If f(x) = begin{cases} ax^2 + bx + 1 & text{i | vedclass

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(A) The function is defined as: $f(x) = \begin{cases} ax^2 + bx + 1 & x \leq \frac{1}{2} \ 3x + 2 & \frac{1}{2} < x < \frac{5}{2} \ ax^2 + bx + 1 &

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$\frac{23}{5}$

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(A) The function is defined as: $f(x) = \begin{cases} ax^2 + bx + 1 & x \leq \frac{1}{2} \ 3x + 2 & \frac{1}{2} For continuity at $x = \frac{1}{2}$: $a(\frac{1}{2})^2 + b(\frac{1}{2}) + 1 = 3(\frac{1}{2}) + 2$ $\frac{a}{4} + \frac{b}{2} + 1 = \frac{3}{2} + 2 = \frac{7}{2}$ $\frac{a}{4} + \frac{b}{2} = \frac{5}{2} \implies a + 2b = 10$ ... $(1)$ For continuity at $x = \frac{5}{2}$: $a(\frac{5}{2})^2 + b(\frac{5}{2}) + 1 = 3(\frac{5}{2}) + 2$ $\frac{25a}{4} + \frac{5b}{2} + 1 = \frac{15}{2} + 2 = \frac{19}{2}$ $\frac{25a}{4} + \frac{5b}{2} = \frac{17}{2} \implies 25a + 10b = 34$ ... $(2)$ Multiplying $(1)$ by $5$: $5a + 10b = 50$ ... $(3)$ Subtracting $(3)$ from $(2)$: $20a = -16 \implies a = -\frac{16}{20} = -\frac{4}{5}$ Substituting $a = -\frac{4}{5}$ in $(1)$: $-\frac{4}{5} + 2b = 10 \implies 2b = 10 + \frac{4}{5} = \frac{54}{5} \implies b = \frac{27}{5}$ Thus,$a + b = -\frac{4}{5} + \frac{27}{5} = \frac{23}{5}$.

If $f(x) = \begin{cases} ax^2 + bx + 1 & \text{if } |2x - 3| \geq 2 \\ 3x + 2 & \text{if } \frac{1}{2} < x < \frac{5}{2} \end{cases}$ is continuous on its domain,then $a + b$ has the value:

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