The positive integer k for which (101)^ k 2 k ! is a maximum is - Vidyadip

Question

The positive integer $k$ for which $\frac{(101)^{k 2}}{k !}$ is a maximum is

✓

Answer

b
(b)

We have, $\frac{(101)^{k / 2}}{k !}$

$\sqrt{10} \overline{1} > 10$

$\frac{(\sqrt{101})^9}{9 !}-\frac{(\sqrt{10}) 1)^{10}}{10 !}$

$\frac{(\sqrt{10} 1)^9}{9 !}\left[1-\frac{\sqrt{10} \overline{0}}{10}\right] < 0 \quad\left[\because \frac{\sqrt{101}}{10} > 1\right]$

$\therefore \quad \frac{(\sqrt{10} \overline{1})^9}{9 !} < \frac{(\sqrt{101})^{10}}{10 !}$

$\frac{(\sqrt{101})^{10}}{10 !}-\frac{ \quad(\sqrt{101})^{11}}{11!}$

$=\frac{(\sqrt{10} \overline{0})^{10}}{10 !}\left[1-\frac{\sqrt{10} \overline{1}}{11} > 0\left[\because \frac{\sqrt{101}}{11} < 1\right]\right.$

$\frac{(\sqrt{101})^{10}}{10 !} > \frac{(\sqrt{101})^{11}}{11 !}$

$=\frac{\sqrt{101}}{10 !}$ $\left[1-\frac{(\sqrt{101})^{91}}{11 \times 12 \ldots 101}\right] > 0$

$\therefore$ For $k=10$, Maximum value of $\frac{(101)^{k / 2}}{k !}$

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