A) 2
B) ?1
C) \[-2+{{2}^{\pi }}{{\log }_{e}}2\]
D) \[-2+{{\log }_{e}}2\]
Correct Answer: B
Solution :
\[f(x)=\sin 2x.\cos 2x.\cos 3x+{{\log }_{2}}{{2}^{x+3}}\] \[f(x)=\frac{1}{2}\sin 4x\cos 3x+(x+3){{\log }_{2}}2\] \[f(x)=\frac{1}{4}[\sin 7x+\sin x]+x+3\] Differentiate w.r.t. x, \[f'(x)=\frac{1}{4}[7\cos 7x+\cos x]+1\] \[f'(x)=\frac{7}{4}\cos 7x+\frac{1}{4}\cos x+1\]. Hence \[f'(\pi )=-2+1=-1\].
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