A) 15
B) 16
C) 63
D) 64
Correct Answer: D
Solution :
\[\frac{d}{dx}F(x)=\frac{{{e}^{\sin x}}}{x}\], \[x>0\] On integrating both sides, we get \[F(x)=\int{\frac{{{e}^{\sin x}}}{x}}\] ... (i) Also, \[\int_{1}^{4}{\frac{3}{x}}\,{{e}^{\sin {{x}^{3}}}}dx=\int_{1}^{4}{\frac{3{{x}^{2}}}{{{x}^{3}}}.\,{{e}^{\sin {{x}^{3}}}}dx}\] \[=F(k)-F(1)\] (given) Let \[{{x}^{3}}=z\,\,\,\,\Rightarrow \,\,\,3{{x}^{2}}dx=dz\] \[\therefore \] \[\int_{1}^{64}{\frac{{{e}^{\sin z}}}{z}dz=F(k)-F(1)}\] [from Eq.(i)] \[\Rightarrow \] \[[F(z)_{1}^{64}=F(k)-F(1)\] \[\Rightarrow \] \[F(64)-F(1)=F(k)-F(1)\] \[\Rightarrow \] \[k=64\]You need to login to perform this action.
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