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JEE Main & Advanced JEE Main Paper (Held On 10-Jan-2019 Evening)

  • question_answer
    The positive value of \[\lambda \] for which the co-efficient of \[{{x}^{2}}\] in the expression \[{{x}^{2}}{{\left( \sqrt{x}+\frac{\lambda }{{{x}^{2}}} \right)}^{10}}\]is 720, is- [JEE Main Online Paper (Held On 10-Jan-2019 Evening]

    A) 4                                 

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

    C) 3                     

    D)                  \[\sqrt{5}\]

    Correct Answer: A

    Solution :

    \[{{T}_{r+1}}{{=}^{10}}Cr(\sqrt{x}){{\,}^{10-r}}\,{{\left( \frac{\lambda }{{{x}^{2}}} \right)}^{r}}\,={{\,}^{10}}Cr\,{{\lambda }^{r}}\,{{x}^{5-\frac{r}{2}-2r}}\,\] \[\Rightarrow \,\,\,5-\frac{r}{2}\,-\,2r=0\,\,\,\Rightarrow \,\,\,r=2\] coefficient of \[{{x}^{2}}\Rightarrow \,{{\,}^{10}}{{C}_{2}}{{\lambda }^{2}}=720\,\,\Rightarrow \,{{\lambda }^{2}}=16\] \[\Rightarrow \,\,\lambda \,\,=\pm \,\text{4}\,\,\,\Rightarrow \,\,\lambda \,\,\Rightarrow \,\,4\]


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