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JEE Main & Advanced JEE Main Paper (Held On 11 April 2015)

  • question_answer
    The term independent of x in the binomial expansion of\[\left( 1-\frac{1}{x}+3{{x}^{5}} \right){{\left( 2{{x}^{2}}-\frac{1}{x} \right)}^{8}}\]is: [JEE Main Online Paper (Held On 11 April 2015)]  

    A)  400

    B)  406

    C)  -400

    D)  -496

    Correct Answer: C

    Solution :

      The general term is second bracket is \[^{8}{{C}_{r}}{{\left( 2{{x}^{2}} \right)}^{8-r}}{{\left( -\frac{1}{x} \right)}^{r}}\] Total exponent of x is \[16-3r\] Term independent of \[x=1x\]exp. of\[{{x}^{0}}+(-1)+\exp .\]of\[x+3\times \exp .\]of\[.\frac{1}{{{x}^{5}}}\] \[=0{{-}^{8}}{{C}^{5}}{{2}^{3}}\left( -1 \right)+3\left( ^{8}{{C}_{7}}2(-1) \right)\]\[=-400\]


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