Q65 / 160 · TS EAMCET 2024 (10 May Shift 2) · Mathematics
If $A$ is the solution set of the equation $\cos ^2 x=\cos ^2 \frac{\pi}{6}$ and $B$ is the solution set of the equation $\cos ^2 x=\log _{16} P$ where, $P+\frac{16}{P}=10$ then, $B-A=$
- A$\left\{x \in \mathbf{R} / x=2 n \pi \pm \frac{\pi}{4}, 2 n \pi \pm \frac{\pi}{3}, n=0,1,2,3, \ldots.\right\}$
- Correct answerB$\left\{x \in \mathbf{R} / x=2 n \pi \pm \frac{\pi}{3}, 2 n \pi \pm \frac{2 \pi}{3}, n=0,1,2,3, \ldots\right\}$
- C$\left\{x \in \mathbf{R} / x=2 n \pi \pm \frac{\pi}{6}, 2 n \pi \pm \frac{\pi}{12}, n=0,1,2,3, \ldots.\right\}$
- D$\left\{x \in \mathbf{R} / x=2 n \pi \pm \frac{\pi}{8}, 2 n \pi \pm \frac{\pi}{16}, n=0,1,2,3, \ldots.\right\}$
Solution
$\begin{gathered}\cos ^2 x=\cos ^2 \frac{\pi}{6} \\ \cos x=\cos \frac{\pi}{6} \Rightarrow x=2 n \pi \pm \frac{\pi}{6}, n \in Z\end{gathered}$
$\begin{aligned} & \cos x=-\cos \frac{\pi}{6} \Rightarrow x=(2 n+1) \pi \pm \frac{\pi}{6}, n \in Z \\ & \text { So, } A=\left\{n \pi \pm \frac{\pi}{6}, n \in Z\right\}\end{aligned}$
Now, $P+\frac{16}{P}=10 \Rightarrow P^2-10 P+16=0$ $\Rightarrow P=8,2$ $\begin{aligned} & \cos ^2 x=\log _{16} P ; \cos ^2 x=\log _{16} p \\ & \Rightarrow \cos ^2 x=\log _{16} 8 ; \cos ^2 x=\frac{1}{4} \log _2 2 \\ & \Rightarrow \cos ^2 x=\frac{1}{4} \log _2 8 ; \cos ^2 x=\frac{1}{4} \\ & \Rightarrow \cos ^2 x=\frac{3}{4} \Rightarrow \cos x= \pm \frac{\sqrt{3}}{2} \Rightarrow \cos x= \pm \frac{1}{2} \\ & \Rightarrow \cos x= \pm \cos \frac{\pi}{6} \Rightarrow \cos x= \pm \cos \frac{\pi}{3} \\ & \Rightarrow x=n \pi \pm \frac{\pi}{6}, x=n \pi \pm \frac{\pi}{3}\end{aligned}$ So, $B=\left\{n \pi \pm \frac{\pi}{6}, n \pi \pm \frac{\pi}{3}, n \in Z\right\}$ $\mathrm{B}-\mathrm{A}\left\{x \in R \left\lvert\, x=2 n \pi \pm \frac{\pi}{3}\right., 2 n \pi \pm \frac{2 \pi}{3}, n=0,1,2, \ldots\right\}$
Now, $P+\frac{16}{P}=10 \Rightarrow P^2-10 P+16=0$ $\Rightarrow P=8,2$ $\begin{aligned} & \cos ^2 x=\log _{16} P ; \cos ^2 x=\log _{16} p \\ & \Rightarrow \cos ^2 x=\log _{16} 8 ; \cos ^2 x=\frac{1}{4} \log _2 2 \\ & \Rightarrow \cos ^2 x=\frac{1}{4} \log _2 8 ; \cos ^2 x=\frac{1}{4} \\ & \Rightarrow \cos ^2 x=\frac{3}{4} \Rightarrow \cos x= \pm \frac{\sqrt{3}}{2} \Rightarrow \cos x= \pm \frac{1}{2} \\ & \Rightarrow \cos x= \pm \cos \frac{\pi}{6} \Rightarrow \cos x= \pm \cos \frac{\pi}{3} \\ & \Rightarrow x=n \pi \pm \frac{\pi}{6}, x=n \pi \pm \frac{\pi}{3}\end{aligned}$ So, $B=\left\{n \pi \pm \frac{\pi}{6}, n \pi \pm \frac{\pi}{3}, n \in Z\right\}$ $\mathrm{B}-\mathrm{A}\left\{x \in R \left\lvert\, x=2 n \pi \pm \frac{\pi}{3}\right., 2 n \pi \pm \frac{2 \pi}{3}, n=0,1,2, \ldots\right\}$
Hence, the correct answer is option: 'B'.