Question

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

$\displaystyle \lim _{n \rightarrow \infty}\left(\cos \frac{k \pi}{4}\right)^{2 n}-\left(\cos \frac{k \pi}{6}\right)^{2 n}=0$
holds good if
Case I : $\cos \frac{k \pi}{4}=\cos \frac{k \pi}{6}=1$
i.e., $\frac{k \pi}{4}=2 m \pi$ and $\frac{k \pi}{6}=2 p \pi, m, p \in Z$
i.e., $k=8 m$ and $k=12 p$
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 : $\cos \frac{k \pi}{4}=-1=\cos \frac{k \pi}{6}$
$\frac{k \pi}{4}=(2 m+1) \pi$ and $\frac{k \pi}{6}=(2 p+1) \pi$
$k=4(2 m+1)$ and $k=6(2 p+1)$
This is not possible.