<s>[INST] All real numbers $(p,q)$ such that $|\sqrt{1-x^{2}}-p x-q| \leq \frac{\sqrt{2}-1}{2}$ holds for every $x \in[0,1]$ Find all pairs of real numbers $(p, q)$ such that the inequality
$|\sqrt{1-x^{2}}-p x-q| \leq \frac{\sqrt{2}-1}{2}$ holds for every $x \in[0,1]$

Originally I thought to rephrase it in geometric terms, seeing one degree terms I think about equation of a line.
I also have that $\sqrt{1-x^{2}}-\frac{\sqrt{2}-1}{2} \leq p x+q \leq \sqrt{1-x^{2}}+\frac{\sqrt{2}-1}{2}$
I think that a geometric solution might be possible, please help me to proceed.
Thanks.
 [/INST] Remark: @VIVID gave a nice solution. I rewrote it in elementary way (without using derivative, extrema). Note that while we used calculus to motivate our solution, we do not need to include any
calculus in the solution!
Letting $x = 0$, we have $|1 - q| \le \frac{\sqrt{2}-1}{2}$ or
$$1 - \frac{\sqrt{2}-1}{2} \le q \le 1 + \frac{\sqrt{2}-1}{2}. \tag{1}$$
Letting $x = 1$, we have $|-p - q| \le \frac{\sqrt{2}-1}{2}$ or
$$-p - \frac{\sqrt{2}-1}{2} \le q \le -p + \frac{\sqrt{2}-1}{2}. \tag{2}$$
From (1) and (2), we know that $p < 0$ (reason: if $p\ge 0$ then $-p + \frac{\sqrt{2}-1}{2} < 1 - \frac{\sqrt{2}-1}{2}$).
Since $-\frac{p}{\sqrt{1+p^2}} \in [0, 1]$, by letting $x = -\frac{p}{\sqrt{1+p^2}}$,
we have $|\sqrt{p^2+1} - q| \le \frac{\sqrt{2}-1}{2}$ or
$$\sqrt{p^2+1} - \frac{\sqrt{2}-1}{2} \le q \le \sqrt{p^2 + 1} + \frac{\sqrt{2}-1}{2}. \tag{3}$$
From (2) and (3), since $\sqrt{p^2 + 1} > -p$, we have
$$\sqrt{p^2+1} - \frac{\sqrt{2}-1}{2} \le -p + \frac{\sqrt{2}-1}{2}$$
that is
$\sqrt{p^2 + 1} + p - \sqrt{2} + 1 \le 0$ or
$\frac{(p^2+1) - 2}{\sqrt{p^2+1} + \sqrt{2}}  + (p+1) \le 0$ or
$\frac{(p -  1 + \sqrt{p^2+1} + \sqrt{2})(p+1)}{\sqrt{p^2+1} + \sqrt{2}} \le 0$ which results in
$p \le -1$.
From (1) and (3), since $1 + \frac{\sqrt{2}-1}{2} \le \sqrt{p^2 + 1} + \frac{\sqrt{2}-1}{2}$, we have
$$\sqrt{p^2+1} - \frac{\sqrt{2}-1}{2} \le 1 + \frac{\sqrt{2}-1}{2}$$
that is $p^2 \le 1$. Thus, we have $p = -1$.
From (1) and (3), we have $q = 1 + \frac{\sqrt{2}-1}{2}$.
Finally, when $p = -1, q = 1 + \frac{\sqrt{2}-1}{2}$, it is easy to prove that $|\sqrt{1-x^2} + x - 1 - \frac{\sqrt{2}-1}{2}| \le \frac{\sqrt{2}-1}{2}, \forall x \in [0, 1]$
that is $1\le \sqrt{1-x^2} + x \le \sqrt{2}, \forall x\in [0, 1]$.
Thus, $p = -1, q = 1 + \frac{\sqrt{2}-1}{2}$ is the only one pair satisfies the statement.
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