Analysis of Piecewise Function Monotonicity Differentiability and Surjectivity

To determine the correctness of the given statements, we analyze the piecewise function $f: \mathbb{R} \rightarrow \mathbb{R}$ section by section.

1. Analysis on $(-\infty, 0)$ and Option A:

For $x < 0$, the function is defined as: $f(x) = x^5 + 5x^4 + 10x^3 + 10x^2 + 3x + 1$

Taking the first derivative with respect to $x$: $f'(x) = 5x^4 + 20x^3 + 30x^2 + 20x + 3$

We can rewrite $f'(x)$ by completing the binomial expansion: $f'(x) = 5(x^4 + 4x^3 + 6x^2 + 4x + 1) - 2 = 5(x + 1)^4 - 2$

For $x = -1 \in (-\infty, 0)$: $f'(-1) = 5(0)^4 - 2 = -2 < 0$

Since $f'(x) < 0$ in a neighborhood of $x = -1$, the function $f(x)$ is not strictly increasing on $(-\infty, 0)$.
Thus, Option A is incorrect.

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2. Continuity and Differentiability at Boundary Points:

• At $x = 0$: $\lim_{x \to 0^-} f(x) = 1$ $\lim_{x \to 0^+} f(x) = f(0) = 0^2 - 0 + 1 = 1$ So, $f(x)$ is continuous at $x = 0$.

• At $x = 1$: $\lim_{x \to 1^-} f(x) = 1^2 - 1 + 1 = 1$ $f(1) = \frac{2}{3}(1)^3 - 4(1)^2 + 7(1) - \frac{8}{3} = 1$ So, $f(x)$ is continuous at $x = 1$.

• At $x = 3$: $\lim_{x \to 3^-} f(x) = \frac{2}{3}(27) - 4(9) + 7(3) - \frac{8}{3} = 18 - 36 + 21 - \frac{8}{3} = \frac{1}{3}$ $f(3) = (3 - 2)\log_e(3 - 2) - 3 + \frac{10}{3} = 0 - 3 + \frac{10}{3} = \frac{1}{3}$ So, $f(x)$ is continuous at $x = 3$.

Since $f(x)$ is continuous on all defined sub-intervals and at all boundary points, $f(x)$ is continuous on the entire domain $\mathbb{R}$.

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3. Differentiability of $f'$ at $x = 1$ and Options B & D:

Let's compute $f'(x)$ near $x = 1$:

$$f'(x) = \begin{cases} 2x - 1, & 0 < x < 1 \\ 2x^2 - 8x + 7, & 1 < x < 3 \end{cases}$$

Checking the derivative value at $x = 1$:

• Left-hand limit of $f'(x)$ at $x = 1$: $\lim_{x \to 1^-} f'(x) = 2(1) - 1 = 1$
• Right-hand limit of $f'(x)$ at $x = 1$: $\lim_{x \to 1^+} f'(x) = 2(1)^2 - 8(1) + 7 = 1$

Since LHD = RHD = 1 at $x = 1$, $f(x)$ is differentiable at $x = 1$ with $f'(1) = 1$.

Now, let's examine the derivative of $f'(x)$ (i.e., the second derivative $f''(x)$) around $x = 1$:

• Left-hand derivative of $f'$ at $x = 1$: $\left(f'\right)'_-(1) = \lim_{x \to 1^-} \frac{f'(x) - f'(1)}{x - 1} = \lim_{x \to 1^-} \frac{(2x - 1) - 1}{x - 1} = 2$
• Right-hand derivative of $f'$ at $x = 1$: $\left(f'\right)'_+(1) = \lim_{x \to 1^+} \frac{f'(x) - f'(1)}{x - 1} = \lim_{x \to 1^+} \frac{(2x^2 - 8x + 7) - 1}{x - 1} = \lim_{x \to 1^+} \frac{2(x - 1)(x - 3)}{x - 1} = -4$

Since $\left(f'\right)'_-(1) = 2 \neq -4 = \left(f'\right)'_+(1)$, $f'$ is NOT differentiable at $x = 1$.
Thus, Option D is correct.

Furthermore:

• For $x \in (0, 1)$, $\left(f'\right)'(x) = 2 > 0 \implies f'(x)$ is increasing towards $f'(1) = 1$.
• For $x \in (1, 3)$, $\left(f'\right)'(x) = 4x - 8$, so $\left(f'\right)'(1^+) = -4 < 0 \implies f'(x)$ is decreasing away from $f'(1) = 1$.

Hence, $f'(x) \le f'(1)$ for all $x$ in a neighborhood around $x = 1$, which means $f'$ achieves a local maximum at $x = 1$.
Thus, Option B is correct.

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4. Surjectivity (Onto Nature) and Option C:

Let's evaluate the behavior of $f(x)$ as $x \to \pm\infty$:

1. As $x \to -\infty$: $\lim_{x \to -\infty} f(x) = \lim_{x \to -\infty} \left(x^5 + 5x^4 + 10x^3 + 10x^2 + 3x + 1\right) = -\infty$
2. As $x \to \infty$: $\lim_{x \to \infty} f(x) = \lim_{x \to \infty} \left[(x - 2)\log_e(x - 2) - x + \frac{10}{3}\right] = \lim_{x \to \infty} (x - 2)\left[\log_e(x - 2) - 1\right] + \frac{4}{3} = +\infty$

Since $f(x)$ is continuous on all of $\mathbb{R}$ and its range spans from $-\infty$ to $+\infty$, by the Intermediate Value Theorem, the range of $f(x)$ is $\mathbb{R}$.
Therefore, $f$ is an onto (surjective) function.
Thus, Option C is correct.

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Conclusion:

The correct statements are (B), (C), and (D).