Set, Relation and FunctionHard
Question
Given function f : R → R is defined by f(x) = 
.
Which of the following holds good ?
.Which of the following holds good ?
Options
A.f(x) is continuous and differentiable at x = 1
B.f(x) is continuous but not differentiable at x = 1
C.f′(x) is continuous everywhere
D.Range of f(x) is (0, ∞)
Solution
f′(1+) = -1, f′(1-) = - 1
∴ f(x) is differentiable at x = 1 & hence continuous.
f′(x) =
Hence, f′(x) is continuous everywhere.
Also Rf∈ (0, ∞)
∴ f(x) is differentiable at x = 1 & hence continuous.
f′(x) =
Hence, f′(x) is continuous everywhere.
Also Rf∈ (0, ∞)
Create a free account to view solution
View Solution FreeMore Set, Relation and Function Questions
The Range of the fucntion y = is : -...Events A, B, C are mutually exclusive events such that P (A) = , P(B) = and P(C) = . The set of possible values of x are...In the set A = {1, 2, 3, 4, 5}, a relation R is defined by R = {(x, y) | x, y ∈ A and x < y}. Then R is -...A = {a, e, i, o, u}and B = {i, o} then the true statement is -...Let R : A→ A, A is set of all children in a family such that x R y ⇒ x is brother of ′y′ (where ...