Q. Let $A=\left\{x\in(0,\pi)-\left\{\frac\pi2\right\}:\log_{(2/\pi)}|\sin x|+\log_{(2/\pi)}|\cos x|=2\right\}$ and $B=\{x\ge0:\sqrt x(\sqrt x-4)-3|\sqrt x-2|+6=0\}$. Then $n(A\cup B)$ is equal to :
A) 4
B) 8
C) 6
D) 2
Explanation
[Solution by Dr. Abubacker] $A$: $|\sin x\cos x|=\left(\frac2\pi\right)^2\Rightarrow|\sin2x|=\frac8{\pi^2}<1$, which has $4$ solutions for $2x\in(0,2\pi)$, so $n(A)=4$ (roots near $0.47,1.10,2.04,2.67$). $B$: put $t=\sqrt x$. For $t\ge2$: $t^2-7t+12=0\Rightarrow t=3,4$. For $t<2$: $t^2-t=0\Rightarrow t=0,1$. So $x\in\{0,1,9,16\}$, $n(B)=4$. The sets are disjoint (no root of $A$ equals $1$), so $n(A\cup B)=8$. Answer (2).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.