Test Index

VITEEE 2014 Paper

Section: Mathematics

© examsnet.com

Question : 101 of 120

Marks: +1, -0

The minimum value of

\frac{x}{\log x}

is

e
$\frac{1}{e}$
$e^2$
$e^3$

Solution:

Let f (x) =

\frac{x}{\log x}

⇒ f' (x) =

\frac{lox - 1}{(\log x)^2}

For maxima and minima, put f (x) = 0 logx- 1 = 0

⇒ x = e

Now , f " (x)

=

\frac{ (\log x)^2 \cdot \frac{1}{x} - (\log x - 1) \cdot \frac{2\log x}{x} }{ (\log x)^4 }

⇒ f " (e) =

\frac{ \frac{1}{e} - 0 }{1}

=

\frac{1}{e}

> 0

∴ f (x) is minimum at x = e.

Hence, minimum value of f (x) at x = e is

f (e) =

\frac{e}{\log e}

= e

© examsnet.com

Go to Question:

Prev Question

Next Question