Question

If $a=(100)^{1 / 100}$ and $b=(101)^{1 / 101}$ then

• A. $a=b$
• B. $a>b$
• C. $a < b$
• D. none of these

Answer 1

Answer: (B)

Assume $f(x)=x^{1 / x}$ and let us examine monotonic nature of $f(x)$
$ f^{\prime}(x)=x^{1 / x} \cdot\left(\frac{1-\ell n x}{x^2}\right) $
$ f^{\prime}(x)>0 \Rightarrow x \in(0, e)$
and $ f^{\prime}(x) < 0 \Rightarrow(e, \infty)$
Hence $f(x) $ is M.D. for $x \geq e$

and since $100 < 101$
$ \Rightarrow f(100)>f(101) $
$ \Rightarrow (100)^{1 / 100}>(101)^{1 / 101}$