What is the cardinality of the set of real solutions of e^x + x = 1?

Concept:

To find the critical points of a function \(f(x)\) we put \(f'(x)\) = 0.

Explanation:  

\(e^x + x = 1\)

We are looking for real values of  \(x \)  that satisfy this equation. Let’s rewrite this as \(e^x = 1 - x\)

To find the real solutions, we need to investigate the points where these two functions intersect, i.e., where,

\(e^x = 1 - x\)

As  \(x \to -\infty\) , \(e^x \to 0 \), and \(1 - x \to \infty \).

So, there is no solution in this region.

For large positive  \(x\) :

As  \(x \to \infty\) , \( e^x \to \infty\) , and  \(1 - x \to -\infty\) .

Again, there is no solution in this region.

Check around  \( x = 0\) :

At \(x = 0 \), \(e^0 + 0 = 1 \), which satisfies the equation.

So,  \(x = 0 \) is a solution.

We can check the behavior of the function \( f(x) = e^x + x\) .

The derivative of this function is \(f'(x) = e^x + 1 \)

Since \(e^x + 1 > 0 \) for all real \(x\), the function is strictly increasing. Therefore, it can only cross  \( y = 1\) at one point,

which implies that the solution  \(x = 0 \) is the only real solution.

Since there is only one real solution, the cardinality of the set of real solutions is 1.

Thus, the correct answer is option 2).