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(D) The given function $f$ is defined as $f(x) = \begin{cases} ax + 1, & 	ext{if } x \le 3 \ bx + 3, & 	ext{if } x > 3 \end{cases}$.For the functio

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$a = b + \frac{2}{3}$

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(D) The given function $f$ is defined as $f(x) = \begin{cases} ax + 1, & 	ext{if } x \le 3 \ bx + 3, & 	ext{if } x > 3 \end{cases}$.For the function $f$ to be continuous at $x = 3$,the left-hand limit,right-hand limit,and the value of the function at $x = 3$ must be equal:$\lim_{x 	o 3^-} f(x) = \lim_{x 	o 3^+} f(x) = f(3)$.$1$. Left-hand limit at $x = 3$:$\lim_{x 	o 3^-} f(x) = \lim_{x 	o 3^-} (ax + 1) = 3a + 1$.$2$. Right-hand limit at $x = 3$:$\lim_{x 	o 3^+} f(x) = \lim_{x 	o 3^+} (bx + 3) = 3b + 3$.$3$. Value of the function at $x = 3$:$f(3) = 3a + 1$.Equating the limits:$3a + 1 = 3b + 3$$3a = 3b + 2$$a = b + \frac{2}{3}$.Thus,the required relationship is $a = b + \frac{2}{3}$.

Find the relationship between $a$ and $b$ so that the function $f$ defined by $f(x) = \begin{cases} ax + 1, & \text{if } x \le 3 \\ bx + 3, & \text{if } x > 3 \end{cases}$ is continuous at $x = 3$.

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