+cx g′(x)=ax2+bx+c Given: ax2+bx+c=0 and 2a+3b+6c=0 Statement-2: (i) g(0)=0 and g(1)=
a
3
+
b
2
+c=
2a+3b+6c
6
=
0
6
=0 ⇒g(0)=g(1) (ii) g is continuous on [0,1] and differentiable on (0,1) ∴ By Rolle's theorem ∃k∈(0,1) such that g′(k)=0 This holds the statement 2 . Also, from statement-2,we can say ax2+bx+c=0 has at least one root in (0,1). Thus statement- 1 and 2 both are true and statement- 2 is a correct explanation for statement-1.