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:
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:
Show Hint
For composite functions \(f(g(x))\), first identify the values of \(g(x)\) for which the outer function takes the required value.
Updated On: Jun 11, 2026
- \(|x|\le \sqrt2\)
- \(|x|\le \sqrt3\)
- \(\frac12\le |x|\le \sqrt3\)
- \(1\le |x|\le \sqrt3\)
Show Solution
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The Correct Option is D
Solution and Explanation
Step 1: Determine when \(f(g(x))=1\).
Since \[ f(y)=1 \] whenever \[ |y|\le 1, \] we require \[ |g(x)|\le 1. \]
Step 2: Use the expression of \(g(x)\).
For \(|x|\le 2\), \[ g(x)=2-x^2. \] Hence \[ |2-x^2|\le 1. \] This gives \[ -1\le 2-x^2\le 1. \]
Step 3: Solve the inequalities.
From \[ 2-x^2\le 1, \] \[ x^2\ge 1. \] From \[ 2-x^2\ge -1, \] \[ x^2\le 3. \] Therefore \[ 1\le x^2\le 3. \] Hence \[ 1\le |x|\le \sqrt3. \] \[ \boxed{\text{Option (D)}} \]
Since \[ f(y)=1 \] whenever \[ |y|\le 1, \] we require \[ |g(x)|\le 1. \]
Step 2: Use the expression of \(g(x)\).
For \(|x|\le 2\), \[ g(x)=2-x^2. \] Hence \[ |2-x^2|\le 1. \] This gives \[ -1\le 2-x^2\le 1. \]
Step 3: Solve the inequalities.
From \[ 2-x^2\le 1, \] \[ x^2\ge 1. \] From \[ 2-x^2\ge -1, \] \[ x^2\le 3. \] Therefore \[ 1\le x^2\le 3. \] Hence \[ 1\le |x|\le \sqrt3. \] \[ \boxed{\text{Option (D)}} \]
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