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Theory of expressions

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Question

If the equation $x^{2} + 5 + 4 \cos (αx + β) = 2 x$ has at least one solution where $α, β \in [2, 5]$ then the value of $(α + β)$ equal to

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Solution

$\begin{aligned} x^{2} + 5 + 4 \cos (αx + β) = 2 x \\ \Rightarrow (x - 1)^{2} + 4 = - 4 \cos (αx + β) \end{aligned}$
Here $(x - 1)^{2} + 4 \geq 4, - 4 \cos (αx + β) \leq 4$
But, the given equation has at least one solution, then
$\begin{array}{l} (x - 1)^{2} + 4 = 4 = - 4 \cos (αx + β) \\ ∴ x = 1 \Rightarrow \cos (α + β) = - 1 \\ \Rightarrow α + β = 3 π = 9.42 (∵ α + β \in [4, 10]) \end{array}$

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If x is real, the expression $\frac{x + 2}{2 x^{2} + 3 x + 6}$ takes all value in the interval

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