Verified Solution • 2025–26 Session moderate

QJEEM25Ma69b1: JEE Main 2025 Mathematics - Class 11 JEEM25Ma69b1 Solution: Let and . Then is equal to :

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 :

Multiple Choice Options

  • A 4
  • B 8
  • C 6
  • D 2

Step-by-Step Verified Solution

amp;A:log⁡2π∣sin⁡x∣+log⁡2π∣cos⁡x∣=2amp;⇒log⁡2π(∣sin⁡x⋅cos⁡x∣)=2amp;⇒∣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}

JEE Main 2025 (Online) 24th January Evening Shift Mathematics - Sets and Relations Question 2 English Explanation

Number of solution 4

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

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

amp;⇒t2−4t+3t−6+6=0amp;⇒t2−t=0,t=0,t=1amp;x=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{x}=t>2

then t2−4t−3t+6+6=0\mathrm{t}^2-4 \mathrm{t}-3 \mathrm{t}+6+6=0

amp;⇒t2−7t+12=0amp;⇒t=3,4amp;x=9,16\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

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

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