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We need to find the number of real roots of the equation ${e^{x - 1}} + x - 2 = 0$.

Let us re – arrange the given equation as:

${e^{x - 1}} + x - 2 = 0$ $ \Rightarrow {e^{x - 1}} = 2 - x$

Now considering ${e^{x - 1}}$ as a function g (x) and 2 – x as another function h (x).

Upon plotting the graphs of both the functions g (x) and h (x), we get

Here, we get only one intersection point between the graph of the curves g (x) = ${e^{x - 1}}$and h (x) = 2 – x.

Hence, we can say that the number of real roots of the given equation ${e^{x - 1}} + x - 2 = 0$ is only one.