$^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{3}^{-}}\text{f(x)} = ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{3}^{-}}(\text{ax + 1}) = \text{3a + 1}$
$^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{3}^{+}}\text{f(x)} = ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{3}^{+}}(\text{bx + 3}) = \text{3b + 3}$
Also f(3) = 3a + 1 Since f is continuous at x = 3 $\therefore \ ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{3}^{-}}\text{f(x)} = ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{3}^{+}}\text{f(x }) = \text{f(3)}$ $\therefore$ 3a + 1 = 3b + 3 = 3a + 1 $\therefore$ 3a = 3b + 2 $\Rightarrow\ \text{a} = \text{b}+\frac{2}{3}$, which is required relation.Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.