Question
Let $\alpha,\beta \in \mathbb{R}$ be such that the function
$$f(x) = \left\{ \begin{matrix} 2\alpha\left( x^{2} - 2 \right) + 2\beta x & ,x < 1 \\ (\alpha + 3)x + (\alpha - \beta) & ,x \geq 1 \end{matrix} \right.\ $$be differentiable at all $x \in \mathbb{R}$. Then $34(\alpha + \beta)$ is equal to
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Solution
$f(x) = \left\{ \begin{matrix} 2\alpha x^{2} + 2\beta x - 4\alpha; & x < 1 \\ (\alpha + 3)x + \alpha - \beta; & x \geq 1 \end{matrix} \right.\ $
$$f\left( 1^{\dagger} \right) = 2\alpha - \beta + 3,f\left( 1^{-} \right) = - 2\alpha + 2\beta$$
$$\begin{array}{r} 2\alpha - \beta + 3 = 2\beta - 2\alpha \Rightarrow 4\alpha - 3\beta + 3 = 0\#(1) \end{array}$$
$$f'\left( 1^{+} \right) = 4\alpha + 2\beta,f'\left( 1^{-} \right) = \alpha + 3$$
$$\begin{array}{r} 4\alpha + 2\beta = \alpha + 3 \Rightarrow 3\alpha + 2\beta - 3 = 0\#(2) \end{array}$$
Solving (1) & (2)
We get $\alpha = \frac{3}{17},\beta = \frac{21}{17}$
$$\Rightarrow 34(\alpha + \beta) = 34 \times \frac{27}{17} = 48$$
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