A) \[{{2}^{20}}(\cos 2x+{{2}^{20}}\cos 4x)\]
B) \[{{2}^{20}}(\cos 2x-{{2}^{20}}\sin 2x)\]
C) \[-6\sin x\sin 3x\]
D) \[6\sin x\sin 3x\]
E) \[6({{2}^{20}}\cos 2x-{{2}^{40}}\cos 4x)\]
Correct Answer: A
Solution :
\[\therefore \]\[2\text{ }cos\text{ }x\text{ }cos\text{ }3x=cos\text{ }4x+cos\text{ }2x\] \[\therefore \]\[\frac{d}{dx}(~2\text{ }cos\text{ }x\text{ }cos\text{ }3x)=\frac{d}{dx}(cos\text{ }4x+cos\text{ }2x)\] \[=-4\text{ }sin\text{ }4x-2\text{ }sin\text{ }2x\] \[\frac{{{d}^{2}}}{d{{x}^{2}}}(2\cos x\cos 3x)=-16\cos 4x-4cos2x\] \[\therefore \] \[\frac{{{d}^{20}}}{d{{x}^{20}}}(2\cos x\cos 3x)\] \[={{4}^{20}}\,\cos 4x\,+{{(2)}^{20}}\cos 2x\] \[={{2}^{20}}({{2}^{20}}\cos 4x+\cos 2x)\]
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