Limits, Continuity and Differentiability Part 3

Concept:For a function to be differentiable at a point, it must be continuous and have equal left-hand and right-hand derivatives at that point.Explanation:Since $f(x)$ is differentiable for all real $x$, it must be continuous and differentiable at $x=1$.First, apply continuity at $x=1$: $\lim_{x\to1^-} f(x) = \lim_{x\to1^+} f(x) = f(1)$.Substitute the expressions: $2\alpha(1^2-2) + 2\beta(1) = (\alpha+3)(1) + (\alpha-\beta)$.Simplify: $-2\alpha + 2\beta = \alpha+3 + \alpha-\beta$ → $-2\alpha+2\beta = 2\alpha-\beta+3$.Rearrange: $3\beta - 4\alpha = 3$.   — (1)Next, differentiate $f(x)$ piecewise:For $x<1$, $f'(x)=4\alpha x + 2\beta$. For $x>1$, $f'(x)=\alpha+3$.Apply differentiability condition: left-hand derivative equals right-hand derivative at $x=1$:$f'(1^-) = f'(1^+)$ → $4\alpha(1)+2\beta = \alpha+3$.Simplify: $4\alpha+2\beta = \alpha+3$ → $3\alpha+2\beta = 3$.   — (2)Solve equations (1) and (2):Multiply (1) by 2 and (2) by 3, then subtract:$(6\beta-8\alpha) - (9\alpha+6\beta) = 6-9$ → $-17\alpha = -3$ → $\alpha = \frac{3}{17}$.Substitute $\alpha$ into (1): $3\beta - 4\cdot\frac{3}{17} = 3$ → $3\beta = 3+\frac{12}{17} = \frac{63}{17}$ → $\beta = \frac{21}{17}$.Compute $34(\alpha+\beta) = 34\left(\frac{3}{17}+\frac{21}{17}\right) = 34\cdot\frac{24}{17} = 2\cdot24 = 48$.Answer:Option A: 48.