Question

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:ex=−1x=0x=2f(x)360g(x)01−1In each of the intervals (-1,0) and (0,2) the function (f−3g) never vanishes. Then the correct statement(s) is (are):

• A. (A) f′(x)−3g′(x)=0 has exactly three solutions in (-1,0)∪(0,2)
• B. (B) f′(x)−3g′(x)=0 has exactly one solution in (-1,0) and exactly one solutions in (0,2)
• C. (C) f′(x)−3g′(x)=0 has exactly two solutions in (-1,0) and exactly two solutions in (0,2)
• D. (D) None of the above

Answer 1

Answer: (B)

(B) f′(x)−3g′(x)=0 has exactly one solution in (-1,0) and exactly one solutions in (0,2)

Answer 2

Explanation:
f,g:[−1,2]→Rf(x) is twice differentiable on (-1,2)f(−1)=3,g(−1)=0f(0)=6,g(0)=1f(2)=0,g(2)=−1(f−3g)′′≠0 on (-1,0) and (0,2)Number of solutions of f′(x)−3g′(x)=0 in (−1,0)∪(0,2)=?Let h(x)=f(x)−3g(x). Then h(−1)=f(−1)−3g(−1)=3h(0)=f(0)−3g(0)=6−3(1)=3Therefore, by Rolle's theorem, h′(x) that is, F′(x)−3 g′(x)=0 has at least one root in (-1,0).Also h(2)=f(2)−3g(2)=0−3(−1)=3Hence, again by Rolle's theorem, f′(x)−3g′(x)=0 has at least one root in (0,2).That is, f′(x)−3g′(x)=0 has at least 2 roots in (-1,2).Since (f−3g)′′≠0 for (-1,0) and (0,2)So, f(x) has no point of inflexion in (-1,0) and (0,2) .Therefore, (f′−3g′)(x)≠0 in (-1,0) and (0,2) , that is, (f′−3g′)(x)≠0 exactly once in (-1,0) and exactly once in (0,2).Hence, the correct option is (B).