JEE Main & Advanced JEE Main Solved Paper-2014

  • question_answer
    If the coefficients of \[{{x}^{3}}\]and \[{{x}^{4}}\] in the expansion of \[(1+ax+b{{x}^{2}}){{(1-2x)}^{18}}\]in powers of x are both zero, then (a, b) is equal to:   JEE Main  Solved  Paper-2014

    A) \[\left( 16,\frac{251}{3} \right)\]                              

    B) \[\left( 14,\frac{251}{3} \right)\]

    C) \[\left( 14,\frac{272}{3} \right)\]                              

    D) \[\left( 16,\frac{272}{3} \right)\]

    Correct Answer: D

    Solution :

    \[(1+ax+b{{x}^{2}})\] \[[1{{-}^{18}}{{C}_{1}}2x{{+}^{18}}{{C}_{2}}{{(2x)}^{2}}{{-}^{18}}{{C}_{3}}{{(2x)}^{3}}{{+}^{18}}{{C}_{4}}{{(2x)}^{4}}...]\]Coefficient of \[{{x}^{3}}\]is \[{{-}^{18}}{{C}_{3}}({{2}^{3}})+a{{(}^{18}}{{C}_{2}}\times 4)-b{{(}^{18}}{{C}_{1}}\times 2)=0\]   ?(i) Coefficient of is \[^{18}{{C}_{4}}({{2}^{4}})+a({{-}^{18}}{{C}_{3}}{{2}^{3}}){{+}^{18}}{{C}_{2}}b{{2}^{2}}=0\]           ?(ii) or solving both these equation a = 16 and b = 272/3.

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