A) Statement 1 is false, Statement 2 is true.
B) Statement 1 is true. Statement 2 is false. Statement 1 is true, Statement 2 is true,
C) Statement 2 is not a correct explanation for Statement 1.
D) Statement 1 is true, Statement 2 is true, Statement 2 is a correct explanation for Statement 1.
Correct Answer: D
Solution :
Let \[g(g)=\frac{a{{x}^{3}}}{3}+b.\frac{{{x}^{2}}}{2}+cx\] \[g'(x)=a{{x}^{2}}+bx+c\] Given: \[a{{x}^{2}}+bx+c=0\]and\[2a+3b+6c=0\] Statement-2: (i) g(0) = 0 and g (1) \[=\frac{a}{3}+\frac{b}{2}+c=\frac{2a+3b+6c}{6}\] \[=\frac{0}{6}=0\]\[\Rightarrow \]\[g(0)=g(1)\] (ii) g is continuous on [0,1] and differentiable on (0,1) \[\therefore \]By Rolle's theorem \[\exists \,k\in (0,1)\] such that) = 0 This holds the statement 2. Also, from statement-2,we can say \[\text{a}{{\text{x}}^{\text{2}}}+\text{bx}+\text{c}=0\]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.You need to login to perform this action.
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