If 0 leq a, b leq 3 and the equation x^2+4+3 | vedclass

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(C) The given equation is $x^2+4+3 \cos (ax+b)=2x$. Rearranging the terms, we get $(x^2-2x+1) + 3 + 3 \cos (ax+b) = 0$. This simplifies to $(x-1)^2 +

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$\pi$

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(C) The given equation is $x^2+4+3 \cos (ax+b)=2x$. Rearranging the terms, we get $(x^2-2x+1) + 3 + 3 \cos (ax+b) = 0$. This simplifies to $(x-1)^2 + 3(1 + \cos (ax+b)) = 0$. Since $(x-1)^2 \geq 0$ and $1 + \cos (ax+b) \geq 0$ for all real $x$, the sum can be zero only if both terms are zero simultaneously. Thus, $(x-1)^2 = 0 \implies x = 1$ and $1 + \cos (ax+b) = 0$. This implies $\cos (ax+b) = -1$, so $ax+b = (2n+1)\pi$ for some integer $n$. Substituting $x=1$, we get $a+b = (2n+1)\pi$. Given $0 \leq a, b \leq 3$, we have $0 \leq a+b \leq 6$. Since $\pi \approx 3.14$, the only possible value for $a+b$ in the range $[0, 6]$ is $\pi$ (where $n=0$).

If $0 \leq a, b \leq 3$ and the equation $x^2+4+3 \cos (ax+b)=2x$ has a real solution, then the value of $(a+b)$ is

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