Let f(x)=bx+c for x>13cx−2b+1 for x<1 then a relation between b and c so that Ltx→1⁡f(x) exists is

Ltx→1fxexists⇒Ltx→1+fx=Ltx→1-fx⇒Ltx→1+bx+c=Ltx→1-3cx−2b+1⇒b+c=3c−2b+1⇒3b−2c−1=0

Let f(x)=bx+c for x>13cx−2b+1 for x<1 then a relation between b and c so that Ltx→1⁡f(x) exists is

Ltx→1fxexists⇒Ltx→1+fx=Ltx→1-fx⇒Ltx→1+bx+c=Ltx→1-3cx−2b+1⇒b+c=3c−2b+1⇒3b−2c−1=0

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$Let f (x) = \{\begin{array}{l} b x + c for x > 1 \\ 3 c x - 2 b + 1 for x < 1 \end{array} then a relation between b and c so that$ ${Lt}_{x \to 1} f (x) exists is$

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$3 b + 2 c - 1 = 0$

$3 b + 2 c + 1 = 0$

$3 b - 2 c - 1 = 0$

$3 b - 2 c + 1 = 0$

answer is C.

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Detailed Solution

$\underset{x \to 1}{L t} f (x) e x i s t s \Rightarrow \underset{x \to 1^{+}}{L t} f (x) = \underset{x \to 1^{-}}{L t} f (x) \Rightarrow \underset{x \to 1^{+}}{L t} \{b x + c\} = \underset{x \to 1^{-}}{L t} \{3 c x - 2 b + 1\}$

$\Rightarrow b + c = 3 c - 2 b + 1 \Rightarrow 3 b - 2 c - 1 = 0$

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Let f(x)=bx+c for x>13cx−2b+1 for x<1 then a relation between b and c so that Ltx→1⁡f(x) exists is

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$Let f (x) = \{\begin{array}{l} b x + c for x > 1 \\ 3 c x - 2 b + 1 for x < 1 \end{array} then a relation between b and c so that$ ${Lt}_{x \to 1} f (x) exists is$