JEE Main · 2024 · 31 Jan 2024 · Shift 2 · Application of Derivatives

$f(x)=e^{x^3-3x+1}$ on $(-\infty ,-1]$. Differentiating,

$f^{\prime }(x)=e^{x^3-3x+1}(3x^2-3)=3e^{x^3-3x+1}(x-1)(x+1)$.

For $x\le -1$, both $(x-1)$ and $(x+1)$ are $\le 0$, so their product is $\ge 0$; hence $f^{\prime }(x)\ge 0$ and $f$ is increasing on $(-\infty ,-1]$.

As $x\to -\infty$, the exponent $x^3-3x+1\to -\infty$, so $f(x)\to 0^{+}$; at $x=-1$, the exponent is $-1+3+1=3$, so $f(-1)=e^3$.

Since $f$ is one-one and onto from $(-\infty ,-1]$ to $(a,b]$, the range is $(0,e^3]$, giving $a=0,b=e^3$.

So $P(2b+4,a+2)=P(2e^3+4,2)$.

Distance from $P$ to the line $x+e^{-3}y-4=0$:

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

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