KVPY Sample Paper KVPY Stream-SX Model Paper-7

  • question_answer
    If the third term in the binomial expansion of \[{{\left( 1+{{x}^{{{\log }_{2}}x}} \right)}^{5}}\]equals 2560, then a possible value  x is:

    A) \[\frac{1}{4}\]

    B)                                 \[4\sqrt{2}\]

    C) \[\frac{1}{8}\]                          

    D) \[2\sqrt{2}\]

    Correct Answer: A

    Solution :

    In the expansion of \[{{\left( 1+{{x}^{{{\log }_{2}}}}^{x} \right)}^{5}}\]
    Third term say \[{{T}_{3}}={}^{5}{{C}_{2}}{{\left( {{x}^{1o{{g}_{2}}x}} \right)}^{2}}=2560\]
    \[\Rightarrow \]   \[{{\left( {{x}^{{{\log }_{2}}x}} \right)}^{2}}=256\]
    Taking logarithm to the base 2 on both sides
    \[\Rightarrow \]   \[2{{\left( {{\log }_{2}}x \right)}^{2}}=8\]
    \[\Rightarrow \]   \[({{\log }_{2}}x)=\pm 2\]
    \[\Rightarrow \]   \[x=4,\frac{1}{4}\]
    Here,     \[x=\frac{1}{4}.\]


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