If x is a real number in [0, 1] then the value oflimn→∞ limm→∞ 1+cos2m⁡(n!πx) is given by

If x is a real number in  then the value of$\underset{n\to \mathrm{\infty }}{lim} \underset{m\to \mathrm{\infty }}{lim} \left[1+{\mathrm{cos}}^{2m}\left(n!\pi x\right)\right]$ is given by

1. A

2 or 1 according as x is rational or irrational

2. B

1 or 2 according as x is rational or irrational

3. C

1 for all $x$

4. D

2 or 1 for all x

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Solution:

If $x\in Q$ then $n!\pi x$ will be an integral multiple of $\pi$

for large values of $n$. Therefore, $\mathrm{cos}\left(n!\pi x\right)$ will be either 1 or

-1 and so ${\mathrm{cos}}^{2m}\left(n!\pi x\right)=1$

$\underset{m\to \mathrm{\infty }}{lim} \underset{n\to \mathrm{\infty }}{lim} \left[1+{\mathrm{cos}}^{2m}\left(n!\pi x\right)\right]=1\mid +1=2$

If  will not be an integral multiple of $\pi$ and so

$\mathrm{cos}\left(n!\pi x\right)$ will lie between -1 and 1

thus, $\underset{m\to \mathrm{\infty }}{lim} {\mathrm{cos}}^{2m}\left(n!\pi x\right)=0$

Thus, option (a) is correct.

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