# Interpretating 1D-motion properties described by $x(t) = t - \sin{t}$ [closed]

A question in the third chapter of Exemplar Problems in Physics for Class XI goes thus:

(3.9) For the one-dimensional motion described by $$x = t - \sin{t}$$

(a) $$x(t) > 0$$ for all $$t>0$$.

(b) $$v(t) > 0$$ for all $$t>0$$.

(c) $$a(t) > 0$$ for all $$t>0$$.

(d) $$v(t)$$ lies between $$0$$ and $$2$$.

The answers are (a) and (d).

I do not quite understand how they've arrived at the answers despite taking the first- and second-order derivatives of $$x(t)$$. A detailed explanation as to why the two options are correct would be helpful.

$$v(t) = x'(t) = 1 - cos \;t$$
Since $$-cos \;t$$ lies between -1 and 1, then $$1 - cos \;t$$ lies between 0 and 2. Therefore $$v(t) = x'(t) \geq 0$$. Note that beacuse $$x(0) = 0$$ and $$x'(t) > 0$$ in $$(0, 2\pi)$$, by a corollary of Lagrange Theorem we conclude that $$x(t) > 0$$ in $$(0, 2\pi)$$. Thus, $$x(t) > 0$$ for all $$t > 0$$ beacuse the derivative is non-negative.
Clearly (b) is false and (c) is false beacuse $$a(t) = v'(t) = sin \;t$$, which can be any value in $$[-1, 1]$$