Question

Consider $f(x) = \begin{cases} \frac{\sin x}{x}, & x \neq 0 \\ 1, & x = 0 \end{cases}$
Number of critical points of $f(x)$ in the interval $(-2\pi, 2\pi)$ is:

• A. 3
• B. 5
• C. 7
• D. 1

Answer 1

The Correct Option is A

Step 1: Understand critical points.
Critical points are values in the domain where the derivative is zero or undefined. Note that the function is continuous at $x=0$.


Step 2: Find the derivative for $x \neq 0$.
Using quotient rule on $f(x) = \frac{\sin x}{x}$:
$f'(x) = \frac{x \cos x - \sin x}{x^2}$


Step 3: Set $f'(x) = 0$.
$x \cos x - \sin x = 0 \implies x \cos x = \sin x \implies \tan x = x$.


Step 4: Find solutions for $\tan x = x$ in $(-2\pi, 2\pi)$.
By looking at the intersection of the graphs of $y = \tan x$ and $y = x$:
1. There is one intersection at $x = 0$. (Check derivative at 0: $\lim_{x \to 0} f'(x) = 0$, so $x=0$ is a critical point).
2. There is one solution in $(\pi, \frac{3\pi}{2})$.
3. There is one solution in $(-\frac{3\pi}{2}, -\pi)$.


Step 5: Total points.
The three points are $x = 0$ and two other symmetric points near $\pm 4.49$ radians.
Total critical points = 3.
The answer is 3 (Option 1).