Question:

Step 1: Determine the greatest value of $f(x)$ on $[0, 5]$. The function $f(x) = -x^2 + 4x + \lambda - 4$ is a downward-opening parabola. Its vertex occurs at $x = -\frac{4}{2(-1)} = 2$. Since $x=2$ lies within the interval $[0, 5]$, the greatest value of $f(x)$ occurs at the vertex. $$f(2) = -(2)^2 + 4(2) + \lambda - 4 = -4 + 8 + \lambda - 4 = \lambda$$ Step 2: Confirm the greatest value of $f(x)$ on $[0, 5]$. Evaluate $f(x)$ at the endpoints of the interval: $$f(0) = -(0)^2 + 4(0) + \lambda - 4 = \lambda - 4$$ $$f(5) = -(5)^2 + 4(5) + \lambda - 4 = -25 + 20 + \lambda - 4 = \lambda - 9$$ Comparing the values, $\lambda > \lambda - 4 > \lambda - 9$. Thus, the greatest value of $f(x)$ on $x \in [0, 5]$ is $\lambda$. Step 3: Determine the least value of $g(x)$ on $x \in \mathbb{R}$. The function $g(x) = x^2 - 2\lambda x + 10 - 2\lambda$ is an upward-opening parabola. Its vertex occurs at $x = -\frac{-2\lambda}{2(1)} = \lambda$. Since the domain is $x \in \mathbb{R}$, the least value of $g(x)$ occurs at its vertex. $$g(\lambda) = (\lambda)^2 - 2\lambda(\lambda) + 10 - 2\lambda = \lambda^2 - 2\lambda^2 + 10 - 2\lambda = -\lambda^2 - 2\lambda + 10$$ Step 4: Establish the inequality. The problem states that the greatest value of $f(x)$ is smaller than the least value of $g(x)$. Therefore, we have the inequality: $$\lambda < -\lambda^2 - 2\lambda + 10$$ Step 5: Solve the inequality for $\lambda$. Rearrange the inequality: $$\lambda^2 + 3\lambda - 10 < 0$$ Factor the quadratic expression: $$(\lambda + 5)(\lambda - 2) < 0$$ This inequality holds when $\lambda$ is between the roots $-5$ and $2$. Thus, the condition for $\lambda$ is: $$-5 < \lambda < 2$$