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Answer :

Given that f: A B defined by f(x) = –x2 + 6x – 8 is a bijection.


f(x) = –x2 + 6x – 8


f(x) = –(x2 - 6x + 8)


f(x) = - (x2 - 6x + 8 + 1 - 1)


f(x) = - (x2 - 6x + 9 - 1)


f(x) = - [(x – 3)2 – 1]


Hence, x ϵ (–∞, 5] and f(x) ϵ (–∞, 1]

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