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 [0, 1] then the value of

limnlimm1+cos2m(n!πx) 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 xQ then n!πx will be an integral multiple of π

    for large values of n. Therefore, cos(n!πx) will be either 1 or

    -1 and so cos2m(n!πx)=1

    limmlimn1+cos2m(n!πx)=1+1=2

    If xQ, n!πx will not be an integral multiple of π and so

    cos(n!πx) will lie between -1 and 1

    thus, limmcos2m(n!πx)=0

     limmlimn1+cos2m(n!πx)=1+0=1.

    Thus, option (a) is correct.

     

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