Question:
What is the length of the longest interval in which the function \(f(x) = 3\sin x – 4\sin^3 x\) is increasing?
उस सबसे लंबे अंतराल की लंबाई क्या है जिसमें फलन \(f(x) = 3\sin x – 4\sin^3 x\) वर्धमान है?
Options:
- \(\pi/3\)
- \(\pi/2\)
- \(3\pi/2\)
- \(\pi\)
- \(\pi/3\)
- \(\pi/2\)
- \(3\pi/2\)
- \(\pi\)
Correct Answer: a)
Explanation:
Using the trigonometric identity \(\sin 3x = 3\sin x – 4\sin^3 x\):
\[f(x) = \sin 3x \implies f'(x) = 3\cos 3x\]
The function is increasing where \(f'(x) \ge 0 \iff \cos 3x \ge 0\).
The principal interval where \(\cos \theta \ge 0\) is \(\theta \in [-\pi/2, \pi/2]\):
\[-\frac{\pi}{2} \le 3x \le \frac{\pi}{2} \implies -\frac{\pi}{6} \le x \le \frac{\pi}{6}\]
The length of this interval is:
\[\frac{\pi}{6} – \left(-\frac{\pi}{6}\right) = \frac{2\pi}{6} = \frac{\pi}{3}\]
Hence, the correct option is (a).
त्रिकोणमितीय सर्वसमिका \(f(x) = \sin 3x\) से:
\[f'(x) = 3\cos 3x \ge 0 \implies -\frac{\pi}{2} \le 3x \le \frac{\pi}{2} \implies -\frac{\pi}{6} \le x \le \frac{\pi}{6}\]
अंतराल की लंबाई:
\[\frac{\pi}{6} – \left(-\frac{\pi}{6}\right) = \frac{\pi}{3}\]
अतः सही विकल्प (a) है।