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

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(B) The given equation is $\frac{x^2 + 5}{2} = x - 2\cos(ax + b)$.Rearranging the terms,we get $2\cos(ax + b) = x - \frac{x^2 + 5}{2}$.Multiplying by

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$\pi$

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(B) The given equation is $\frac{x^2 + 5}{2} = x - 2\cos(ax + b)$.Rearranging the terms,we get $2\cos(ax + b) = x - \frac{x^2 + 5}{2}$.Multiplying by $-\frac{1}{2}$,we obtain $-\cos(ax + b) = \frac{x^2 - 2x + 5}{4}$.This can be written as $-\cos(ax + b) = \frac{(x - 1)^2 + 4}{4} = \frac{(x - 1)^2}{4} + 1$.We know that $-\cos(ax + b) \in [-1, 1]$.Also,$\frac{(x - 1)^2}{4} + 1 \geq 1$ for all real $x$.For the equation to have a solution,both sides must be equal to $1$.Thus,$\frac{(x - 1)^2}{4} + 1 = 1 \Rightarrow x = 1$.Substituting $x = 1$ into $-\cos(ax + b) = 1$,we get $\cos(ax + b) = -1$.This implies $ax + b = (2k + 1)\pi$ for some integer $k$.For $x = 1$,we have $a + b = (2k + 1)\pi$.Taking $k = 0$,we get $a + b = \pi$.

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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