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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 .

1132021436_m2.JPG

 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 $(-1,0)\cup (0,2)$

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,C

Explanation:

Let F(x)=f(x)-3g(x)

   $\therefore$    F(-1)=3, F(0)=3 and F(2)=3

 So, F'(x) will vanish atleast twice in   $(-1,0)\cup (0,2)$

$\therefore$     $F"(x)>0 $  or $<0,\forall x \epsilon(-1,0)\cup(0,2)$

 Hence, f'(x) -3g'(x)=0 has exactly  one solution in (-1,0) and one solution in (0,2)