Question

The equation $e^{x-1}+x-2=0$ has:

• A. one real root
• B. two real roots
• C. three real roots
• D. four real roots

Answer 1

Answer: (A)

one real root

Answer 2

Explanation of the solution:

1. Define the function: Let $f(x) = e^{x-1} + x - 2$. We are looking for the number of real roots of $f(x) = 0$.

2. Find the derivative: Calculate the first derivative of $f(x)$: $f'(x) = \frac{d}{dx}(e^{x-1} + x - 2) = e^{x-1} \cdot \frac{d}{dx}(x-1) + \frac{d}{dx}(x) - \frac{d}{dx}(2)$ $f'(x) = e^{x-1} \cdot 1 + 1 - 0$ $f'(x) = e^{x-1} + 1$

3. Analyze the derivative: For any real number $x$, $e^{x-1}$ is always positive ($e^{x-1} > 0$). Therefore, $f'(x) = e^{x-1} + 1 > 0 + 1 = 1$. Since $f'(x) > 0$ for all $x \in \mathbb{R}$, the function $f(x)$ is strictly increasing over its entire domain.

4. Determine the number of roots from monotonicity: A strictly increasing continuous function can intersect the x-axis at most once. If it does intersect, it intersects exactly once.

5. Check for existence of a root (using limits):

   • As $x \to -\infty$: $e^{x-1} \to 0$ $x-2 \to -\infty$ So, $f(x) = e^{x-1} + x - 2 \to 0 + (-\infty) = -\infty$.
   • As $x \to +\infty$: $e^{x-1} \to +\infty$ $x-2 \to +\infty$ So, $f(x) = e^{x-1} + x - 2 \to +\infty + (+\infty) = +\infty$.

   Since $f(x)$ is continuous and its value ranges from $-\infty$ to $+\infty$, by the Intermediate Value Theorem, there must be at least one real root.

6. Conclusion: Combining the facts that $f(x)$ is strictly increasing (at most one root) and it spans from $-\infty$ to $+\infty$ (at least one root), we conclude that there is exactly one real root.

7. Find the root (optional): By inspection, if we substitute $x=1$ into the equation: $e^{1-1} + 1 - 2 = e^0 + 1 - 2 = 1 + 1 - 2 = 0$. Thus, $x=1$ is the unique real root.

The equation has one real root.