A) \[0\]
B) \[1\]
C) \[2\]
D) None of these
Correct Answer: D
Solution :
We have, \[f(x)=\left\{ \begin{matrix} {{\tan }^{-1}}a-3{{x}^{2}} & ,0<x<1 \\ -6x & ,x\ge 1 \\ \end{matrix} \right.\] If \[f(x)\] attains a maximum at\[x=1\], then \[f'(1)\] must exist and should be zero. This means that\[f(x)\] must be continuous and differentiable at\[x=1\]. We observe that \[f(x)\] will be continuous at\[x=1\], if\[{{\tan }^{-1}}a=-3\] But, \[(LHD\,\,at\,\,x=1)=(RHD\,\,at\,\,x=1)=-6\ne 0\]for any value of\[a\] Hence, there is no value of \[a\] for which \[f(x)\] has a minimum at\[x=1\].
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