MathematicsMedium129×since 2002Q5045The number of real roots of the equation e6x−e4x−2e3x−12e2x+ex+1=0{e^{6x}} - {e^{4x}} - 2{e^{3x}} - 12{e^{2x}} + {e^x} + 1 = 0e6x−e4x−2e3x−12e2x+ex+1=0 is :A2B4C6D1Check answerSkip