Let f , g : - 1 , 2 → R be continuous functions which are twice differentiable on the interval - 1 , 2 . Let the values of f and g at the points - 1 , 0 and 2 be as given in the following table: x = - 1 x = 0 x = 2 f ( x ) 3 6 0 g x 0 1 -1 In each of the intervals - 1 , 0 and 0 , 2 the function ( f - 3 g ) " never vanishes. Then the correct statement(s) is(are)

Step-by-step solution

h x = f x - 3 g x h - 1 = 3 - 0 = 3 h 0 = 6 - 3 = 3 h 2 = 0 + 3 = 3 h ′ ( x ) has at least one root is - 1,0 And h ′ x has at least one in ( 0,2 ) Hence h ′ ′ x atleast one root in ( - 1,2 ) but Also, f - 3 g ′ ′ ≠ 0 i n - 1 , 0 a n d ( 0 , 2 ) h ′ ′ x = 0 o n l y a t x = 0 , s o f ′ x - 3 g ′ x has exactly one solution in (-1, 0) and (0, 2)