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a
{x} + cos πx
b
tanπ2[x]
c
sin x + {x}
d
sin (cos x)
answer is A.
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Detailed Solution
(A) The period of cosπx is 2ππ = 2, and period of {x} is 1 Hence, period of the given function is L.C.M. of (1, 2) = 2(B) Solving tanπ2[x+T]=tanπ2[x] i.e., [x + T] – [x] = 2n gives a value of T independent of x only if T is an integer.In that case, the above equation reduces to[x] + T – [x] = 2ni.e., T = 2nHence, period of f (x), is the smallest positive value of T, i.e., 2.(C) We have period of sin x = 2π and period of {x} = 1 Hence, period of the given functions is L.C.M. of (2p, 1) which does not exist since 2π is an irrational number. Hence, the function is not periodic(D) Let us solve sin{cos (x + T)} = sin{cos x} i.e., cos (x + T) = np + (– 1) n cos x, n ∈ I Putting n = 0, gives cos (x + T) = cos x, which gives T = 2π as the smallest positive value. For no other value of n can a value of T be found independent of x.Hence, the required fundamental period is 2p. The correct option is (1) and (2)