Let f.g:[-1,2]\to\mathbb{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-3g)" never vanishes.Then the correct statement(s) is (are) [JEE Adv 2015 P2]

f'(x)-3g'(x)=0 has exactly three solutions in (-1,0)\cup(0,2)

f'(x)-3g'(x)=0 has exactly one solutions in (-1,0)

f'(x)-3g'(x)=0 has exactly one solutions in (0,2)

f'(x)-3g'(x)=0 has exactly two solutions in (-1,0) and exactly two solutions in 0,2

Application of Derivatives

Section: Mathematics

Question : 15 of 32

Marks: +1, -0

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-3g)" never vanishes.Then the correct statement(s) is (are)

[JEE Adv 2015 P2]

f′(x)−3g′(x)=0 has exactly three solutions in (−1,0)∪(0,2)
f′(x)−3g′(x)=0 has exactly one solutions in (−1,0)
f′(x)−3g′(x)=0 has exactly one solutions in (0,2)
f′(x)−3g′(x)=0 has exactly two solutions in (−1,0) and exactly two solutions in 0,2

Validate

Go to Question:

Prev Question

Next Question