The sixth term in the expansion of
`[sqrt({2^(log(10-3^(x)))})+5sqrt({2^((x-2)log3)})]^(m)` is equal to 21, if it is know that the binomial coefficient of the `2^(nd), 3^(nd)` and `4^(th)` terms in the expansion represents, respectively, the first, third and fifth terms of an A.P. (the symbol log stand for logarithm tothe base 10).
The sum of possible values of x is

Options

A. 6
B. 7
C. 8
D. 9

Answer

B

Explanation

The coefficient of the `2^(nd), 3^(rd)` and `4^(th)` terms in expansion are `.^(m)C_(1), .^(m)C_(2)` and `.^(m)C_(3)` which are given in A.P. Hence,
`2 .^(m)C_(2) = .^(m)C_(1) xx .^(m)C_(3)`
or `(2m(m-1))/(2!) = m+(m(m+1)(m-2))/(3!)`
or `m(m^(2)-9m+14)=0`
or `m(m-2)(m-7)=0`
or `m = 7` (`:'m ne 0` or `2` as `6^(th)` term is given equal to 21)
Now, `6^(th)` term in the expansion, when `m = 7`,is
`.^(7)C_(5)p[sqrt({2^(log(10-3^(n))}}]]^(7-5) xx [5sqrt({2^((x-2)log3)})]^(5)=21`
`rArr (7xx6)/(2!) 2^(log(10-3^(x)))xx2^((x-2)log3)= 21`
or `2^(log(10-3^(x))+(x-2)log3)=1=2^(0)`
or `log(10-3^(x))+(x-2)log3=0`
or `log(10-3^(x))(3)^((x-2)) = 0`
or `(10-3^(x)) xx 3^(x) xx 3^(-2) = 1`
or `10 xx 3^(x) - (3^(x))^(2) = 9`
or `(3^(x))^(2) - 10 xx 3^(x) + 9 = 0`
or `3^(x) = 1,9`
`rArr x = 0, 2`
When `x = 2`,
`[sqrt({2^(log(10-3^(x)))})+5sqrt({2^((x-2)log3)})]^(m) = [1+1]^(7) = 128`
When `x = 0`,
`[sqrt({2^(log(10-3^(x)))})+5sqrt({2^((x-2)log3)}]]^(m)`
`= [sqrt({2^(log9)})+5sqrt({2^(-2log3)})]^(7)`
`= [2^((log9)/(2))+(1)/(2^((log9)/(5)))]^(7) gt 2^(7)`
Hence, the minimum value is 128.