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 ?

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 pa
sin pa = - 1
πa = 2nπ + ⇒ a = 2n +
Also option (C) is subset of option (A)

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