Question

Let $f, g: (1, \infty) \rightarrow \mathbb{R}$ be defined as $f(x) = \frac{2x + 3}{5x + 2}$ and $g(x) = \frac{2 - 3x}{1 - x}$. If the range of the function $fog: [2, 4] \rightarrow \mathbb{R}$ is $[\alpha, \beta]$, then $\frac{1}{\beta - \alpha}$ is equal to

• A. 68
• B. 29
• C. 2
• D. 56

Hint:

The range of a composite function can be determined by evaluating the function at the endpoints of the domain.

Answer 1

The Correct Option is D

To find $\frac{1}{\beta - \alpha}$, we first need to determine the range of the function $fog(x) = f(g(x))$. 
1. Calculate $fog(x)$: \[ fog(x) = f\left(g(x)\right) = f\left(\frac{2 - 3x}{1 - x}\right) \] Substitute $g(x)$ into $f(x)$: \[ fog(x) = \frac{2\left(\frac{2 - 3x}{1 - x}\right) + 3}{5\left(\frac{2 - 3x}{1 - x}\right) + 2} \] Simplify the expression: \[ fog(x) = \frac{\frac{4 - 6x + 3 - 3x}{1 - x}}{\frac{10 - 15x + 2 - 2x}{1 - x}} = \frac{7 - 9x}{12 - 17x} \] 
2. Determine the range of $fog(x)$ for $x \in [2, 4]$: - Calculate $fog(2)$: \[ fog(2) = \frac{7 - 9(2)}{12 - 17(2)} = \frac{7 - 18}{12 - 34} = \frac{-11}{-22} = \frac{1}{2} \] - Calculate $fog(4)$: \[ fog(4) = \frac{7 - 9(4)}{12 - 17(4)} = \frac{7 - 36}{12 - 68} = \frac{-29}{-56} = \frac{29}{56} \] 
3. Identify $\alpha$ and $\beta$: 
- The range of $fog(x)$ is $\left[\frac{1}{2}, \frac{29}{56}\right]$. 
- Therefore, $\alpha = \frac{1}{2}$ and $\beta = \frac{29}{56}$. 
4. Calculate $\frac{1}{\beta - \alpha}$: \[ \beta - \alpha = \frac{29}{56} - \frac{1}{2} = \frac{29}{56} - \frac{28}{56} = \frac{1}{56} \] \[ \frac{1}{\beta - \alpha} = \frac{1}{\frac{1}{56}} = 56 \] Therefore, the correct answer is (4) 56.

Answer 2

We are given two functions:

\[ f(x) = \frac{2x + 3}{5x + 2}, \quad g(x) = \frac{2 - 3x}{1 - x}. \] We must find the range of \( f(g(x)) \) for \( x \in [2, 4] \), and then compute \( \dfrac{1}{\beta - \alpha} \), where the range is \([\alpha, \beta]\).

 

Step-by-Step Solution:

Step 1: Compute \( f(g(x)) \).

\[ f(g(x)) = \frac{2g(x) + 3}{5g(x) + 2}. \] Substitute \( g(x) = \frac{2 - 3x}{1 - x} \): \[ f(g(x)) = \frac{2\left(\frac{2 - 3x}{1 - x}\right) + 3}{5\left(\frac{2 - 3x}{1 - x}\right) + 2}. \]

Step 2: Simplify the numerator and denominator.

Numerator: \[ 2\left(\frac{2 - 3x}{1 - x}\right) + 3 = \frac{4 - 6x + 3(1 - x)}{1 - x} = \frac{4 - 6x + 3 - 3x}{1 - x} = \frac{7 - 9x}{1 - x}. \] Denominator: \[ 5\left(\frac{2 - 3x}{1 - x}\right) + 2 = \frac{10 - 15x + 2(1 - x)}{1 - x} = \frac{10 - 15x + 2 - 2x}{1 - x} = \frac{12 - 17x}{1 - x}. \] So, \[ f(g(x)) = \frac{7 - 9x}{12 - 17x}. \]

Step 3: Find the range of \( f(g(x)) \) for \( x \in [2, 4] \).

Let \( y = \frac{7 - 9x}{12 - 17x} \).

We’ll find \(y\) at \(x = 2\) and \(x = 4\).

At \(x = 2\): \[ y = \frac{7 - 18}{12 - 34} = \frac{-11}{-22} = \frac{1}{2}. \] At \(x = 4\): \[ y = \frac{7 - 36}{12 - 68} = \frac{-29}{-56} = \frac{29}{56}. \]

Step 4: Check if \(f(g(x))\) is increasing or decreasing.

Compute derivative sign (optional quick check): \[ f(g(x)) = \frac{7 - 9x}{12 - 17x}, \] Differentiate: \[ \frac{d}{dx} = \frac{(-9)(12 - 17x) - (7 - 9x)(-17)}{(12 - 17x)^2} = \frac{-108 + 153x + 119 - 153x}{(12 - 17x)^2} = \frac{11}{(12 - 17x)^2} > 0. \] So the function is **increasing**. Hence, range = \([f(g(2)), f(g(4))] = [\frac{1}{2}, \frac{29}{56}]\).

 

Step 5: Compute \(\frac{1}{\beta - \alpha}\).

\[ \beta - \alpha = \frac{29}{56} - \frac{1}{2} = \frac{29}{56} - \frac{28}{56} = \frac{1}{56}. \] \[ \frac{1}{\beta - \alpha} = 56. \]

Final Computation & Result:

\[ \boxed{56} \]