1) 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 . In each of the intervals (-1,0) and (0,2), the function (f-3g)" never vanishes. Then the correct statement(s) is/are A) f'(x)-3g'(x)=0 has exactly three solutions in (−1,0)∪(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 (0,2) D) f'(x)-3g'(x)=0 has exactly two solution in (-1,0) and exactly two solutions in (0,2) Answer: Option B,CExplanation:Let F(x)=f(x)-3g(x) ∴ F(-1)=3, F(0)=3 and F(2)=3 So, F'(x) will vanish atleast twice in (−1,0)∪(0,2) ∴ F"(x)>0 or <0,∀xϵ(−1,0)∪(0,2) Hence, f'(x) -3g'(x)=0 has exactly one solution in (-1,0) and one solution in (0,2)