Let a,b,c,d are non − zero real numbers such that 6a+4b+3c+3d=0, then the equation ax3+bx2+cx+d=0 has

We have ax3+bx2+cx+d=0 Let f(x)=ax44+bx33+cx22+dx+e∴f(0)=ef(2)=4a+8b3+2c+2d+e=(12a+8b+6c+6d)3+e=23(6a+4b+3c+3d)+e=0⇒f(2)=e∴ By Rolle's theorem, there exist at least one value of x∈(0,2) such that f′(x)=0