The function ⨍ x=ax+1+bx-1, a≠0, b≠0 where x is the greatest integer function is continuous at x=1 if

fx=ax+1+b[x-1]fx=ax+1+bx-1]=a+bx+(a-b) is continuous at x = 0⇒limx→1-fx=limx→1+fx⇒limx→1-a+bx+a-b=limx→1+a+bx+a-b⇒a+b0+a-b=a+b1+a-b ⇒a+b=0Option (2) is correct

The function ⨍ x=ax+1+bx-1, a≠0, b≠0 where x is the greatest integer function is continuous at x=1 if

fx=ax+1+b[x-1]fx=ax+1+bx-1]=a+bx+(a-b) is continuous at x = 0⇒limx→1-fx=limx→1+fx⇒limx→1-a+bx+a-b=limx→1+a+bx+a-b⇒a+b0+a-b=a+b1+a-b ⇒a+b=0Option (2) is correct

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The function $⨍ (x) =a [x + 1] +b [x - 1], (a \neq 0, b \neq 0)$ where $[x]$ is the greatest integer function is continuous at $x = 1$ if

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$a = 2$

$a = b$

$a + 2 b = 0$

$a + b = 0$

answer is B.

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Detailed Solution

$f (x) = a [x + 1] + b [x - 1]$

$f (x) = a [[x] + 1] + b [x] - 1]$

$= (a + b) [x] + (a - b)$ is continuous at x = 0

$\Rightarrow \lim_{x \to 1^{-}} f (x) = \lim_{x \to 1^{+}} f (x) \Rightarrow \lim_{x \to 1^{-}} (a + b) [x] + a - b = \lim_{x \to 1^{+}} (a + b) [x] + (a - b)$ $\Rightarrow (a + b) 0 + (a - b) = (a + b) 1 + (a - b) \Rightarrow a + b = 0$