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Application of Derivatives question

2019 · 12 Jan · Shift 2 · Q41
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  5. /2019 · 12 Jan · Shift 2 · Q41

Application of Derivatives question

2019 · 12 Jan · Shift 2 · Q41

JEE MainMathematicsApplication of DerivativesMCQ+4 / −1
If the function f given by f(x) = x3 – 3(a – 2)x2 + 3ax + 7, for some a ∈\in∈ R is increasing in (0, 1] and decreasing in [1, 5), then a root of the equation, f(x)−14(x−1)2=0(xe1){{f\left( x \right) - 14} \over {{{\left( {x - 1} \right)}^2}}} = 0\left( {x e 1} \right)(x−1)2f(x)−14​=0(xe1) is :
  1. A
    −-− 7
  2. B
    5
  3. C
    7
  4. D
    6
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