Set, Relation and FunctionEasy
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
Let R be a relation defined on N × N by (a, b) R(c, d) ⇔ a(b + c) = c(a + d). Then R is...Two sets A and B are as under A = {(a, b) ∈R × R : |a - 5| < 1 and |b - 5| < 1}; B = {(a, b)∈R &#...Let A = {x1, x2,......, x7} and B = {y1, y2, y3} be two sets containing seven and three distinct elements respectively.T...If the two roots of the equation, (a − 1)(x4 + x2 + 1) + (a + 1)(x2 + x + 1)2 = 0 are real and distinct, then the ...If A = {x | x/2 ∈ Z, 0 ≤ x ≤ 10}, B = {x | x is one digit prime} C = {x | x/3 ∈ N, x ≤ 12}...