let f(x)={(ax+1, if x le 1),(3, if x=1)...

let `f(x)={(ax+1, if x le 1),(3, if x=1),(bx^2+1,if x > 1))` if `f(x)` is continuous at `x=1` then value of `a-b` is (A) `0` (B) `1` (C) `2` (D) `4`

A

0

B

1

C

2

D

4

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A

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