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(B) For $f(x)$ to be differentiable at $x = 1$, it must be continuous at $x = 1$ and $f'(1^-) = f'(1^+)$.First, for continuity at $x = 1$, we have $f(

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$48$

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(B) For $f(x)$ to be differentiable at $x = 1$, it must be continuous at $x = 1$ and $f'(1^-) = f'(1^+)$.First, for continuity at $x = 1$, we have $f(1^-) = f(1^+)$.$f(1^-) = \lim_{x 	o 1^-} (2 \alpha (x^2 - 2) + 2 \beta x) = 2 \alpha (1 - 2) + 2 \beta (1) = -2 \alpha + 2 \beta$.$f(1^+) = (\alpha + 3)(1) + (\alpha - \beta) = 2 \alpha - \beta + 3$.Equating them: $-2 \alpha + 2 \beta = 2 \alpha - \beta + 3 \Rightarrow 4 \alpha - 3 \beta = -3$ $\quad (1)$.Next, for differentiability, $f'(1^-) = f'(1^+)$.$f'(x) = 4 \alpha x + 2 \beta$ for $x  1$.$f'(1^-) = 4 \alpha + 2 \beta$.$f'(1^+) = \alpha + 3$.Equating them: $4 \alpha + 2 \beta = \alpha + 3 \Rightarrow 3 \alpha + 2 \beta = 3$ $\quad (2)$.Solving equations $(1)$ and $(2)$:From $(2)$, $2 \beta = 3 - 3 \alpha \Rightarrow \beta = \frac{3 - 3 \alpha}{2}$.Substitute into $(1)$: $4 \alpha - 3(\frac{3 - 3 \alpha}{2}) = -3 \Rightarrow 8 \alpha - 9 + 9 \alpha = -6 \Rightarrow 17 \alpha = 3 \Rightarrow \alpha = \frac{3}{17}$.Then $\beta = \frac{3 - 3(3/17)}{2} = \frac{3 - 9/17}{2} = \frac{42/17}{2} = \frac{21}{17}$.Thus, $34(\alpha + \beta) = 34(\frac{3}{17} + \frac{21}{17}) = 34(\frac{24}{17}) = 2 	imes 24 = 48$.

Let $\alpha, \beta \in R$ be such that the function $f(x) = \begin{cases} 2 \alpha (x^2 - 2) + 2 \beta x, & x < 1 \\ (\alpha + 3) x + (\alpha - \beta), & x \ge 1 \end{cases}$ is differentiable at all $x \in R$. Then $34(\alpha + \beta)$ is equal to

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