\[\frac{{{d}^{2}}}{d{{x}^{2}}}(2\cos x\,\cos 3x)=\] [RPET 2003]

\[{{2}^{2}}(\cos 2x+{{2}^{2}}\cos 4x)\]

\[{{2}^{2}}(\cos 2x-{{2}^{2}}\cos 4x)\]

\[{{2}^{2}}(-\cos 2x+{{2}^{2}}\cos 4x)\]

\[-{{2}^{2}}(\cos 2x+{{2}^{2}}\cos 4x)\]

\[\frac{{{d}^{2}}}{d{{x}^{2}}}(2\cos x\,\cos 3x)=\] �

1.	\[\frac{{{d}^{2}}}{d{{x}^{2}}}(2\cos x\,\cos 3x)=\] [RPET 2003]
A.	\[{{2}^{2}}(\cos 2x+{{2}^{2}}\cos 4x)\]
B.	\[{{2}^{2}}(\cos 2x-{{2}^{2}}\cos 4x)\]
C.	\[{{2}^{2}}(-\cos 2x+{{2}^{2}}\cos 4x)\]
D.	\[-{{2}^{2}}(\cos 2x+{{2}^{2}}\cos 4x)\]
Answer» E.

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\[\frac{{{d}^{2}}}{d{{x}^{2}}}(2\cos x\,\cos 3x)=\] [RPET 2003]

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