Discussion
Let \(f(x) = |\sin 3x| - |\cos 3x|\), where \(\frac{\pi}{6} \leq x \leq \frac{\pi}{3}\). Then the value of \(f'\left(\frac{\pi}{4}\right)\) is equal to
Answer: A
\(f'\left(\frac{\pi}{4}\right) = \) \(-3\sqrt{2}\).
For \(\frac{\pi}{6} \leq x \leq \frac{\pi}{3}\), \(3x\) lies between \(\frac{\pi}{2}\) and \(\pi\), where \(\sin 3x \geq 0\) and \(\cos 3x \leq 0\). So the moduli open as
\[ \begin{aligned} f(x) &= \sin 3x + \cos 3x \\ f'(x) &= 3\cos 3x - 3\sin 3x \end{aligned} \]
At \(x = \frac{\pi}{4}\), \(3x = \frac{3\pi}{4}\), with \(\cos = -\frac{1}{\sqrt{2}}\) and \(\sin = \frac{1}{\sqrt{2}}\):
\[ \begin{aligned} f'\left(\tfrac{\pi}{4}\right) &= 3\left(-\tfrac{1}{\sqrt{2}} - \tfrac{1}{\sqrt{2}}\right) \\ &= -\tfrac{6}{\sqrt{2}} = -3\sqrt{2} \end{aligned} \]
Ignoring the moduli (\(f = \sin 3x - \cos 3x\)) gives 0 at this point; \(\frac{\pm 3}{\sqrt{2}}\) drops one of the two terms.
Post your comment
Please log in or register to post a comment.
Examination Categories
- SSC Exams
- Banking Exams
- Teaching Exams
- Civil Services Exams
- Railways Exams
- Engineering Recruitment Exams
- Defence Exams
- State Government Exams
- Police Exams
- Insurance Exams
- Nursing Exams
- Other Government Exams
- NRA CET
- PG Entrance Exams
- Campus Placement
- MBA Entrance Exams
- CUET and UG Entrance
- Judiciary Exams
- Regulatory Body Exams
- Software Development