Sets, Relations & Functions
One-One and Onto — Exponential Function, Distance from Line
nta_pyq_2024_jan
Grade 11
Question:
If the function $f:(-\infty,-1]\to(a,b]$ 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 $x+e^{-3}y=4$ is:
$2\sqrt{1+e^6}$
$4\sqrt{1+e^6}$
$3\sqrt{1+e^6}$
$\sqrt{1+e^6}$
Step-by-Step Solution
Key Concept: $f'(x)=e^{x^3-3x+1}(3x^2-3)=3e^{\cdots}(x-1)(x+1)$. On $(-\infty,-1]$: $x\leq-1$, so $(x+1)\leq0$ and $(x-1)<0$: $f'\geq0$, so $f$ is increasing. $a=\lim_{x\to-\infty}f(x)=0$ (excluded), $b=f(-1)=e^3$. Compute distance of $P(2e^3+4,2)$ from $x+e^{-3}y=4$.
$a=0,b=e^3$. $P=(2e^3+4,2)$. Distance from $x+e^{-3}y=4$: $d=\frac{|2e^3+4+2e^{-3}-4|}{\sqrt{1+e^{-6}}}=\frac{2(e^3+e^{-3})}{\sqrt{1+e^{-6}}}=2\sqrt{1+e^6}$.
Correct Answer: 1