fx=bx+c for x>13cx−2b+1 for x<1 then relation between b and c so that limx→1 fx exist

limx→1− f(x)=limx→1+ f(x)limx→1− 3cx−2b+1=limx→1+ bx+c3c-2b+1=b+c3b-2c-1=0

fx=bx+c for x>13cx−2b+1 for x<1 then relation between b and c so that limx→1 fx exist

limx→1− f(x)=limx→1+ f(x)limx→1− 3cx−2b+1=limx→1+ bx+c3c-2b+1=b+c3b-2c-1=0

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$f (x) = \{\begin{cases} bx + c for x > 1 \\ 3 cx - 2 b + 1 for x < 1 \end{cases}$ then relation between b and c so that $lim_{x \to 1} f (x)$ exist

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3b-2c+1=0

3b+2c-1=0

3b+2c+1=0

3b-2c-1=0

answer is C.

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Detailed Solution

$lim_{x \to 1^{-}} f (x) = lim_{x \to 1^{+}} f (x)$
$lim_{x \to 1^{-}} 3 cx - 2 b + 1 = lim_{x \to 1^{+}} bx + c$

3c-2b+1=b+c

3b-2c-1=0

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fx=bx+c for x>13cx−2b+1 for x<1 then relation between b and c so that limx→1 fx exist

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$f (x) = \{\begin{cases} bx + c for x > 1 \\ 3 cx - 2 b + 1 for x < 1 \end{cases}$ then relation between b and c so that $lim_{x \to 1} f (x)$ exist