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

The Correct Option is A

Given:

The function is: \[ f(x) = e^{x^3 - 3x + 1} \]

Step 1: Finding the derivative of \( f(x) \):

The derivative \( f'(x) \) is calculated using the chain rule: \[ f'(x) = e^{x^3 - 3x + 1} \cdot (3x^2 - 3) \] Simplifying: \[ f'(x) = e^{x^3 - 3x + 1} \cdot 3(x - 1)(x + 1) \]

Step 2: Determining the behavior of the function:

For \( f'(x) \geq 0 \), we conclude that: \[ f(x) \text{ is an increasing function.} \]

Step 3: Solving for \( a \) and \( b \):

The values of \( a \) and \( b \) are: \[ a = e^{-\infty} = 0, \quad f(-\infty) = 0 \] \[ b = e^{-1 + 3 + 1} = e^3, \quad f(-1) = e^3 \]

Step 4: Point \( P \):

We are given that: \[ P(2b + 4, a + 2) \] Thus: \[ P(2e^3 + 4, 2) \]

Step 5: Distance calculation:

The distance \( d \) from the point \( P \) to the curve is given by: \[ d = \frac{(2e^3 + 4) + 2e^{-3} - 4}{\sqrt{1 + e^{-6}}} \] Simplifying: \[ d = 2 \sqrt{1 + e^{-6}} \]

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:

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