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Solve it.

Q). The least value of $a\in R$ for which $4a{x}^{2}+\frac{1}{x}\ge 1$ for all $x>0$, is

(a) $\frac{1}{64}$

(b) $\frac{1}{32}$

(c) $\frac{1}{27}$

(d) $\frac{1}{25}$

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