Let f ( x ) = ⌊ x − 1 ⌋ x 2 + ( a 2 + ab + 2 ) sgn ( x − 2 ) + ( a + b ) { x } for all real x . If lim x → 2 f ( x ) exists for some real a , find the smallest positive integral value of b .

Question

Let f ( x ) = ⌊ x − 1 ⌋ x 2 + ( a 2 + ab + 2 ) sgn ( x − 2 ) + ( a + b ) { x } for all real x . If lim x → 2 ​ f ( x ) exists for some real a , find the smallest positive integral value of b .

TestHub · Accepted Answer

Equating the one-sided limits gives 2 a 2 + ( 2 b − 1 ) a + 8 − b = 0 . A real a exists iff the discriminant is nonnegative: ( 2 b − 1 ) 2 − 8 ( 8 − b ) = 4 b 2 + 4 b − 63 ≥ 0 . The smallest positive integer satisfying this is b = 4 .

Mathematics - Limits Question with Solution | TestHub

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