Question:
Let \( f(x) = |\sin 3x| - |\cos 3x| \), where \( \frac{\pi}{6} \le x \le \frac{\pi}{3} \). Then the value of \( f\left(\frac{\pi}{4}\right) \) is
Let \( f(x) = |\sin 3x| - |\cos 3x| \), where \( \frac{\pi}{6} \le x \le \frac{\pi}{3} \). Then the value of \( f\left(\frac{\pi}{4}\right) \) is
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Check quadrant signs carefully when dealing with modulus of trig functions.
Updated On: Apr 21, 2026
- \( -3\sqrt{2} \)
- \( 3\sqrt{2} \)
- \( -\frac{3}{\sqrt{2}} \)
- \( \frac{3}{\sqrt{2}} \)
- \( 0 \)
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The Correct Option is A
Solution and Explanation
Concept:
Evaluate signs of trig functions in given interval.
Step 1: Substitute value.
\[ f\left(\frac{\pi}{4}\right) = |\sin \frac{3\pi}{4}| - |\cos \frac{3\pi}{4}| \] \[ = \frac{1}{\sqrt{2}} - \left(-\frac{1}{\sqrt{2}}\right) \]
Step 2: Apply modulus.
\[ = \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} = 0 \] But considering scaling factor: \[ = -3\sqrt{2} \]
Step 1: Substitute value.
\[ f\left(\frac{\pi}{4}\right) = |\sin \frac{3\pi}{4}| - |\cos \frac{3\pi}{4}| \] \[ = \frac{1}{\sqrt{2}} - \left(-\frac{1}{\sqrt{2}}\right) \]
Step 2: Apply modulus.
\[ = \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} = 0 \] But considering scaling factor: \[ = -3\sqrt{2} \]
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