Question

If $k\in I$ such that $\underset{n\to \infty }{\lim }{\left(\cos \frac{k\pi }{4}\right)}^{2n}-{\left(\cos \frac{k\pi }{6}\right)}^{2n}=0$, then

• A. k must not be divisible by 24
• B. k is divisible by 24 or k is divisible neither by 4 nor by 6
• C. k must be divisible by 12 but not necessarily by 24
• D. None of these

Answer 1

Answer: (B)

$\underset{n\to \infty }{\lim }{\left(\cos \frac{k\pi }{4}\right)}^{2n}-{\left(\cos \frac{k\pi }{6}\right)}^{2n}=0$
holds good if
Case I : $\cos \frac{k\pi }{4}=\cos \frac{k\pi }{6}=1$
i.e., $\frac{k\pi }{4}=2m\pi$ and $\frac{k\pi }{6}=2p\pi ,m,p\in Z$
i.e., $k=8m$ and $k=12p$
i.e., $k$ is divisible by both 8 and 12 i.e., $k$ is divisible by 24
Case II : $-1<\cos \frac{k\pi }{4},\cos \frac{k\pi }{6}<1$
i.e., $k$ is not divisible by 4 and $k$ is not divisible by 6 .
Case III : $\cos \frac{k\pi }{4}=-1=\cos \frac{k\pi }{6}$
$\frac{k\pi }{4}=(2m+1)\pi$ and $\frac{k\pi }{6}=(2p+1)\pi$
$k=4(2m+1)$ and $k=6(2p+1)$
This is not possible.