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$a_{1} < {(28)}^{1 / 3} - 3 < b_{1}$ . Then $(a_{1}, b_{1})$ =

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$(\frac{1}{28}, \frac{1}{27})$

$(\frac{1}{27}, \frac{1}{28})$

(27,28)

(0,1)

answer is A.

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Detailed Solution

$f (x) = x^{\frac{1}{3}}$ on $\to$ (27, 28)

Lagrange’s Theorem $\to$ $\frac{f (28) - f (27)}{28 - 27} = \frac{1}{3} x^{- \frac{2}{3}}$
$\Rightarrow 28^{- \frac{1}{3}} - 3 = \frac{1}{3 \cdot x^{\frac{2}{3}}}$ and 27 < x < 28 $\Rightarrow \frac{1}{27} > \frac{1}{x} > \frac{1}{28}$

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