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, ${{f\left( x \right) - 14} \over {{{\left( {x - 1} \right)}^2}}} = 0\left( {x \ne 1} \right)$ is :

Question

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, ${{f\left( x \right) - 14} \over {{{\left( {x - 1} \right)}^2}}} = 0\left( {x \ne 1} \right)$ is :

$-$ 7

Solution

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