Question

If the function $f : (-\infty, -1] \rightarrow (a, b)$ 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:

• A. $2 \sqrt{1 + e^6}$
• B. $4 \sqrt{1 + e^6}$
• C. $3 \sqrt{1 + e^6}$
• D. $\sqrt{1 + e^6}$

Answer 1

Answer: (A)

$2 \sqrt{1 + e^6}$

Answer 2

Analyze the function $f(x) = e^{x^3 - 3x + 1}$. To determine if $f(x)$ is one-one, we need to check if $f(x)$ is strictly increasing or decreasing. Calculate the derivative $f'(x)$:

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

Since $e^{x^3 - 3x + 1} > 0$ for all $x \in (-\infty, -1]$, the sign of $f'(x)$ depends on $(x - 1)(x + 1)$. For $x \leq -1$, $f'(x) \geq 0$, indicating that $f(x)$ is an increasing function on $(-\infty, -1]$. Thus, $f(x)$ is one-one.

Determine the range of $f(x)$. Since $f(x)$ is one-one and increasing:

As $x \to -\infty$, $x^3 - 3x + 1 \to -\infty$, so $f(x) \to 0$. At $x = -1$,

$f(-1) = e^{(-1)^3 - 3(-1) + 1} = e^{1 + 3 + 1} = e^3.$

Thus, $a = 0$ and $b = e^3$, so the range of $f(x)$ is $(0, e^3]$.

Define point $P$ and line equation. The point $P$ is given by $P(2b + 4, 0 + 2)$. Substitute $a = 0$ and $b = e^3$:

$P = (2e^3 + 4, 0 + 2) = (2e^3 + 4, 2).$

The line equation is:

$x + e^{-3}y = 4$

Find the distance from $P$ to the line. The distance $d$ from a point $(x_1, y_1)$ to a line $Ax + By + C = 0$ is given by:

$d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$

Rewrite the line equation in standard form:

$x + e^{-3}y - 4 = 0$

Here, $A = 1$, $B = e^{-3}$, $C = -4$, and $(x_1, y_1) = (2e^3 + 4, 2)$.

Substitute these values into the distance formula:

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

Multiply the numerator and the denominator by $e^3$ to simplify:

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