JEE MAIN - Mathematics (2024 - 31st January Evening Shift - No. 13)
If the function $$f:(-\infty,-1] \rightarrow(a, b]$$ defined by $$f(x)=e^{x^3-3 x+1}$$ is one - one and onto, then the distance of the point $$P(2 b+4, a+2)$$ from the line $$x+e^{-3} y=4$$ is :
$$2 \sqrt{1+e^6}$$
$$\sqrt{1+e^6}$$
$$3 \sqrt{1+e^6}$$
$$4 \sqrt{1+e^6}$$
Explanation
$$\begin{aligned} & f(x)=e^{x^3-3 x+1} \\ & f^{\prime}(x)=e^{x^3-3 x+1} \cdot\left(3 x^2-3\right) \\ & =e^{x^3-3 x+1} \cdot 3(x-1)(x+1) \end{aligned}$$
For $$\mathrm{f}^{\prime}(\mathrm{x}) \geq 0$$
$$\therefore \mathrm{f}(\mathrm{x})$$ is increasing function
$$\begin{aligned} & \therefore \mathrm{a}=\mathrm{e}^{-\infty}=0=\mathrm{f}(-\infty) \\ & \mathrm{b}=\mathrm{e}^{-1+3+1}=\mathrm{e}^3=\mathrm{f}(-1) \\ & \mathrm{P}(2 \mathrm{~b}+4, \mathrm{a}+2) \\ & \therefore \mathrm{P}\left(2 \mathrm{e}^3+4,2\right) \end{aligned}$$
_31st_January_Evening_Shift_en_13_1.png)
$$\mathrm{d}=\frac{\left(2 \mathrm{e}^3+4\right)+2 \mathrm{e}^{-3}-4}{\sqrt{1+\mathrm{e}^{-6}}}=2 \sqrt{1+\mathrm{e}^6}$$
Comments (0)
