Let `f,g:[-1,2]vecR` be continuous functions which are twice differentiable on the interval `(-1,2)dot` Let the values of `fa n dg` at the points `-1,0a n d2` 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)a n d(0,2)` the function `(f-3g)' '` never vanishes. Then the correct statement(s) is (are) `f^(prime)(x)-3g^(prime)(x)=0` has exactly three solutions in `(-1,0)uu(0,2)dot` `f^(prime)(x)-3g^(prime)(x)=0` has exactly one solutions in `(-1,0)dot` `f^(prime)(x)-3g^(prime)(x)=0` has exactly one solutions in `(-1,2)dot` `f^(prime)(x)-3g^(prime)(x)=0` has exactly two solutions in `(-1,0)` and exactly two solutions in `(0,2)dot`

Options

A. f(x) -3g(x)=0 has exactly three solutions in `(-1,0)uu (0,2)`
B. f(x)-3g(x)=0 has exactly one solution in `(-1,0)`
C. f(x) -3g(x) =0 has exactly one solution in (0.2)
D. f(x) -3g (x)=0 has exactly two solutions in (-1,0) and exactly two solutions in (0.2)

Answer

B, C