If the function f:(-∞ ,-1]→ (a,b)] defined by f(x)=e^x^3-3x+1 is one-one and onto, then the distance of the point P(2b+4,a+2) from the line x+e^-3y=4…

Given: $f(x)={e}^{{x}^{3}-3x+1}$f(x)=ex3−3x+1

$$\Rightarrow {f}^{'}(x)={e}^{{x}^{3}-3x+1}\times (3{x}^{2}-3)$$⇒f′(x)=ex3−3x+1×(3x2−3)

$$\Rightarrow {f}^{'}(x)=3\times {e}^{{x}^{3}-3x+1}\times (x-1)\times (x+1)$$⇒f′(x)=3×ex3−3x+1×(x−1)×(x+1)

So, for $x\in (-\infty ,-1]$x∈(−∞,−1] ${f}^{'}(x)>0$f′(x)>0, hence $f(x)$f(x) is increasing function in given domain,

Now, finding $a$a and $b$b we get,

$$\Rightarrow a=\underset{x\rightarrow -\infty }{\mathrm{lim}}({e}^{{x}^{3}-3x+1})$$⇒a=x→−∞lim​(ex3−3x+1)

$$\Rightarrow a=0$$⇒a=0

And $b=f(-1)$b=f(−1)

$$\Rightarrow b={e}^{-1+3+1}$$⇒b=e−1+3+1

$$\Rightarrow b={e}^{3}$$⇒b=e3

So, $P(2b+4,a+2)$P(2b+4,a+2)

$$\Rightarrow P\equiv (2{e}^{3}+4,2)$$⇒P≡(2e3+4,2)

Now, finding the distance from the given line we get,

$$\Rightarrow d=\frac{(2{e}^{3}+4)+2{e}^{-3}-4}{\sqrt{1+{e}^{-6}}}$$⇒d=1+e−6​(2e3+4)+2e−3−4​

$$\Rightarrow d=\frac{2({e}^{3}+{e}^{-3})}{\sqrt{1+{e}^{-6}}}$$⇒d=1+e−6​2(e3+e−3)​

$$\Rightarrow d=\frac{1}{{e}^{-3}}\frac{2(1+{e}^{-6})}{\sqrt{1+{e}^{-6}}}$$⇒d=e−31​1+e−6​2(1+e−6)​

$$\Rightarrow d=\frac{2\sqrt{1+{e}^{-6}}}{{e}^{-3}}$$⇒d=e−321+e−6​​

$$\Rightarrow d=2\sqrt{\frac{1}{{e}^{-6}}+1}$$⇒d=2e−61​+1​

$$\Rightarrow d=2\sqrt{1+{e}^{6}}$$⇒d=21+e6​