The function f(x) = e^x - x , x ∈ R , is strictly decreasing in

The correct answer is Option (2) → \((-\infty, 0)\)

Differentiating,

    \(f'(x) = e^{x} - 1\)

Since \(e^{x}\) is increasing and \(e^{0} = 1\), the sign of the derivative is

    \(f'(x) \lt 0\) for \(x \lt 0\)
    \(f'(x) \gt 0\) for \(x \gt 0\)

So f is strictly decreasing in \((-\infty, 0)\) and strictly increasing in \((0, \infty)\).

The intervals \((-\infty, 1)\) and R both contain points between 0 and 1, where f is increasing (for example f(0) = 1 < f(0.5) = 1.149…).