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:

Limit, Continuity and Differentiability: JEE Main 2019 Mathematics question

JEE Main 2019 · 12 Jan, Evening shift · Q2709

JEE Main 201912 Jan, Evening shiftMathematicsMediumLimit, Continuity and Differentiability420× since 2002

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

JEE Main 201912 Jan, Evening shiftMathematicsMediumLimit, Continuity and Differentiability420× since 2002

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 :