f(x)=(x−2)(x−1)(x−3)∀x>3. The minimum value of f(x) is equal to
A
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B
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C
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D
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Solution
The correct option is A Let x-3 = t ⇒ x-2 = (t+1), x-1 = t+2 ⇒ f(x) = (x−2)(x−1)(x−3) i.e f(x) = (t+1)(t+2)t f(t)=t2+3t+2t⇒f′(t)=t2−2t2f′(t)=0⇒t=√2[∵t≠−√2]∴Minimumvalue=f(√2)=3+2√2