JEE Main & Advanced JEE Main Paper (Held on 10-4-2019 Morning)

  • question_answer If the coefficients of \[{{x}^{2}}\]and \[{{x}^{3}}\]are both zero, in the expansion of the expression \[(1+ax+b{{x}^{2}})\]\[{{(1-3x)}^{15}}\]in powers of x, then the ordered pair (a, b) is equal to: [JEE Main 10-4-2019 Morning]  

    A) (28, 315)

    B) (-54, 315)

    C) (-21, 714)                   

    D) (24, 861)

    Correct Answer: A

    Solution :

    Coefiicient of \[{{x}^{2}}={}^{15}{{C}_{2}}\times 9-3a\left( {}^{15}{{C}_{1}} \right)+b=0\]       \[\Rightarrow -45a+b+{}^{15}{{C}_{2}}\times 9=0\]                 ....(i) Also, \[-27\times {}^{15}{{C}_{3}}+9a\times {}^{15}{{C}_{2}}-3b\times {}^{15}{{C}_{1}}=0\] \[\Rightarrow 9\times {}^{15}{{C}_{2}}\,a-45\,b-27\times {}^{15}{{C}_{3}}=0\] \[\Rightarrow 21\,a-\,b-273=0\]                                  ?(ii) ...(ii) + (ii) \[24a+672=0\]\[~\Rightarrow a=28\] So, \[b=315\]             


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