Key Concepts and Formulas

• One-one and Onto Functions: A function $f: A \rightarrow B$ is one-one (injective) if $f(x_1) = f(x_2)$ implies $x_1 = x_2$. For a differentiable function, this often means the function is strictly monotonic (either increasing or decreasing). A function $f: A \rightarrow B$ is onto (surjective) if for every $y \in B$, there exists an $x \in A$ such that $f(x) = y$.
• Derivative of Exponential Functions: The derivative of $e^{g(x)}$ with respect to $x$ is $e^{g(x)} \cdot g'(x)$ using the chain rule.
• Distance of a Point from a Line: The perpendicular distance $d$ of a point $(x_1, y_1)$ from a line $Ax+By+C=0$ is given by the formula: $d = \frac{|Ax_1+By_1+C|}{\sqrt{A^2+B^2}}$

Step-by-Step Solution

1. Analyze the Monotonicity of the Function $f(x)$

To determine if $f(x)$ is one-one on the given domain $(-\infty, -1]$, we need to check its monotonicity. We will find the first derivative $f'(x)$ and analyze its sign.

Given function: $f(x)=e^{x^3-3 x+1}$

Let $g(x) = x^3-3x+1$. Then $f(x) = e^{g(x)}$. Using the chain rule, $f'(x) = e^{g(x)} \cdot g'(x)$.

First, find $g'(x)$: $g'(x) = \frac{d}{dx}(x^3-3x+1) = 3x^2-3$

Now, substitute this back into the expression for $f'(x)$: $f'(x)=e^{x^3-3 x+1} \cdot (3x^2-3)$

Factor out 3 from the quadratic term: $f'(x)=e^{x^3-3 x+1} \cdot 3(x^2-1)$ $f'(x)=3e^{x^3-3 x+1} (x-1)(x+1)$

Next, we analyze the sign of $f'(x)$ in the domain $(-\infty, -1]$.

• The term $e^{x^3-3 x+1}$ is an exponential term, which is always positive for any real $x$.
• The term $3$ is a positive constant.
• We need to determine the sign of $(x-1)(x+1)$ for $x \in (-\infty, -1]$.

Consider $x \in (-\infty, -1]$:

• If $x \le -1$, then $x-1 < 0$ (e.g., if $x=-2$, $x-1=-3$).
• If $x \le -1$, then $x+1 \le 0$ (e.g., if $x=-2$, $x+1=-1$; if $x=-1$, $x+1=0$).

Therefore, for $x \in (-\infty, -1]$:

• $(x-1)$ is negative.
• $(x+1)$ is non-positive.
• The product $(x-1)(x+1)$ will be non-negative (negative $\times$ negative = positive; negative $\times$ zero = zero).

So, for $x \in (-\infty, -1]$, $f'(x) = (\text{positive}) \times (\text{positive or zero}) = \text{positive or zero}$. This means $f'(x) \ge 0$ for all $x \in (-\infty, -1]$. Since $f'(x) \ge 0$ on the interval and is strictly positive for $x < -1$ (it's zero only at $x=-1$), the function $f(x)$ is increasing on $(-\infty, -1]$. Because $f(x)$ is strictly increasing, it is indeed a one-one function on its domain.

2. Determine the Range of the Function and Parameters $a, b$

Since the function $f(x)$ is increasing and onto the interval $(a, b]$, its range must be exactly $(a, b]$. For an increasing function on $(-\infty, -1]$, the lower bound of the range will be the limit as $x \rightarrow -\infty$, and the upper bound will be the function value at $x=-1$.

• Lower bound of the range ($a$): As $x \rightarrow -\infty$, the exponent $x^3-3x+1 \rightarrow (-\infty)^3 - 3(-\infty) + 1 \rightarrow -\infty + \infty + 1$. We need to be careful with indeterminate forms. For large negative $x$, $x^3$ dominates. So, $x^3-3x+1 \rightarrow -\infty$. Therefore, $\lim_{x \to -\infty} f(x) = \lim_{x \to -\infty} e^{x^3-3x+1} = e^{-\infty} = 0$. Since the codomain is $(a,b]$, $a=0$.

• Upper bound of the range ($b$): At the right endpoint of the domain, $x=-1$: $f(-1) = e^{(-1)^3 - 3(-1) + 1} = e^{-1 + 3 + 1} = e^3$. Since the codomain is $(a,b]$, $b=e^3$.

Thus, the codomain is $(a, b] = (0, e^3]$.

3. Determine the Coordinates of Point P

The point $P$ is given by $(2b+4, a+2)$. Substitute the values $a=0$ and $b=e^3$: $P(2e^3+4, 0+2)$ $P(2e^3+4, 2)$

4. Calculate the Distance of Point P from the Line

The given line equation is $x+e^{-3} y=4$. To use the distance formula, we rewrite it in the standard form $Ax+By+C=0$: $x+e^{-3}y-4=0$ Here, $A=1$, $B=e^{-3}$, and $C=-4$.

The point is $P(x_1, y_1) = (2e^3+4, 2)$. So, $x_1 = 2e^3+4$ and $y_1=2$.

Now, apply the distance formula: $d = \frac{|Ax_1+By_1+C|}{\sqrt{A^2+B^2}}$ $d = \frac{|1(2e^3+4) + e^{-3}(2) - 4|}{\sqrt{1^2+(e^{-3})^2}}$ $d = \frac{|2e^3+4 + 2e^{-3} - 4|}{\sqrt{1+e^{-6}}}$ $d = \frac{|2e^3 + 2e^{-3}|}{\sqrt{1+e^{-6}}}$ $d = \frac{2|e^3 + e^{-3}|}{\sqrt{1+e^{-6}}}$ Since $e^3 + e^{-3}$ is always positive, we can drop the absolute value: $d = \frac{2(e^3 + e^{-3})}{\sqrt{1+e^{-6}}}$ Multiply the numerator and denominator by $e^3$: $d = \frac{2(e^6 + 1)}{\sqrt{e^6(1+e^{-6})}}$ $d = \frac{2(e^6 + 1)}{e^3\sqrt{1+e^{-6}}}$ $d = \frac{2(e^6 + 1)}{\sqrt{e^6(1+e^{-6})}} = \frac{2(e^6+1)}{\sqrt{e^6+1}} = \frac{2(e^6+1)}{\sqrt{1+e^{-6}}}e^{-3}$ $d = \frac{2(e^3 + e^{-3})}{\sqrt{1 + e^{-6}}} = \frac{2e^3(1+e^{-6})}{\sqrt{1 + e^{-6}}} = 2e^3\sqrt{1+e^{-6}} = 2\sqrt{e^6(1+e^{-6})} = 2\sqrt{e^6+1}$

Therefore, the distance is $2\sqrt{1+e^6}$.

Common Mistakes & Tips

• When finding the range of the function, remember to consider the limit as $x$ approaches $-\infty$ for the lower bound, especially when dealing with exponential functions.
• Be careful with signs when analyzing the monotonicity of the function using the first derivative.
• Double-check the distance formula and ensure that the line equation is in the standard form before plugging in the values.

Summary

We first determined the monotonicity of the function $f(x)$ on the interval $(-\infty, -1]$, finding that it is strictly increasing. Then, we found the range of the function, which allowed us to determine the values of $a$ and $b$. Next, we found the coordinates of the point $P$ and finally calculated the distance of the point $P$ from the given line using the distance formula. The final distance is $2\sqrt{1+e^6}$.

Final Answer

The final answer is \boxed{2 \sqrt{1+e^6}}, which corresponds to option (A).