∣+log⁡(2/π)∣cos⁡x∣=2}\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 B={x⩾0:x(x−4)−3∣x−2∣+6=0}\mathrm{B}=\{x \geqslant 0: \sqrt{x}(\sqrt{x}-4)-3|\sqrt{x}-2|+6=0\}. Then n(A∪B)\mathrm{n}(\mathrm{A} \cup \mathrm{B}) is equal to :

Options

  1. A4
  2. B8
  3. C6
  4. D2

Correct answer

B. 8

Step-by-step solution

A:log⁡2π∣sin⁡x∣+log⁡2π∣cos⁡x∣=2⇒log⁡2π(∣sin⁡x⋅cos⁡x∣)=2⇒∣sin⁡2x∣=8π2\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 x=t<2\text{ B : let } \sqrt{\mathrm{x}}=\mathrm{t} < 2 Then x(x−4)+3(x−2)+6=0\sqrt{x}(\sqrt{x}-4)+3(\sqrt{x}-2)+6=0⇒t2−4t+3t−6+6=0⇒t2−t=0,t=0,t=1x=0,x=1\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}$$TotalnumberofsolutionsTotal number of solutions\mathrm{n}(\mathrm{ A} \cup \mathrm{ B})=4+4=8$
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