VITEEE 2014 Paper

Let f (x) =

x

logx

⇒ f' (x) =

lox−1

(logx)2

For maxima and minima, put f (x) = 0 logx- 1 = 0
⇒ x = e
Now , f " (x)
=

(logx)2. 1 x −(logx−1). 2logx x

(logx)4

⇒ f " (e) =

1 e −0

1

=

1

e

> 0
∴ f (x) is minimum at x = e.
Hence, minimum value of f (x) at x = e is
f (e) =

e

loge

= e