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  • The maximum value of the term independent of 't' in the expansion of <img alt="\left ( tx^{\frac{1}{5}}+\frac{(1-x)^{\frac{1}{10}}}{t} \right )^{10}" src="https://entrancecorner.codecogs.com/gif.latex?%5Cleft%20%28%20tx%5E%7B%5Cfrac%7B1%7D%7B5%

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\; \epsilon \; \left ( 0,1 \right ) is :
 
Option: 1 \frac{2.10!}{3\sqrt{3}(5!)^{2}}
Option: 2 \frac{10!}{3(5!)^{2}}
Option: 3 \frac{10!}{\sqrt{3}(5!)^{2}}  
Option: 4 \frac{2.10!}{3(5!)^{2}}

Answers (1)

best_answer

The term independent of t will be the middle term due to exact same magnitude but opposite sign powers of t in the binomial expression given

\\\text { so } \mathrm{T}_{6}={ }^{10} \mathrm{C}_{5}\left(\mathrm{tx}^{2} 5\right)^{5}\left(\frac{(1-\mathrm{x})^{\frac{1}{10}}}{\mathrm{t}}\right)^{5}\\T_6=f(x)={ }^{10} \mathrm{C}_{5}(\mathrm{x} \sqrt{1-\mathrm{x}})\\f'(x)=^{10}C_5\left ( \sqrt{1-x}-\frac{x}{2\sqrt{1-x}} \right )=^{10}C_5\left ( -\frac{3 x-2}{2 \sqrt{1-x}} \right )

for maximum ƒ'(x) = 0

\Rightarrow x=\frac{2}{3}

\\f''(x)=\frac{2-3 x}{4(1-x)^{\frac{3}{2}}}-\frac{3}{2 \sqrt{1-x}}=\frac{3 x-4}{4(1-x)^{\frac{3}{2}}}\\f''\left (\frac{2}{3} \right )<0

\text { so } f(x)_{\max .}={ }^{10} C_{5}\left(\frac{2}{3}\right) \cdot \frac{1}{\sqrt{3}}

 

Posted by

himanshu.meshram

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