At x=1,f′(x)=0∴f′(x)=3x2−6(a−2)x+3a
Now, f′(1)=0
⇒3−6(a−2)+3a=0
⇒a=5
∴(x−1)2f(x)−14=0
⇒(x−1)2x3−9x2+15x−7=0
⇒(x−1)2(x−7)(x2−2x+1)=0
⇒x−7=0⇒x=7
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, (x−1)2f(x)−14=0,(x=1) is :
Held on 12 Jan 2019 · Verified 6 Jul 2026.
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