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Correct option is C
case 1 : Let fx ≠2 be true and fy=2,fz≠1 are false then fx≠2,fy≠2,fz=1 ⇒ fx=3 ,fy=3 ,fz=1 , which is not one-one case 2 : let fy =2 be true and fx≠2,fz≠1 be false then fy=2 , fx =2 ,fz=1 , which is not one-one case 3 : let fz≠1 be true and fx≠2 ,fy=2 are false then fx =2 ,fy ≠2, fz ≠1 are true f(x)=2,f(y)=1,f(z)=3, one to oneTalk to our academic expert!
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