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JEE Main & Advanced Sample Paper JEE Main Sample Paper-38

  • question_answer
    Let\[{{(1-x-2{{x}^{2}})}^{6}}=1+{{a}_{1}}x+{{a}_{2}}{{x}^{2}}+...+{{a}_{12}}{{x}^{12}}\]. Then\[\frac{{{a}_{2}}}{{{2}^{2}}}+\frac{{{a}_{4}}}{{{2}^{4}}}+\frac{{{a}_{6}}}{{{2}^{6}}}+...+\frac{{{a}_{12}}}{{{2}^{12}}}\]is equal

    A) \[-1\]                           

    B) \[-1/2\]

    C) \[0\]                             

    D) \[1/2\]

    Correct Answer: B

    Solution :

    \[{{(1-x-2{{x}^{2}})}^{6}}={{(1+x)}^{6}}{{(1-2x)}^{6}}\]\[=1+{{a}_{1}}x+{{a}_{2}}{{x}^{2}}+.....+{{a}_{12}}{{x}^{12}}\] Putting \[x=1/2\], we have \[0=1+{{a}_{1}}/2+{{a}_{2}}/{{2}^{2}}+{{a}_{3}}/{{2}^{3}}+\]\[{{a}_{4}}/{{2}^{4}}+...+{{a}_{12}}/{{2}^{2}}\]                                    ? (1) Putting \[x=-1/2\], we have \[1=1-{{a}_{1}}/2+{{a}_{2}}/{{2}^{2}}-{{a}_{3}}/{{2}^{3}}+{{a}_{4}}/{{2}^{4}}-...{{a}_{12}}/{{2}^{12}}\]                                                            ? (2) Adding (1) and (2), we have \[1=2(1+{{a}_{2}}/{{2}^{2}}+{{a}_{4}}/{{2}^{4}}+...+{{a}_{12}}/{{2}^{12}})\] \[\Rightarrow \]\[{{a}_{2}}/{{2}^{2}}+{{a}_{4}}/{{2}^{4}}+{{a}_{6}}/{{2}^{6}}+...+{{a}_{12}}/{{2}^{12}}=-1/2\]


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