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(B) Given the equation $\int_{0}^{\alpha} \cos(x + \alpha^2) \, dx = \sin \alpha$. Evaluating the integral: $[\sin(x + \alpha^2)]_{0}^{\alpha} = \sin

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$2$

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(B) Given the equation $\int_{0}^{\alpha} \cos(x + \alpha^2) \, dx = \sin \alpha$. Evaluating the integral: $[\sin(x + \alpha^2)]_{0}^{\alpha} = \sin \alpha$. This simplifies to $\sin(\alpha + \alpha^2) - \sin(\alpha^2) = \sin \alpha$. Using the formula $\sin C - \sin D = 2 \cos(\frac{C+D}{2}) \sin(\frac{C-D}{2})$,we get: $2 \cos(\frac{\alpha^2 + \alpha + \alpha^2}{2}) \sin(\frac{\alpha + \alpha^2 - \alpha^2}{2}) = \sin \alpha$. $2 \cos(\alpha^2 + \frac{\alpha}{2}) \sin(\frac{\alpha}{2}) = 2 \sin(\frac{\alpha}{2}) \cos(\frac{\alpha}{2})$. Since $\alpha \in (2, 3)$,$\sin(\frac{\alpha}{2}) 
eq 0$,so we can divide by $2 \sin(\frac{\alpha}{2})$: $\cos(\alpha^2 + \frac{\alpha}{2}) = \cos(\frac{\alpha}{2})$. This implies $\alpha^2 + \frac{\alpha}{2} = 2n\pi \pm \frac{\alpha}{2}$. Case $1$: $\alpha^2 + \frac{\alpha}{2} = 2n\pi + \frac{\alpha}{2} \Rightarrow \alpha^2 = 2n\pi \Rightarrow \alpha = \sqrt{2n\pi}$. For $\alpha \in (2, 3)$,$4 Case $2$: $\alpha^2 + \frac{\alpha}{2} = 2n\pi - \frac{\alpha}{2} \Rightarrow \alpha^2 + \alpha - 2n\pi = 0$. Using the quadratic formula: $\alpha = \frac{-1 \pm \sqrt{1 + 8n\pi}}{2}$. For $\alpha \in (2, 3)$,we check $n=1$: $\alpha = \frac{-1 + \sqrt{1 + 8\pi}}{2} \approx \frac{-1 + 5.12}{2} \approx 2.06$. Both values lie in $(2, 3)$. Thus,there are $2$ solutions.

If $\alpha \in (2, 3)$,then the number of solutions of the equation $\int_{0}^{\alpha} \cos(x + \alpha^2) \, dx = \sin \alpha$ is:

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