Consider the functions I. $e^{-x}$ II. $x^2 - \sin x$ III. $\sqrt{x^3 + 1}$ Which of the above functions is/are increasing everywhere in $[0,1]$?

A differentiable function $f(x)$ is increasing on an interval if its first derivative $f^{\prime}(x) \ge 0$ everywhere on that interval:

1. Function I:$f_1(x) = e^{-x}$

   $$f_1'(x) = -e^{-x} < 0 \quad \text{for all } x \in [0, 1]$$

   Thus, $f_1(x)$ is strictly decreasing.

2. Function II:$f_2(x) = x^2 - \sin x$

   $$f_2'(x) = 2x - \cos x$$

   Evaluating at $x = 0$:

   $$f_2'(0) = 2(0) - \cos(0) = -1 < 0$$

   Since the derivative is negative near $x = 0$, $f_2(x)$ is not increasing everywhere in $[0, 1]$.

3. Function III:$f_3(x) = \sqrt{x^3 + 1}$

   $$f_3'(x) = \frac{3x^2}{2\sqrt{x^3 + 1}} \ge 0 \quad \text{for all } x \in [0, 1]$$

   In fact, $f_3'(x) > 0$ for all $x \in (0, 1]$, so $f_3(x)$ is strictly increasing on $[0, 1]$.

Hence, only III is increasing everywhere in $[0, 1]$, making (A) the correct option.