Given that the equation `ax+4=3x-b` has more than 1 solution for `x`. Find the value of `(4a+3b)^(2021)`

0 le a le 3, 0 le b le 3 and the equation, x^(2)+4+3 cos ( ax+b) = 2x has atleast one solution, then find the value of (a+b) .

If a, b ∈ [0,2pi] and the equation x^(2) + 4+3sin(ax+b)=2x has atleast one solution, and the least (+ve) . Value of a+b is k then,

x^(2)+4+3cos(ax+b)=2x has atleast on solution then the value of a+b is :

If the equation x^(4)-4x^(3)+ax^(2)+bx+1=0 has four positive roots, fond the values of a and b.

If x=-1,\ y=2 is a solution of the equation 3x+4y=k , find the value of k

If the equation x^(4)-4x^(3)+ax^(2)+bx+1=0 has four positive roots,then the value of (a+ b) is :

If 3sin x+4cos ax=7 has at least one solution,then find the possible values of a.