1. Given function and domain

We have $f(x)=e^{x^3-3x+1}, \, x\in(-\infty,-1].$

It is given that $f:(-\infty,-1]\to (a,b]$ is one-one and onto. So we need the exact range of $f$ on $(-\infty,-1]$.

2. Check monotonicity on the given domain

Let $g(x)=x^3-3x+1.$ Then $f(x)=e^{g(x)}.$ Since the exponential function is strictly increasing, the monotonicity of $f$ is same as that of $g$.

Now, $g'(x)=3x^2-3=3(x^2-1)=3(x-1)(x+1).$

For $x\in(-\infty,-1)$, we have $x^2>1$, so $g'(x)>0.$ Also at $x=-1$, $g'(-1)=0.$

Hence $g(x)$, and therefore $f(x)$, is increasing on $(-\infty,-1]$. So $f$ is one-one on the given domain.

3. Find the range of $f$

Because $f$ is increasing on $(-\infty,-1]$:

• the maximum value occurs at $x=-1$,
• the minimum is approached as $x\to -\infty$.

First, $f(-1)=e^{(-1)^3-3(-1)+1}=e^{-1+3+1}=e^3.$ So $b=e^3.$

Now as $x\to -\infty$, $x^3-3x+1\to -\infty,$ therefore $f(x)=e^{x^3-3x+1}\to 0^+.$ Since $0$ is not attained, $a=0.$

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

4. Coordinates of point $P$

Given $P(2b+4, a+2).$ Substitute $a=0,\ b=e^3$:

$P=(2e^3+4,\ 2).$

5. Distance from point to line

The line is $x+e^{-3}y=4,$ which we write as $x+e^{-3}y-4=0.$

Distance of point $(x_1,y_1)$ from line $Ax+By+C=0$ is $d=\frac{|Ax_1+By_1+C|}{\sqrt{A^2+B^2}}.$

Here, $A=1,\quad B=e^{-3},\quad C=-4,$ $(x_1,y_1)=(2e^3+4,2).$

So, $d=\frac{|(2e^3+4)+e^{-3}(2)-4|}{\sqrt{1+e^{-6}}}$ $=\frac{|2e^3+2e^{-3}|}{\sqrt{1+e^{-6}}}.$

Factor $2$: $d=\frac{2(e^3+e^{-3})}{\sqrt{1+e^{-6}}}.$

Now, $e^3+e^{-3}=e^{-3}(e^6+1).$ So $d=\frac{2e^{-3}(e^6+1)}{\sqrt{1+e^{-6}}}.$

Also, $\sqrt{1+e^{-6}}=\sqrt{\frac{e^6+1}{e^6}}=\frac{\sqrt{e^6+1}}{e^3}.$

Hence $d=\frac{2e^{-3}(e^6+1)}{\frac{\sqrt{e^6+1}}{e^3}}=2\sqrt{e^6+1}.$

Therefore, $d=2\sqrt{1+e^6}.$

6. Compare with options

This matches Option A.

7. Verification with stored answer

Stored correct answer: A

Our derived answer: A

So the answer agrees with the stored correct answer.