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, … is :

Option C. f '(x) = 3x² … 6(a … 2)x + 3a

Application of Derivatives

JEE Main 2019 · 12 Jan, Evening shift · Q2709

MathematicsMedium175×since 2002Q2709

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 :

JEE Main 2019 · 12 Jan, Evening shift · Q2709

MathematicsMedium175×since 2002Q2709

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 :