Multiple choice

The number of real roots of the equation ${ e }^{ x-1 }+x-2=0$ is

  1. $1$
  2. $2$
  3. $3$
  4. $4$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let f(x) = exp(x-1) + x - 2. f'(x) = exp(x-1) + 1, which is always positive. Thus, f(x) is strictly increasing. f(1) = 1 + 1 - 2 = 0. Since it is strictly increasing, x=1 is the only real root.

AI explanation

Let f(x) = e^(x-1) + x - 2. The derivative f'(x) = e^(x-1) + 1 is always positive for all real x, meaning the function is strictly increasing. At x = 1, f(1) = e^0 + 1 - 2 = 0, which means x = 1 is the only root. The equation has 1 real root.

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