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  • The interval in which x must lie so that the numerically greatest term in the expansion of  h

The interval in which x must lie so that the numerically greatest term in the expansion of (1-x)^{21} has the numerically greatest coefficient is
 

Option: 1

\left[\frac{5}{6}, \frac{6}{5}\right]


Option: 2

\left(\frac{5}{6}, \frac{6}{5}\right)


Option: 3

\left(\frac{4}{5}, \frac{5}{4}\right)


Option: 4

\left[\frac{4}{5}, \frac{5}{4}\right]


Answers (1)

best_answer

If n is odd, then numerically the greatest coefficient in the expansion of (1-x)^n is { }^n C_{(n-1) / 2} or { }^n C_{(n+1) / 2} , Therefore in case of (1-x)^{21}, the numerically greatest coefficient is { }^{21} C_{10} \text { or }{ }^{21} C_{11} \text {. }

Therefore the numerically greatest term ={ }^{21} C_{11} x^{11} or { }^{21} C_{10} x^{10}

\begin{aligned} & \therefore{ }^{21} C_{11} x^{11}>{ }^{21} C_{12} x \text { and }{ }^{21} C_{10} x x^{10}>{ }^{21} C_9 x^9 \\ & \Rightarrow \frac{21 !}{10 ! 11 !}>\frac{21 !}{9 ! 12 !} x \text { and } \frac{21 !}{11 ! 10 !} x>\frac{21 !}{9 ! 12 !} \\ & \Rightarrow \frac{6}{5}>x \text { and } x<\frac{5}{6} \Rightarrow x \in\left(\frac{5}{6}, \frac{6}{5}\right) \end{aligned}

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