If f(x) = begin{cases} 3ax - 2b, & x 1 ax | vedclass

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(B) Given,$f(x) = \begin{cases} 3ax - 2b, & x > 1 \ ax + b + 1, & x < 1 \end{cases}$ Since $\lim_{x ightarrow 1} f(x)$ exists,the left-hand limit m

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$2a - 3b = 1$

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(B) Given,$f(x) = \begin{cases} 3ax - 2b, & x > 1 \ ax + b + 1, & x Since $\lim_{x ightarrow 1} f(x)$ exists,the left-hand limit must equal the right-hand limit: $\lim_{x ightarrow 1^{-}} f(x) = \lim_{x ightarrow 1^{+}} f(x)$ Substituting the given functions: $\lim_{x ightarrow 1^{-}} (ax + b + 1) = \lim_{x ightarrow 1^{+}} (3ax - 2b)$ Evaluating the limits at $x = 1$: $a(1) + b + 1 = 3a(1) - 2b$ $a + b + 1 = 3a - 2b$ Rearranging the terms to find the relation: $1 = 3a - a - 2b - b$ $1 = 2a - 3b$ Thus,the relation is $2a - 3b = 1$.

If $f(x) = \begin{cases} 3ax - 2b, & x > 1 \\ ax + b + 1, & x < 1 \end{cases}$ and $\lim_{x \rightarrow 1} f(x)$ exists,then the relation between $a$ and $b$ is

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