For all $x \in(0,1)$

$\log _e(1+x)<x$

Consider the function

$f(x)=e^x-(1+x)$

$\Rightarrow f'(x)=e^x-1$

$\Rightarrow f'(x)>0 \text { for } x \in(0,1)$

⇒ f(x) is increasing on (0, 1)

$\Rightarrow f(x)>f(0) $ for all $ x \in(0,1)$

$\Rightarrow e^x-(1+x)>$ for all $ x \in(0,1)$

$\Rightarrow e^x>1+x $ for all $ x \in(0,1)$

Thus, option (a) is not correct.

Let $g(x)=\log _e(1+x)-x, x \in(0,1)$. Then,

$g'(x)=\frac{1}{1+x}-1=-\frac{x}{1+x}<0$ for all $x \in(0,1)$

⇒  g(x) is decreasing on (0, 1)

$\Rightarrow g(x)<g(0)$ for all $x \in(0,1)$

$\Rightarrow \log _e(1+x)-x<0$ for all $x \in(0,1)$

$\Rightarrow \log _e(1+x)<x$ for all $x \in(0,1)$

Thus, option (b) is correct.

Similarly, it can be shown that options (c) and (d) do not hold.