If `F :(- oo, -1] rightarrow (a, b]` is defined as `f(x) = e^(x^3-3x+1)` such that F is both onel-one and onto thenthe distance from a point P(2a + 4, b + 2) to curve `x + ye^(-3)-4 = 0` is

Options

A. `sqrt (e^3 +2)`
B. `frac {e^3+2}{sqrt (e^3 +2)}`
C. `frac {e^3+2}{sqrt (e^6 +1)}`
D. e

Answer

To solve the problem step by step, we need to analyze the function \( f(x) = e^{x^3 - 3x + 1} \) defined on the interval \( (-\infty, -1] \) and find the distance from the point \( P(2a + 4, b + 2) \) to the curve defined by the equation \( x + ye^{-3} - 4 = 0 \).

### Step 1: Analyze the function \( f(x) \)

The function is given by:
\[ f(x) = e^{x^3 - 3x + 1} \]
We need to determine the behavior of this function on the interval \( (-\infty, -1] \).

### Step 2: Check if \( f(x) \) is one-one and onto

To check if \( f(x) \) is one-one, we can find its derivative:
\[ f'(x) = e^{x^3 - 3x + 1} \cdot (3x^2 - 3) \]
Setting \( f'(x) = 0 \):
\[ 3x^2 - 3 = 0 \implies x^2 = 1 \implies x = \pm 1 \]
Since we are only interested in the interval \( (-\infty, -1] \), \( f'(x) < 0 \) for \( x < -1 \). Thus, \( f(x) \) is decreasing on this interval, confirming that it is one-one.

Next, we find the range of \( f(x) \) as \( x \) approaches \( -\infty \) and at \( x = -1 \):
- As \( x \to -\infty \), \( f(x) \to 0 \).
- At \( x = -1 \):
\[ f(-1) = e^{(-1)^3 - 3(-1) + 1} = e^{-1 + 3 + 1} = e^{3} \]
Thus, the function maps \( (-\infty, -1] \) onto \( (0, e^3] \).

### Step 3: Determine values of \( a \) and \( b \)

From the range, we have:
- \( a = 0 \)
- \( b = e^3 \)

### Step 4: Find the coordinates of point \( P \)

Using the values of \( a \) and \( b \):
\[ P(2a + 4, b + 2) = P(2(0) + 4, e^3 + 2) = P(4, e^3 + 2) \]

### Step 5: Rewrite the curve equation

The curve is given by:
\[ x + ye^{-3} - 4 = 0 \implies y = 4e^3 - xe^3 \]

### Step 6: Calculate the perpendicular distance from point \( P \) to the curve

The distance \( d \) from a point \( (x_0, y_0) \) to the line \( Ax + By + C = 0 \) is given by:
\[ d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \]
For our line:
- \( A = 1 \)
- \( B = e^{-3} \)
- \( C = -4 \)

Substituting \( P(4, e^3 + 2) \):
\[ d = \frac{|1 \cdot 4 + e^{-3} \cdot (e^3 + 2) - 4|}{\sqrt{1^2 + (e^{-3})^2}} \]
Calculating the numerator:
\[ = |4 + 1 + 2e^{-3} - 4| = |1 + 2e^{-3}| \]
The denominator:
\[ = \sqrt{1 + e^{-6}} \]
Thus, the distance \( d \) becomes:
\[ d = \frac{1 + 2e^{-3}}{\sqrt{1 + e^{-6}}} \]

### Final Answer

The distance from point \( P \) to the curve is:
\[ d = \frac{1 + 2e^{-3}}{\sqrt{1 + e^{-6}}} \]