Question

If \[ f(x)= \begin{cases} 1,& |x|\le 1\\ 0,& |x|>1 \end{cases} \] \[ g(x)= \begin{cases} 2-x^2,& |x|\le 2\\ 2,& |x|>2 \end{cases} \] and \(h(x)=f(g(x))\), then the interval in which \(h(x)=1\) for all values of \(x\) is:

• A. \(|x|\le \sqrt2\)
• B. \(|x|\le \sqrt3\)
• C. \(\frac12\le |x|\le \sqrt3\)
• D. \(1\le |x|\le \sqrt3\)

Hint:

For composite functions \(f(g(x))\), first identify the values of \(g(x)\) for which the outer function takes the required value.

Answer 1

The Correct Option is D

Step 1: Read what $h=f(g(x))=1$ demands.
The outer function $f$ equals $1$ exactly when its input has absolute value at most $1$. So we need $|g(x)|\le 1$.
Step 2: Use the relevant branch of $g$.
For the interesting region $|x|\le 2$, $g(x)=2-x^2$. So the condition is $|2-x^2|\le 1$.
Step 3: Unfold the absolute value.
This means $-1\le 2-x^2\le 1$.
Step 4: Solve the right inequality.
From $2-x^2\le 1$ we get $x^2\ge 1$.
Step 5: Solve the left inequality.
From $2-x^2\ge -1$ we get $x^2\le 3$.
Step 6: Combine the bounds.
Together $1\le x^2\le 3$, that is $1\le|x|\le\sqrt 3$, which is option 4.
\[ \boxed{1\le |x|\le \sqrt 3} \]