Question

If $$2a+3b+6c=0( a, b, c \in R)$$ then the quadratic equation $$x^2+bx+c=0$$ has

Answer 1

Correct option is A. At least one in $$[0, 1]$$
$$\textbf{Step-1: Apply the concept of polynomial to get the requied unknown}$$

$$\text{Rolle's theorem states that if a function is continuous on the closed interval [a,b] &}$$

$$\text{ differentiable on the open interval (a,b) such that f(a) = f(b) }$$

$$\text{then f'(x) = 0 for some x with}$$ $$a< x < b$$

$$\text{Let f(x) = }$$ $$\dfrac{a}{3} x^3+ \dfrac{b}{2} x^2+cx$$

$$f'(x) = ax^2 +bx+c$$

$$f(1) = \dfrac{a}{3} + \dfrac{b}{2}+c = \dfrac{1}{6} (2a+3b+6c) = 0$$

$$\text{According to Rolle's Theorem, there exists at least one root in [0,1] such that f'(x) =0}$$

$$\textbf{Hence, answer is A,B,C}$$