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Let us see through counter examples :

For A) :

Let f(n) = 2n , g(n) = n and h(n)  =  2n..So

f(n) is O(g(n)) is fine..But 

h(f(n))  = h(2n) = 22n   and

h(g(n))  = h(n)  = 2n

So 22n is not O(2n)..

Hence A) option is false.. 

For B) :

f(n)  !=  O(g(n)) meaning that f(n) > g(n) and hence g(n) < f(n)..So strict inequality exists..So g(n) = o(f(n)) not O(f(n)) as for O(f(n)) the tight upper bound need also be considered.

.Hence option B) is also false.

Hence D) is the correct answer ..

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You can neither say f(n) = O(g(n)) nor say f(n)≠O(g(n)) when f(n),g(n) are asymptotically unbounded functions because ≠ also compares. sin,cos are not asymptotically comparable.
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