The sixth term in the expansion of `( sqrt(2^(log(10-3^x))) + (2^((x-2)log3))^(1/5))^m` is equal to 21, if it is known that the binomial coefficient of the 2nd 3rd and 4th terms in the expansion represent, respectively, the first, third and fifth terms of an A.P. (the symbol log stands for logarithm to the base 10) The value of m 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.