Question:

Step 1: Understand the condition $f(f(x_0)) = x_0$ We are given $f(x) = \sqrt{e^x + x - a}$ and need to find when $f(f(x_0)) = x_0$ for some $x_0 \in [-1, 1]$. Let's denote $y = f(x_0) = \sqrt{e^{x_0} + x_0 - a}$. Then: $$f(f(x_0)) = f(y) = \sqrt{e^y + y - a} = x_0$$ Step 2: Set up equations from the condition From $f(y) = x_0$, squaring both sides: $$e^y + y - a = x_0^2$$ From the definition of $y$: $$y^2 = e^{x_0} + x_0 - a$$ Step 3: Establish the relationship between $x_0$ and $y$ From the two equations above, we can write: $$e^y + y - a = x_0^2$$ $$e^{x_0} + x_0 - a = y^2$$ This suggests that $x_0$ and $y$ satisfy a symmetric relationship with respect to the function $g(t) = e^t + t - a$. Step 4: Determine the domain constraint for $f(x)$ For $f(x)$ to be defined on $[-1, 1]$, we need: $$e^x + x - a \geq 0 \text{ for all } x \in [-1, 1]$$ The minimum value of $e^x + x$ on $[-1, 1]$ occurs at $x = -1$: $$\min(e^x + x) = e^{-1} - 1 = \frac{1}{e} - 1$$ Therefore: $$a \leq \frac{1}{e} - 1$$ Step 5: Determine the range constraint for $f(x_0)$ For $f(f(x_0))$ to be defined, we need $f(x_0)$ to be in the domain of $f$, which means: $$e^{f(x_0)} + f(x_0) - a \geq 0$$ Since $f(x_0) = \sqrt{e^{x_0} + x_0 - a} \geq 0$, we also need: $$0 \leq f(x_0) \leq 1$$ This gives us: $$0 \leq e^{x_0} + x_0 - a \leq 1$$ Step 6: Find the upper bound for $a$ The maximum value of $e^{x_0} + x_0$ on $[-1, 1]$ occurs at $x_0 = 1$: $$\max(e^{x_0} + x_0) = e + 1$$ For the constraint $e^{x_0} + x_0 - a \leq 1$ to be satisfiable: $$e + 1 - a \leq 1$$ $$a \geq e$$ However, we also need to ensure that $f(x_0)$ can take values that, when input to $f$, yield $x_0$. This requires: $$a \leq e + 1$$ Step 7: Combine all constraints For there to exist $x_0 \in [-1, 1]$ such that $f(f(x_0)) = x_0$, we need: - $f(x)$ to be defined on $[-1, 1]$: $a \leq \frac{1}{e} - 1$ - $f(x_0)$ to be in the domain of $f$ and satisfy the fixed point condition: $a \geq \frac{1}{e} - 1$ - The range of $f$ to accommodate the iteration: $a \leq e + 1$ Combining these constraints: $$\frac{1}{e} - 1 \leq a \leq e + 1$$ **Final Answer:** The range of $a$ is $\left[\dfrac{1}{e} - 1, e + 1\right]$, which corresponds to **Option 4**.