The number of critical points of the function in the interval () is equal to: [2026]

Question

The number of critical points of the function $f (x) = {\begin{matrix} | \frac{\sin x}{x} |, & x \neq 0 \\ 1, & x = 0 \end{matrix}$ in the interval ( $- 2 π, 2 π$ ) is equal to: [2026]

Smart Achievers · Accepted Answer

(3)

We have, $f (x) = {\begin{matrix} | \frac{\sin x}{x} |, & x \neq 0 \\ 1, & x = 0 \end{matrix}; x \in (- 2 π, 2 π)$

$f (x) = {\begin{matrix} - \frac{\sin x}{x}, & x \in (- 2 π, - π] \cup [π, 2 π) \\ \frac{\sin x}{x}, & x \in (- π, π) - {0} \\ 1, & x = 0 \end{matrix}$

From graph, clearly f(x) is non-differentiable at $x = - π, π$

$\frac{d}{d x} (\frac{\sin x}{x}) = \frac{x \cos x - \sin x}{x^{2}}$

$= \frac{\cos x (x - \tan x)}{x^{2}}$

Also, $f' (x) = 0$ at $x = - \frac{3 π}{2}, 0, \frac{3 π}{2}$

Hence, the number of critical points is 5.

Solution

Q.
The number of critical points of the function $f (x) = {\begin{matrix} | \frac{\sin x}{x} |, & x \neq 0 \\ 1, & x = 0 \end{matrix}$ in the interval ( $- 2 π, 2 π$ ) is equal to: [2026]

1 1

2 3

3 5

4 7

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Solution

Q. The number of critical points of the function f(x)={|sinxx|,x≠01,x=0 in the interval (–2π, 2π) is equal to: [2026]

1 1

2 3

3 5

4 7

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Q.
The number of critical points of the function $f (x) = {\begin{matrix} | \frac{\sin x}{x} |, & x \neq 0 \\ 1, & x = 0 \end{matrix}$ in the interval ( $- 2 π, 2 π$ ) is equal to: [2026]