Question

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:

• A. \(1\)
• B. \(3\)
• C. \(5\)
• D. \(7\)

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Critical points are points in the domain where the function's derivative is zero or the derivative does not exist. For absolute value functions, we check where the expression inside the absolute value is zero (potential non-differentiability) and where its derivative is zero.
Step 3: Detailed Explanation:
Let \(g(x) = \frac{\sin x}{x}\). The function is \(f(x) = |g(x)|\).
1. Points where \(f'(x) = 0\):
This occurs where \(g'(x) = 0\).
\[ g'(x) = \frac{x \cos x - \sin x}{x^2} \]
\(g'(x) = 0 \implies x \cos x = \sin x \implies x = \tan x\).
In the interval \((-2\pi, 2\pi)\), the graphs of \(y=x\) and \(y=\tan x\) intersect at 3 points:
- \(x = 0\) (The limit is 1, and the derivative of \((\sin x)/x\) at 0 is 0 by expansion).
- One point in \((\pi, 3\pi/2)\).
- One point in \((-3\pi/2, -\pi)\).
Total 3 points where the derivative is zero.
2. Points where \(f'(x)\) does not exist:
This occurs where \(g(x) = 0\) (where the function crosses the x-axis).
\(\frac{\sin x}{x} = 0 \implies \sin x = 0\) for \(x \neq 0\).
In \((-2\pi, 2\pi)\), this occurs at \(x = \pi\) and \(x = -\pi\).
At these 2 points, the absolute value creates a sharp corner (cusp), so the derivative does not exist.
Total critical points = \(3 + 2 = 5\).
Step 4: Final Answer:
The number of critical points is 5.