Continuity and DifferentiabilityHard
Question
Given f(x) = b ([x]2 + [x]) + 1 for x ≥ -1
= sin (π(x + a)) for x < -1
where [x] denotes the integral part of x, then for what values of a, b the function is con tenuous at x = -1 ?
= sin (π(x + a)) for x < -1
where [x] denotes the integral part of x, then for what values of a, b the function is con tenuous at x = -1 ?
Options
A.a = 2n + (3 /2) ; b ∈ R ; n ∈ I
B.a = 4n + 2 ; b ∈ R ; n ∈ I
C.a = 4n + (3/2) ; b ∈ R+ ; n ∈ I
D.a = 4n + 1 ; b ∈ R+ ; n ∈ I
Solution
f(x) = 
b([x]2 + [x]) + 1=
b([- 1 + h]2 + [-1 + h]) + 1⇒ b ∈ R
f(x) = 
sin (π (x + a))=
sin (π(- 1 - h + a)) = - sin pasin pa = - 1
πa = 2nπ +
⇒ a = 2n + 
Also option (C) is subset of option (A)
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