Continuity and DifferentiabilityHard
Question
Let [x] denote the greatest integer less than or equal to x. If f(x) = [x sin πx], then f(x) is
Options
A.continuous at x =0
B.continuous in (-1,0)
C.differentiable at x = 1
D.differentiable in (-1,1)
Solution
We have, for -1< x <1
⇒ 0 ≤ x sin π x ≤ 1/2
∴ [x sin π x] = 0
Alos, x sin π x becomes negative and numberically less than 1 when x is slightly greater than 1 and so by definition of [x] f(x) = [x sin π x] = -1, when 1< x <1+ h
Thus., f(x) is constant and equal to 0 in the closed interval [-1,1] and so f(x) is continu ous and differentiable in the open Interval (-1,1)
At x =1, f(x) discontinuous, since
(1 - h) = 0
and (1 + h) = - 1
∴ f (x) is not differentiable at x = 1
Hence, (a) (b) and (d) are correct answers.
⇒ 0 ≤ x sin π x ≤ 1/2
∴ [x sin π x] = 0
Alos, x sin π x becomes negative and numberically less than 1 when x is slightly greater than 1 and so by definition of [x] f(x) = [x sin π x] = -1, when 1< x <1+ h
Thus., f(x) is constant and equal to 0 in the closed interval [-1,1] and so f(x) is continu ous and differentiable in the open Interval (-1,1)
At x =1, f(x) discontinuous, since
(1 - h) = 0and
(1 + h) = - 1 ∴ f (x) is not differentiable at x = 1
Hence, (a) (b) and (d) are correct answers.
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