JEE Mains · Maths · STD 12 - 6. Application of derivatives
The number of critical points of the function \(f(x) = \begin{cases} \left|\dfrac{\sin x}{x}\right|, & x \neq 0 \\ 1, & x = 0 \end{cases}\) in the interval \((-2\pi, 2\pi)\) is equal to :
- A \(1\)
- B \(3\)
- C \(5\)
- D \(7\)
Answer & Solution
Correct Answer
(C) \(5\)
Step-by-step Solution
Detailed explanation
The critical points of a function are the points in its domain where the derivative is zero or does not exist. First, consider \(x = 0\). The function is continuous at \(x = 0\) since \(\lim_{x \to 0} \left|\dfrac{\sin x}{x}\right| = 1 = f(0)\). For \(x \in (-\pi, \pi)\),…
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