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JEE Main & Advanced AIEEE Solved Paper-2003

  • question_answer
    Let \[\frac{d}{dx}F(x)=\left( \frac{{{e}^{\sin x}}}{x} \right),x>0\]. If \[\int_{1}^{4}{\,\,\,\frac{3}{x}{{e}^{\sin {{x}^{3}}}}dx=F(k)-F(1)}\]. then one of the possible value of k, is     AIEEE  Solved  Paper-2003

    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)] \[\Rightarrow \]   \[[F(z)_{1}^{64}=F(k)-F(1)\] \[\Rightarrow \]   \[F(64)-F(1)=F(k)-F(1)\] \[\Rightarrow \]   \[k=64\]


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