1.

If x2+4+3sin(ax+b)−2x=0 has atleast one real solution, where a,b∈[0,2π], then the value of a+b can be

Answer»

If x2+4+3sin(ax+b)2x=0 has atleast one real solution, where a,b[0,2π], then the value of a+b can be



Discussion

No Comment Found