Question:medium
The number of critical points of the function}
\[
f(x)=
\begin{cases}
\dfrac{|\sin x|}{x}, & x\neq0\\
1, & x=0
\end{cases}
\]
in the interval \((-2\pi,2\pi)\) is equal to:
The number of critical points of the function}
\[
f(x)=
\begin{cases}
\dfrac{|\sin x|}{x}, & x\neq0\\
1, & x=0
\end{cases}
\]
in the interval \((-2\pi,2\pi)\) is equal to:
Updated On: Apr 10, 2026
- \(1\)
- \(3\)
- \(5\)
- \(7\)
Show Solution
The Correct Option is C
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