VITEEE 2015 Paper

Let f (x) = log (1 + x) -

x

1+x

∴ f ' (x) =

1

1+x

-

(1+x).1−x.1

(1+x)2

=

1

1+x

-

1

(1+x)2

=

x

(1+x)2

which is positive. [Since x > 0]
∴ f(x) is monotonic increasing, when x > 0.
⇒ f (x) > f (0)
Now, f(0) = log 1 - 0 = 0
∴ f (x) > 0
⇒ log (1 + x) -

x

x+1

> 0
⇒

x

x+1

< log (1 + x) ... (i)
Also, for x > 0,
x2 > 0 ⇒ x2 x > x
⇒ x (x + 1) > x
⇒ x >

x

x+1

... (ii)
From eqs. (i) and (ii), we get

x

x+1

< log (1 + x) < x
[Since log (1 + x) < x for x > 0]