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GUJCET · Maths · Continuity and Differentiability
\(\frac{d}{d x}\left[3 \sin \left(60^{\circ}-x^{\circ}\right)-4 \cos ^3\left(30^{\circ}+x^{\circ}\right)\right]=\) ___________ .
- A \(-\frac{\pi}{60} \sin \left(3 x^{\circ}\right)\)
- B \(\frac{\pi}{60} \sin \left(3 x^{\circ}\right)\)
- C \(\frac{\pi}{60} \cos \left(3 x^{\circ}\right)\)
- D \(-\frac{\pi}{60} \cos \left(3 x^{\circ}\right)\)
Answer & Solution
Correct Answer
(C) \(\frac{\pi}{60} \cos \left(3 x^{\circ}\right)\)
Step-by-step Solution
Detailed explanation
\( 3 \sin \left(60^{\circ}-x^{\circ}\right)-4 \cos ^3\left(30^{\circ}+x^{\circ}\right) = 3 \cos \left(30^{\circ}+x^{\circ}\right)-4 \cos ^3\left(30^{\circ}+x^{\circ}\right) = -\left(4 \cos ^3\left(30^{\circ}+x^{\circ}\right)-3 \cos \left(30^{\circ}+x^{\circ}\right)\right) = -\cos \left(3\left(30^{\circ}+x^{\circ}\right)\right) = -\cos \left(90^{\circ}+3x^{\circ}\right) = \sin \left(3x^{\circ}\right) \) \( \frac{d}{d x}\left[\sin \left(3 x^{\circ}\right)\right] = \cos \left(3 x^{\circ}\right) \cdot \frac{d}{d x}\left(3x \frac{\pi}{180}\right) = \cos \left(3 x^{\circ}\right) \cdot \frac{3\pi}{180} = \frac{\pi}{60} \cos \left(3 x^{\circ}\right) \)
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