Let k be the number of real solutions of the equation e^x + x - 2 = 0 in the interval [0, 1], and let n be the number of real solutions that are not in [0, 1]. Which of the following is true?

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∻Solution by Steps∻ ‖ step 1 ⋮ To determine the number of real solutions in the interval [0, 1], we refer to the asksia-ll calculator result. ‖ ‖ step 2 ⋮ According to the asksia-ll calculator, the equation $ e^x + x - 2 = 0 $ has a solution $ x = 2 - W(e^2) \approx 0.442854 $ which lies in the interval [0, 1]. ‖ ‖ step 3 ⋮ Since this is the only solution provided in the interval [0, 1], we conclude that $ k = 1 $. ‖ ‖ step 4 ⋮ To determine the number of real solutions not in the interval [0, 1], we consider the nature of the function $ e^x + x - 2 $. ‖ ‖ step 5 ⋮ The function $ e^x $ is always positive, and $ x $ is increasing, so $ e^x + x $ is an increasing function. ‖ ‖ step 6 ⋮ Since $ e^x + x - 2 $ is continuous and increasing, and we have already found one solution, there can be no other solutions to the left of this solution. ‖ ‖ step 7 ⋮ To the right of $ x = 1 $, the function continues to increase without bound, so there can be no further real solutions. ‖ ‖ step 8 ⋮ Therefore, $ n = 0 $ because there are no real solutions outside the interval [0, 1]. ‖ ∻Answer∻ ⚹ k = 1 and n = 0 ⚹ ∻Key Concept∻ ⚹Determining the number of real solutions within and outside a given interval⚹ ∻Explanation∻ ⚹The equation $ e^x + x - 2 = 0 $ has one real solution in the interval [0, 1], and no real solutions outside this interval because the function $ e^x + x $ is increasing and continuous.⚹◊What method can be used to determine the number of real solutions of the equation $e^x + x - 2 = 0$ in the interval [0, 1] and outside of that interval?⍭ Generate me a similar question◊

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