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  • Help me please, - If the third term in the bionomial expansion of equals 2560, then a possible value of x is: - Binomial theorem and its simple applications - JEE Main

If the third term in the bionomial expansion of (1+x^{log_{2} x})^{5} equals 2560, then a possible value of x is: 

  • Option 1)

     

    1/4

  • Option 2)

     

    4\sqrt2

  • Option 3)

     

    1/8

  • Option 4)

     

    2\sqrt2

Answers (1)

best_answer

 

General Term in the expansion of (x+a)^n -

T_{r+1}= ^{n}c_{r}\cdot x^{n-r}\cdot a^{r}
 

- wherein

Where r\geqslant 0 \, and \, r\leqslant n

r= 0,1,2,----n

 

Binomial \left ( 1 + x^{log_{2}x} \right )^{5}

from the concept , general term

T_{3} = ^{5}C_{2} \left ( x^{log_{2}x} \right )^{2} = 2560

\Rightarrow 10. \left ( x^{log_{2}x} \right )^{2} = 2560

\Rightarrow \left ( x^{log_{2}x} \right )^{2} = \frac{2560}{10} = 256

take log both side.

2\left ( log_{2}x \right )^{2} = log_{2}256 = log_{2}2^{8} = 8

\left ( log_{2}x \right )^{2} = 4

log_{2}x = 2 \ or -2

x = 4 \ or \ \frac{1}{4}


Option 1)

 

1/4

Option 2)

 

4\sqrt2

Option 3)

 

1/8

Option 4)

 

2\sqrt2

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