If `(x^(2) + 5)/(2) = x - 2` cos (m + nx)` has at least one real root, the

Options

A. number of possible values of x is two
B. number of possible values of x is one
C. the value of ` m + n is (2n + 1 )pi`
D. the value of m +n is ` 2npi`

Answer

2,3,

Explanation

Given equation is
` x^(2) + 5 - 2x = -4 cos (m + mx)`
`rArr (x - 1)^(2) + 4 = - 4 cos (m + nx)`
LHS ` ge ` 4 and RHS ` le ` 4
So, solution is possible only if LHS =RHS = 4
` therefore x = 1 ` and cos (m + n) = - 1
` therefore m+n = (2n + 1) pi, n in 1 `