An algebra problem by Vilakshan Gupta

We note that $\sqrt [2]2 = \sqrt[4]4$ and that from $\sqrt[1]1$ to $\sqrt[2]2$ is increasing while from $\sqrt[4]4$ to $\sqrt[n]n$ for $n>4$ as $\displaystyle \lim_{n \to \infty} n^\frac 1n = 1$ (see note) is decreasing, the maximum must be between $\sqrt[2]2$ to $\sqrt[4]4$ , which is only $\sqrt[3]3$ . Therefore, $a=\boxed{3}$ .

Note:

$\small \begin{aligned} \lim_{n \to \infty} n^\frac 1n & = \lim_{n \to \infty} \exp \left( \log n^\frac 1n \right) = \exp \left({\color{#3D99F6}\lim_{n \to \infty} \frac {\log n}n}\right) = e^{\color{#3D99F6}0} = 1 & \color{#3D99F6} \text{By L'Hôpital's rule}\end{aligned}$

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Alternatively , it can also be shown using calculus as @Christian Daang 's solution. Consider the continuous function of $y = x^\frac 1x$ , then $\dfrac {dy}{dx} =$ $\dfrac d{dx} x^\frac 1x =$ $\dfrac d{dx} e^{\frac 1x \ln x} =$ $\dfrac {1-\ln x}{x^2} x^\frac 1x$ . Equating $\dfrac {dy}{dx} = 0$ , $\implies \ln x = 1$ , $\implies x = e$ . It can be shown that $\dfrac {d^2y}{dx^2} < 0$ (see note), therefore, $\max(y) = e^\frac 1e$ . Since $e \approx 2.718$ , the nearest integer is $\boxed{3}$ .

Note:

$\small \begin{aligned} \frac {d^2y}{dx^2} & = \left(\frac {1-\ln x}{x^2}\right)^2 x^\frac 1x - \left( \frac 1{x^3} + \frac {2(1-\ln x)}{x^3}\right) x^\frac 1x \\ \frac {d^2y}{dx^2}\bigg|_{x=e} & = 0 - \left( \frac 1{e^3} + 0\right) e^\frac 1e < 0 \end{aligned}$

let $f(x) = y = x^{1/x}$

$\implies f'(x) = y' = -x^{1/x \ - \ 2} ( \ln x - 1) = 0$ to find the value of x that will maximize $f (x) \cdot$

Upon solving, we will get $x = e \approx 2.718281828459045$

But, as $x \rightarrow 3 \ ,$ hence, the answer then.

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The $f' (x)$ was solved by the aid of differentiation using logarithms.

$\begin{aligned} y = x^{1/x} \implies \ln (y) = \dfrac{1}{x} ( \ln x) \\ & \implies \dfrac{y'}{y} = \dfrac{1}{x} \left( \dfrac{1}{x} \right) - ( \ln x) \left( \dfrac{1}{x^2} \right) \\ & f'(x) = y' = x^{1/x} \left( \dfrac{1}{x^2} \cdot (1 - \ln x) \right) \\ & f'(x) = -x^{1/x \ - \ 2} \cdot ( \ln x - 1) \end{aligned}$