Let f ,   g   : - 1 ,   2 →   R be continuous functions which are twice…

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
gx 0 1 -1

In each of the intervals -1, 0 and 0, 2 the function (f-3g)" never vanishes. Then the correct statement(s) is(are)
  1. f′x-3g′x=0 has exactly three solutions in -1, 0 ∪(0, 2)
  2. f′x-3g′x=0 has exactly one solution in (-1, 0)
  3. f′x-3g′x=0 has exactly one solution in (0, 2)
  4. f′x-3g′x=0 has exactly two solutions in (-1, 0) and exactly two solutions in (0, 2)

Solution

hx=fx-3gx
h-1=3-0=3
h0=6-3=3
h2=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-3g′′≠0 in -1, 0 and (0, 2)
h′′x=0 only at x=0, so
f′x-3g′x has exactly one solution in (-1, 0) and (0, 2)

Asked in: JEE Advanced 2015 (Paper 2)

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