JEE Main & Advanced JEE Main Paper (Held on 12-4-2019 Afternoon)

  • question_answer
    The term independent of x in the expansion of\[\left( \frac{1}{60}-\frac{{{x}^{8}}}{81} \right).{{\left( 2{{x}^{2}}-\frac{3}{{{x}^{2}}} \right)}^{6}}\]is equal to : [JEE Main 12-4-2019 Afternoon]

    A) 36                               

    B) - 108

    C) - 72                 

    D) - 36

    Correct Answer: D

    Solution :

    \[\frac{1}{60}{{\left( 2{{x}^{2}}-\frac{3}{{{x}^{2}}} \right)}^{6}}-\frac{1}{81}.{{x}^{8}}{{\left( 2{{x}^{2}}-\frac{3}{{{x}^{2}}} \right)}^{6}}\] its general term \[\frac{1}{60}{{\,}^{6}}{{C}_{1}}{{2}^{6-r}}{{\left( -3 \right)}^{r}}{{x}^{12-r}}-\frac{1}{81}{{\,}^{6}}{{C}_{1}}{{2}^{6-r}}{{\left( -3 \right)}^{2}}{{12}^{20-4r}}\] for term independent of x, r for Ist expression is 3 and r for second expression is 5 \[\therefore \]term independent of \[x=36\]


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