\(\frac {(169π)}{4} - \frac {65}{2} + \frac {169}{2} sin^{-1}(\frac {12}{13})\)
\(\frac {(169π)}{4} + \frac {65}{2} - \frac {169}{2} sin^{-1}(\frac {12}{13})\)
\(\frac {(169π)}{4} - \frac {65}{2} + \frac {169}{2} sin^{-1}(\frac {13}{14})\)
\(\frac {(169π)}{4} + \frac {65}{2} + \frac {169}{2} sin^{-1}(\frac {13}{14})\)
The correct option is (A): \(\frac {(169π)}{4} - \frac {65}{2} + \frac {169}{2} sin^{-1}(\frac {12}{13})\).
If the area of the region \[ \{(x, y) : |4 - x^2| \leq y \leq x^2, y \leq 4, x \geq 0\} \] is \( \frac{80\sqrt{2}}{\alpha - \beta} \), where \( \alpha, \beta \in \mathbb{N} \), then \( \alpha + \beta \) is equal to:
Let the area of the region \( \{(x, y) : 2y \leq x^2 + 3, \, y + |x| \leq 3, \, y \geq |x - 1|\} \) be \( A \). Then \( 6A \) is equal to:
Let $\alpha,\beta\in\mathbb{R}$ be such that the function \[ f(x)= \begin{cases} 2\alpha(x^2-2)+2\beta x, & x<1 \\ (\alpha+3)x+(\alpha-\beta), & x\ge1 \end{cases} \] is differentiable at all $x\in\mathbb{R}$. Then $34(\alpha+\beta)$ is equal to}

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