Functions Question 13
Question 13 - 2024 (31 Jan Shift 2)
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 $\mathrm{P}(2 \mathrm{~b}+4, \mathrm{a}+2)$ from the line $x+e^{-3} y=4$ is :
(1) $2 \sqrt{1+\mathrm{e}^{6}}$
(2) $4 \sqrt{1+\mathrm{e}^{6}}$
(3) $3 \sqrt{1+\mathrm{e}^{6}}$
(4) $\sqrt{1+\mathrm{e}^{6}}$
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Answer (1)
Solution
$f(x)=e^{x^{3}-3 x+1}$
$f^{\prime}(x)=e^{x^{3}-3 x+1} \cdot\left(3 x^{2}-3\right)$
$=e^{x^{3}-3 x+1} \cdot 3(x-1)(x+1)$
For $f^{\prime}(x) \geq 0$
$\therefore \mathrm{f}(\mathrm{x})$ is increasing function
$\therefore a=e^{-\infty}=0=f(-\infty)$
$b=e^{-1+3+1}=e^{3}=f(-1)$
$P(2 b+4, a+2)$
$\therefore P\left(2 e^{3}+4,2\right)$
$\mathrm{d}=\frac{\left(2 \mathrm{e}^{3}+4\right)+2 \mathrm{e}^{-3}-4}{\sqrt{1+\mathrm{e}^{-6}}}=2 \sqrt{1+\mathrm{e}^{6}}$
