Let f(x)=5−|x−2| and g(x)=|x+1|,xϵR. If f(x) attains maximum value of α and g(x) attains minimum value at β, then limx→αβ(x−1)(x2−5x+6)x2−6x+8 is equal to
A
1/2
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B
−3/2
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C
3/2
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D
−1/2
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Solution
The correct option is A1/2 Maxima of f(x) occured at x=2 i.e. α=2 Minima of g(x) occured at x=−1 i.e. β=−1 ∴limx→2(x−1)(x−2)(x−3)(x−2)(x−4)=12.