If the function f given by f(x)=x3−3(a−2)x2+3ax+7, for some a∈R is increasing in (0,1] and decreasing in [1,5), then a root of the equation, f(x)−14(x−1)2=0(x≠1) is :
A
−7
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B
5
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C
6
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D
7
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Solution
The correct option is D7 f(x)=x3−3(a−2)x2+3ax+7 ∴f′(x)=3x2−6(a−2)x+3a f′(x)≥0∀x∈(0,1] f′(x)≤0∀x∈[1,5) ⇒x=1 is a critical point. f′(1)=0 ⇒3−6a+12+3a=0 ⇒a=5 ∴f(x)−14(x−1)2=0 ⇒x3−9x2+15x−7(x−1)2=0 ⇒(x−1)2(x−7)(x−1)2=0 ⇒x=7 is a root.