JEE Main 2024 — Functions Question with Solution

Given: $f\left(x\right)=e^{x^3-3x+1}$

$\Rightarrow f^{\text{'}}\left(x\right)=e^{x^3-3x+1}\times \left(3x^2-3\right)$

$\Rightarrow f^{\text{'}}\left(x\right)=3\times e^{x^3-3x+1}\times \left(x-1\right)\times \left(x+1\right)$

So, for $x\in (-\infty ,-1]$ $f^{\text{'}}\left(x\right)>0$, hence $f\left(x\right)$ is increasing function in given domain,

Now, finding $a$ and $b$ we get,

$\Rightarrow a=\underset{x\to -\infty }{\lim }\left(e^{x^3-3x+1}\right)$

$\Rightarrow a=0$

And $b=f\left(-1\right)$

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

$\Rightarrow b=e^3$

So, $P\left(2b+4, a+2\right)$

$\Rightarrow P\equiv \left(2e^3+4, 2\right)$

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

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

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

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

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

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

$\Rightarrow d=2\sqrt{1+e^6}$