LimitsHard
Question
f(x) = (sin-1x)2. cos (1/x) if x ≠ 0; f(0) = 0, f(x) is :
Options
A.continuous no where in -1 ≤ x ≤ 1
B.continuous everywhere in -1 ≤ x ≤ 1
C.differentiable no where in -1 ≤ x ≤ 1
D.differentiable everywhere in -1 < x < 1
Solution
y = f{x} = 
f(x) can be discontinuous only at x = 0 in [-1, 1]
So we check only at x = 0
LHD (x = 0) =
= -1. 0. [finite quantity between [-1,1]] = 0
RHD (x = 0) is
Hence f(x) is differentiable as well as continuous in [-1,1]
f(x) can be discontinuous only at x = 0 in [-1, 1]
So we check only at x = 0
LHD (x = 0) =
= -1. 0. [finite quantity between [-1,1]] = 0RHD (x = 0) is
Hence f(x) is differentiable as well as continuous in [-1,1]
Create a free account to view solution
View Solution FreeMore Limits Questions
For a ∈ R (the set of all real numbers), a ≠ 1. Then a =...If f(x) = [x2] + , where [.] and {.} denote the greatest integer and fractional part functions respectively, then-...If[f(x) + g(x)] = 2 and[f(x) - g(x)] = 1, thenf(x)g(x) -...The value of is-...Let [x] denote the integral part of x ∈ R and g(x) = x - [x]. Let f(x) be any continuous function with f(0) = f(1)...