$f(x)=$

{	$1$	if	$|x|\le 1$
	$0$	if	$|x|>1$

,  $g(x)=$

{	$2-x^2$	if	$|x|\le 2$
	$2$	if	$|x|>2$

If $h(x)=f[g(x)]$, then an interval in which $h(x)=1$ for all values of $x$ in that interval is:

Given,

$h(x)=f[g(x)]$

Now, $f(t)=1$ when

$|t|\le 1$

So, $h(x)=1$ when

$|g(x)|\le 1$

For $|x|\le 2$,

$g(x)=2-x^2$

So,

$|2-x^2|\le 1$

This gives

$-1\le 2-x^2\le 1$

Subtract $2$ from all sides:

$-3\le -x^2\le -1$

Multiplying by $-1$ reverses the inequalities:

$1\le x^2\le 3$

So,

$1\le |x|\le \sqrt{3}$

For $|x|>2$,

$g(x)=2$

So,

$|g(x)|=2>1$

Hence, this case is not valid.

Therefore, the required interval is

$1\le |x|\le \sqrt{3}$