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(B) The general term $T_{r+1}$ in the expansion of $\left( tx^{\frac{1}{5}} + \frac{(1-x)^{\frac{1}{10}}}{t} \right)^{10}$ is given by $T_{r+1} = {}^{

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$\frac{2 \cdot 10!}{3\sqrt{3}(5!)^2}$

Vedclass · Answer

(B) The general term $T_{r+1}$ in the expansion of $\left( tx^{\frac{1}{5}} + \frac{(1-x)^{\frac{1}{10}}}{t} \right)^{10}$ is given by $T_{r+1} = {}^{10}C_r (tx^{\frac{1}{5}})^{10-r} \left( \frac{(1-x)^{\frac{1}{10}}}{t} \right)^r$.For the term to be independent of $t$,the power of $t$ must be zero: $(10-r) - r = 0 \Rightarrow 2r = 10 \Rightarrow r = 5$.Thus,the term independent of $t$ is $T_6 = {}^{10}C_5 (x^{\frac{1}{5}})^5 ((1-x)^{\frac{1}{10}})^5 = {}^{10}C_5 x^{\frac{1}{2}} (1-x)^{\frac{1}{2}} = {}^{10}C_5 \sqrt{x(1-x)}$.Let $f(x) = {}^{10}C_5 \sqrt{x-x^2}$. To find the maximum,we differentiate $f(x)$ with respect to $x$: $f'(x) = {}^{10}C_5 \cdot \frac{1-2x}{2\sqrt{x-x^2}}$.Setting $f'(x) = 0$,we get $1-2x = 0 \Rightarrow x = \frac{1}{2}$.The maximum value is $f\left(\frac{1}{2}\right) = {}^{10}C_5 \sqrt{\frac{1}{2}(1-\frac{1}{2})} = {}^{10}C_5 \sqrt{\frac{1}{4}} = \frac{1}{2} \cdot {}^{10}C_5 = \frac{1}{2} \cdot \frac{10!}{5!5!} = \frac{10!}{2(5!)^2}$.Wait,re-evaluating the expression: $T_6 = {}^{10}C_5 x (1-x)^{1/2}$ is incorrect based on the provided expression. Let's re-calculate: $T_6 = {}^{10}C_5 (x^{1/5})^5 ((1-x)^{1/10})^5 = {}^{10}C_5 x (1-x)^{1/2}$.Let $f(x) = {}^{10}C_5 x(1-x)^{1/2}$. $f'(x) = {}^{10}C_5 [(1-x)^{1/2} + x \cdot \frac{1}{2}(1-x)^{-1/2}(-1)] = {}^{10}C_5 (1-x)^{-1/2} [1-x - \frac{x}{2}] = {}^{10}C_5 \frac{1 - \frac{3}{2}x}{\sqrt{1-x}}$.Setting $f'(x) = 0 \Rightarrow x = \frac{2}{3}$.$f(\frac{2}{3}) = {}^{10}C_5 (\frac{2}{3}) \sqrt{1-\frac{2}{3}} = {}^{10}C_5 \cdot \frac{2}{3} \cdot \frac{1}{\sqrt{3}} = \frac{2 \cdot 10!}{3\sqrt{3}(5!)^2}$.Thus,the correct option is $B$.

The maximum value of the term independent of $t$ in the expansion of $\left( tx^{\frac{1}{5}} + \frac{(1-x)^{\frac{1}{10}}}{t} \right)^{10}$ where $x \in (0, 1)$ is

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