IIT JEE Advanced 2010 Paper 1

We have
f' (x) = 1x+√1+sinx
f ′(x) is not differentiable at sin x = - 1 or x = 2nπ -

π

2

, ∊ N
In x ∈(1,∞): f (x) > 0, f ′(x) > 0.
Consider f (x) - f ′(x). Therefore,
ln x +

x

∫

0

√1+sintdt−

1

x

−√1+sinx
=
Consider g (x) =

x

∫

0

√1+sintdt−√1sinx
It can be proved that g(x) ≥ 2√2−√10 ∀ x ∈(0, ∞).
Now, there exists some α > 1 such that

1

x

- ln x ≤ 2√2−√10 ∀ x ∈(a, ∞)
Since 1x−lnx is strictly decreasing function. Therefore,
g (x) ≥

1

x

−ln‌x