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  • Given that 4th term in the expansion of <img alt="\left(2+\frac{3}{8} x\right)^{10}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cleft%282&plus;%5Cfrac%7B3%7D%7B8%7D%20x%5Cright%29%5

Given that 4th term in the expansion of \left(2+\frac{3}{8} x\right)^{10} has the maximum numerical value, the range of value of x for which this will be true is given by 

Option: 1

-\frac{64}{21}<x<-2


Option: 2

-\frac{64}{21}<x<2


Option: 3

\frac{64}{21}<x<4


Option: 4

None of these


Answers (1)

best_answer

T_3, T_4, T_5 in the given expansion are respectively 

\begin{aligned} &{ }^{10} \mathrm{C}_2 2^8\left(\frac{3 x}{8}\right)^2,{ }^{10} \mathrm{C}_3 2^7\left(\frac{3 x}{8}\right)^3,{ }^{10} \mathrm{C}_4 2^6\left(\frac{3 x}{8}\right)^4\\ &1620 x^2, 810 x^3, \frac{8505}{32} x^4 \end{aligned}

We are given that  is numerically the greatest term so that \left|T_4\right|>|T_3\mid and \left|T_4\right|>|T_5\mid

\begin{aligned} & \therefore|x|>2 \text { and } \frac{64}{21}>x|x| \\ & \quad 2<|x|<\frac{64}{21} \end{aligned}    .....(i)

The above inequality (i) is equivalent to two inequalities 2<x<\frac{64}{21} \text { and }-\frac{64}{21}<x<-2

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