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The term independent of \mathrm{x} in the expansion of \mathrm{\left(1-x^{2}+3 x^{3}\right)\left(\frac{5}{2} x^{3}-\frac{1}{5 x^{2}}\right)^{11}, x \neq 0}     is:

Option: 1

\frac{7}{40}


Option: 2

\frac{33}{200}


Option: 3

\frac{39}{200}


Option: 4

\frac{11}{50}


Answers (1)

best_answer

\mathrm{\text { General } \operatorname{term} \text { in }\left(\frac{5}{2} x^{3}-\frac{1}{5 x^{2}}\right)^{11}} \\

\mathrm{={ }^{11} C_{r}\left(\frac{5}{2}\right)^{11-r}\left(x^{3}\right)^{11-r}\left(-\frac{1}{5}\right)^{r}\left(x^{-2}\right)^{r}} \\

\mathrm{={ }^{11} C_{r} \frac{5^{11-2 r}(-1)^{r} }{2^{11-r}}x^{33-3 r-2 r}}

Constant term in  \mathrm{\left(1-x^{2}+3 x^{3}\right)\left(\frac{5}{2} x^{3}-\frac{1}{5 x^{2}}\right)^{11}}

\mathrm{=1\cdot constant\: term\: in \: second\: bracket -1. coefficient\: of\: x^{-2} in \: second \: bracket }\\

     \mathrm{+3\: coefficient\: of\: x^{-3}in\: second \: bracket}        .........(i)

For constant term in second bracket \mathrm{33-5r=0}

\mathrm{\Rightarrow r=\frac{33}{5} \Rightarrow \text { no such term }}

\mathrm{For \: x^{-2}\: term \: \Rightarrow 33-5 r=-2 \Rightarrow r=7}\\

\mathrm{For \: x^{-3}\: term \: \Rightarrow 33-5 r=-3 \Rightarrow r=\frac{36}{5}}

\mathrm{\Rightarrow no\: such\: term}

\therefore Required answer  \mathrm{=-1 \times{ }^{11} C_{7} \frac{5^{-3}}{2^{4}}(-1)}

                              \mathrm{\begin{aligned} &=\frac{{ }^{11} C_{7}}{5^{3} \cdot 2^{4}} \\ & \end{aligned}}

                              \mathrm{=\frac{33}{200}}

Hence answer is option 2

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vinayak

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