Set, Relation and FunctionHard
Question
Let f: (-1, 1) → R be a differentiable function with f(0) = - 1 and f′(0) = 1. Let g(x) = [f(2f(x) + 2)]2. Then g′(0) =
Options
A.-4
B.0
C.-2
D.4
Solution
g′(x) = 2(f(2f(x) + 2)) 
= 2f(2f(x) + 2) f′(2f(x) + 2) . (2f′(x))
⇒ g′(0) = 2f(2f(0) + 2) . f′(2f(0) + 2) . 2(f′(0) = 4f(0) f′(0)
= 4(-1) (1) = - 4
= 2f(2f(x) + 2) f′(2f(x) + 2) . (2f′(x))⇒ g′(0) = 2f(2f(0) + 2) . f′(2f(0) + 2) . 2(f′(0) = 4f(0) f′(0)
= 4(-1) (1) = - 4
Create a free account to view solution
View Solution FreeMore Set, Relation and Function Questions
Consider two sets A = {1,2,3} & B = {1,2,3,4,5}. A number is selected each from set A & set B. Let p denotes the probabi...A function f from the set of natural numbers to integers defined by f(n) = is...The set of points where f(x)= is differentiable is...Let A and B be two sets containing four and two elements respectively. Then the number of subsets of the set A × B,...Let f be a twice differentiable function such that f″(x) = - f(x) and f′(x) = g(x). If h(x) = (f(x))2 + (g(x...