A) \[a\in [8,\,64]\]
B) \[a\in (0,\,8]\]
C) \[a\in [64,\,\infty )\]
D) None of these
Correct Answer: C
Solution :
[c] Since \[[{{x}^{3}}]\] is not continuous and differentiable at integral points. So \[f(x)\] is continuous and differentiable in [4, 6] if \[\left[ \frac{{{(x-2)}^{3}}}{a} \right]=0\Rightarrow a\ge 64\]
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