Let the function $f\left(x\right)$ be defined as $f\left(x\right)=\{\begin{array}{c}{\tan }^{-1}\alpha -3x^2,0<x<1\\ -6x,x\ge 1\end{array}$

$f\left(x\right)$ can have a maximum at $x=1$ if the value of $\alpha$ is

The correct option is B None of these
We have,

$f^{\prime }\left(x\right)=\{\begin{array}{c}\begin{array}{cc}-6x,& 0<x<1\end{array}\\ \begin{array}{cc}-6,& x\ge 1\end{array}\end{array}$

$\therefore f^{\prime }(1-h)=-6(1-h)<0$

and, $f^{\prime }(1+h)=-6<0.$

Since $f^{\prime }\left(x\right)$ does not change sign as $x$ passes through $1,$

therefore, $f\left(x\right)$ does not have a maximum or minimum at $x=1,$

whatever be the value of $\alpha .$