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
Correct Answer: B
Solution :
| \[\underset{n\to \infty }{\mathop{\lim }}\,{{\left( \cos \frac{k\pi }{4} \right)}^{2n}}-{{\left( \cos \frac{k\pi }{6} \right)}^{2n}}=0\] 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\[\frac{k\pi }{6}=2p\pi ,m,p\in \operatorname{Z}\] |
| 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: \[\cos \frac{k\pi }{4}=-1=\cos \frac{k\pi }{6}\] |
| \[\frac{k\pi }{4}=\left( 2m+1 \right)\pi \]and\[\frac{k\pi }{6}=\left( 2p+1 \right)\pi \] |
| \[k=4\left( 2m+1 \right)\]and\[k=6\left( 2p+1 \right)\] |
| This is not possible. |
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