If the fourth term in the expansion of `{sqrt(1/(""_"x^log(x+1)"}'+1/(x^12)}^6 is equal to `200 and x >1,` then find x

Options

A. `10sqrt(2)`
B. 10
C. `10^(4)`
D. `(10)/(sqrt(2))`

Answer

b

Explanation

We have , `T_(4) = ""^(6)C_(3) (sqrt(x^(((1)/(1+ log x)))))^(3) (x^(1//12))^(3) = 200 " " ` [given]
` rArr 20^((1)/(x^(2(1 + log x))))x^(1//4) = 200`
` rArr x^((3)/(2(1 + log x)) + (1)/(4))= 10`
On taking logarithm on base 10 , we get
`[(3)/(2(1 + logx))+ (1)/(4)] log x= 1`
` rArr ((6 +1 + logx )logx)/(4(1+ log x))= 1`
` rArr (log x)^(2) + 3 log x - 4 = 0 `
`rArr (logx + 4 )(log x - 1) = 0`
` rArr log x = - 4,1`
` therefore x = 10^(-4),10`
But ` x gt 1`
` therefore x = 10 `