For all x ∈ (0,1).

$\log_e(1 + x)<x$

(A) Let $f(x)=e^x-1-x$

$⇒f'(x)=e^x-1>0,∀\,x∈(0,1)$

So, f(x) is increasing, when 0 < x < 1

⇒ f(x) > f(0) or $e^x-1-x>0⇒e^x-1+x$

(B) Let $g(x)=\log_e(1+x)-x$

$⇒g'(x)=\frac{1}{1+x}-1=\frac{-x}{1+x}<0,∀\,x∈(0,1)$

So, g(x) is decreasing, when 0 < x < 1

$⇒g(0)>g(x)⇒\log_e(1+x)<x$

(C) Let h(x) = sin x - x

$⇒h'(x)=\cos x-1<0,∀\,x∈(0,1)$

So, h(x) is decreasing, where 0 < x < 1

⇒ h(x) < h(0)

∵ sin x < x

(D) Let $g(x)=\log_e\,x-x⇒g'(x)=\frac{1}{x}-1$

$∴g'(x)>0,∀\,x∈(0,1)⇒\log_e\,x-x<-1⇒x-1>\log_e\,x$

$⇒x>\log_e\,x$