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This is a Test of Mathematics Solution Subjective 150 (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.

Also visit: I.S.I. & C.M.I. Entrance Course of Cheenta

Find the maximum among $ \mathbf { 1 , 2^{1/2} , 3^{1/3} , 4^{1/4} , ... }$ .

Consider the function $ \mathbf { f(x) = x^{1/x} }$ . We employ standard techniques to compute the maxima.

Take logarithm on both sides we have $ \mathbf { \log f(x) = \frac{1}{x} \log x }$ . Next find out the derivative:

$ \mathbf {\frac {1}{f(x)} f'(x) = \frac{-1}{x^2} \log x + \frac{1}{x}\cdot\frac{1}{x} implies f'(x) = f(x) \cdot \frac{1}{x^2} (1 - \log x) }$

Since $ \mathbf { f(x) = x^{1/x} }$ is always positive for positive x and so is $ \mathbf {\frac{1}{x^2}}$ sign of the derivative depends only on (1-logx). Hence the derivative is 0 at x = e (2.71 approximately), positive before that and negative after that. Hence the function has a maxima at x = e.

We check the values at x=2 and x=3 and easy computations show that $ \mathbf { 3^{1/3} > 2^{1/2} }$. Hence $ \mathbf {3^{1/3} }$ is the largest value.

**Special Note**

One may ask for a non calculus proof of this problem. The basic idea is to understand that the inequality

$ \mathbf { n^{1/n} > (n+1)^{1/n+1}\Rightarrow n^{n+1} > (n+1)^n \Rightarrow n\cdot n^n > (n+1)^n \\ \Rightarrow n > \frac{(n+1)^n}{n^n}\Rightarrow n > (1+ \frac{1}{n})^n }$

It is easy to show that the quantity $ \mathbf { (1+ \frac{1}{n})^n }$ lies within 2 and 3 for all values of n. Hence the inequality $ \mathbf {n > (1+ \frac{1}{n})^n }$ is true for n > 3. The result follows.

This is a Test of Mathematics Solution Subjective 150 (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.

Also visit: I.S.I. & C.M.I. Entrance Course of Cheenta

Find the maximum among $ \mathbf { 1 , 2^{1/2} , 3^{1/3} , 4^{1/4} , ... }$ .

Consider the function $ \mathbf { f(x) = x^{1/x} }$ . We employ standard techniques to compute the maxima.

Take logarithm on both sides we have $ \mathbf { \log f(x) = \frac{1}{x} \log x }$ . Next find out the derivative:

$ \mathbf {\frac {1}{f(x)} f'(x) = \frac{-1}{x^2} \log x + \frac{1}{x}\cdot\frac{1}{x} implies f'(x) = f(x) \cdot \frac{1}{x^2} (1 - \log x) }$

Since $ \mathbf { f(x) = x^{1/x} }$ is always positive for positive x and so is $ \mathbf {\frac{1}{x^2}}$ sign of the derivative depends only on (1-logx). Hence the derivative is 0 at x = e (2.71 approximately), positive before that and negative after that. Hence the function has a maxima at x = e.

We check the values at x=2 and x=3 and easy computations show that $ \mathbf { 3^{1/3} > 2^{1/2} }$. Hence $ \mathbf {3^{1/3} }$ is the largest value.

**Special Note**

One may ask for a non calculus proof of this problem. The basic idea is to understand that the inequality

$ \mathbf { n^{1/n} > (n+1)^{1/n+1}\Rightarrow n^{n+1} > (n+1)^n \Rightarrow n\cdot n^n > (n+1)^n \\ \Rightarrow n > \frac{(n+1)^n}{n^n}\Rightarrow n > (1+ \frac{1}{n})^n }$

It is easy to show that the quantity $ \mathbf { (1+ \frac{1}{n})^n }$ lies within 2 and 3 for all values of n. Hence the inequality $ \mathbf {n > (1+ \frac{1}{n})^n }$ is true for n > 3. The result follows.

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