If the equation frac{x^2 + 5}{2} = x - 2cos(a | vedclass

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(B) Given the equation: $\frac{x^2 + 5}{2} = x - 2\cos(ax + b)$.Rearranging the terms: $2\cos(ax + b) = x - \frac{x^2 + 5}{2}$.Multiply by $-1/2$: $-\

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$\pi$

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(B) Given the equation: $\frac{x^2 + 5}{2} = x - 2\cos(ax + b)$.Rearranging the terms: $2\cos(ax + b) = x - \frac{x^2 + 5}{2}$.Multiply by $-1/2$: $-\cos(ax + b) = \frac{x^2 + 5}{4} - \frac{x}{2} = \frac{x^2 - 2x + 5}{4}$.This simplifies to: $-\cos(ax + b) = \frac{(x - 1)^2}{4} + 1$.We know that the range of $-\cos(ax + b)$ is $[-1, 1]$.Also,for the right side,since $(x - 1)^2 \ge 0$,we have $\frac{(x - 1)^2}{4} + 1 \ge 1$.For the equation to have a solution,both sides must be equal to $1$.Thus,$\frac{(x - 1)^2}{4} + 1 = 1 \Rightarrow x = 1$.And $-\cos(a(1) + b) = 1 \Rightarrow \cos(a + b) = -1$.Therefore,$a + b = (2k + 1)\pi$ for $k \in I$. For $k=0$,$a + b = \pi$.

If the equation $\frac{x^2 + 5}{2} = x - 2\cos(ax + b)$ has at least one solution,then $(b + a)$ can be equal to

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