Correct Answer - a
We have,
`T_(3) = 1000`
`rArr ""^(5)C_(2) ((1)/(x))^(5-2) (x^(log 10x))^(2) = 1000`
`rArr x^(2) log_(10)^(x-3)= 100`
`rArr 2 log_(10) x-3 = log_(x) 10^(2)`
`rArr 2y - 3 = (2)/(y)` , where y `log_(10) x`
`rArr 2y^(2) - 3y - 2 = 0`
`rArr (2y+1)(y-2)= 0`
`rArr y = 2 or y = 11//2`
Now , ` y = 2 rArr log_(10) x = 2 rArr x = 10^(2) = 100`
and , `y = - 1//2 rArr log_(10) x = - 1//2 rArr x = 10 ^(-1//2)`
But , x `gt` 11, So, x = 100