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(A) The general term $T_{r+1}$ in the expansion of $\left( t^{2} x^{\frac{1}{5}} + \frac{(1-x)^{\frac{1}{10}}}{t} \right)^{15}$ is given by:$T_{r+1} =

Vedclass · Accepted Answer

$6006$

Vedclass · Answer

(A) The general term $T_{r+1}$ in the expansion of $\left( t^{2} x^{\frac{1}{5}} + \frac{(1-x)^{\frac{1}{10}}}{t} ight)^{15}$ is given by:$T_{r+1} = {}^{15}C_{r} (t^{2} x^{\frac{1}{5}})^{15-r} \cdot \left( \frac{(1-x)^{\frac{1}{10}}}{t} ight)^{r}$$T_{r+1} = {}^{15}C_{r} t^{30-2r} x^{\frac{15-r}{5}} \cdot t^{-r} (1-x)^{\frac{r}{10}}$$T_{r+1} = {}^{15}C_{r} t^{30-3r} x^{\frac{15-r}{5}} (1-x)^{\frac{r}{10}}$For the term to be independent of $t$,the exponent of $t$ must be zero:$30 - 3r = 0 \implies r = 10$Substituting $r = 10$ into the expression,the term independent of $t$ is:$T_{11} = {}^{15}C_{10} x^{\frac{15-10}{5}} (1-x)^{\frac{10}{10}} = {}^{15}C_{10} x(1-x)$We know ${}^{15}C_{10} = {}^{15}C_{5} = \frac{15 	imes 14 	imes 13 	imes 12 	imes 11}{5 	imes 4 	imes 3 	imes 2 	imes 1} = 3003$Let $f(x) = x(1-x) = x - x^{2}$. To find the maximum value,we set $f'(x) = 1 - 2x = 0$,which gives $x = \frac{1}{2}$.The maximum value $K = 3003 	imes \left( \frac{1}{2} ight) \left( 1 - \frac{1}{2} ight) = 3003 	imes \frac{1}{4} = 750.75$However,the question asks for $8K$:$8K = 8 	imes 750.75 = 6006$

If the maximum value of the term independent of $t$ in the expansion of $\left( t^{2} x^{\frac{1}{5}} + \frac{(1-x)^{\frac{1}{10}}}{t} \right)^{15}$,$x \geq 0$,is $K$,then $8K$ is equal to $....$

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