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Q 2 : Given a PRF E : K { 0 , 1 } n { 0 , 1 } m , we construct ,

Q2: Given a PRF E:K{0,1}n{0,1}m, we construct ,F:K
{0,1}n-1{0,1}2m as follows:
F(K,x):=E(K,x||0)||E(K,x||1).
Is F(K,*) still a PRF? If yes, give a brief proof, if not, give a brief
attack showing the violation of PRF definition. (10 pts) please no chatgpt
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