Let f, g : (1, ∞) → ℝ be defined as f(x) = (2x+3)/(5x+2) and g(x) = (2-3x)/(1-x). If the range of th

Q: Why can't we just plug in the endpoints without checking the derivative?

That is a dangerous trap! If a function has a turning point (a local maximum or minimum) inside the interval, the range is not just the values at the endpoints. You must always check for monotonicity or critical points first to ensure the function is strictly increasing or decreasing on that interval.

Q: What if the vertical asymptote was inside the interval $[2, 4]$?

If the vertical asymptote $x = \frac{12}{17}$ were inside $[2, 4]$, the function would be discontinuous. The range would then be $(-\infty, \infty)$ or split into two infinite branches, making the concept of a simple range $[\alpha, \beta]$ invalid. Always check the domain of your composite function first!

Q: Is there a way to avoid the quotient rule when differentiating?

You could use the chain rule: $h'(x) = f'(g(x)) \cdot g'(x)$. However, for rational functions, simplifying the expression first is often less prone to calculation errors than applying the chain rule to complex fractions. The quotient rule is your best friend here.

Question

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

Options:

(A) 68

(B) 29

(C) 2

(D) 56

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineers! Today, we are embarking on a journey to master the composition of functions. It is not just about plugging one equation into another; it is about understanding the geometric behavior of the resulting machine.

We are given two rational functions, $f(x) = \frac{2x+3}{5x+2}$ and $g(x) = \frac{2-3x}{1-x}$. Our mission is to find the range of their composition, $h(x) = f(g(x))$, on the interval $[2, 4]$.

### The Algebraic Transformation

Imagine you are building a machine where the output of $g(x)$ becomes the input for $f(x)$. To find $h(x)$, we substitute $g(x)$ into $f(x)$:

$$h(x) = \frac{2\left(\frac{2-3x}{1-x}\right) + 3}{5\left(\frac{2-3x}{1-x}\right) + 2}$$

This looks intimidating, but let us simplify it. By multiplying the numerator and denominator by $(1-x)$, we clear the fractions.

The numerator becomes:
$$N = 2(2-3x) + 3(1-x) = 4 - 6x + 3 - 3x = 7 - 9x$$

The denominator becomes:
$$D = 5(2-3x) + 2(1-x) = 10 - 15x + 2 - 2x = 12 - 17x$$

Thus, our composite function is:
$$h(x) = \frac{7-9x}{12-17x}$$

### The Power of Monotonicity

Now, we must ask: how does this function behave? Is it climbing or falling? We use the derivative to find out.

Applying the quotient rule:
$$h'(x) = \frac{-9(12-17x) - (-17)(7-9x)}{(12-17x)^2}$$

Expanding this, we get:
$$h'(x) = \frac{-108 + 153x + 119 - 153x}{(12-17x)^2} = \frac{11}{(12-17x)^2}$$

Since the square in the denominator is always positive, $h'(x) > 0$. This means our function is **strictly increasing**. This is a massive relief because it tells us that the minimum value occurs at the start of our interval and the maximum at the end.

### The Final Calculation

Since $h(x)$ is strictly increasing, the range on $[2, 4]$ is simply $[h(2), h(4)]$.

Evaluating at $x=2$:
$$\alpha = h(2) = \frac{7-9(2)}{12-17(2)} = \frac{7-18}{12-34} = \frac{-11}{-22} = \frac{1}{2}$$

Evaluating at $x=4$:
$$\beta = h(4) = \frac{7-9(4)}{12-17(4)} = \frac{7-36}{12-68} = \frac{-29}{-56} = \frac{29}{56}$$

Finally, we calculate the difference:
$$\beta - \alpha = \frac{29}{56} - \frac{1}{2} = \frac{29-28}{56} = \frac{1}{56}$$

The problem asks for the reciprocal of this value, $\frac{1}{\beta-\alpha}$, which is:
$$\frac{1}{1/56} = \mathbf{56}$$

You have just conquered a complex composite function problem with elegance and precision. Keep this mindset, and no JEE problem will ever stand in your way!

Animated Solution for Mathematics - Functions: Let $f, g : (1, \infty) \to R$ be defined as $f (x) = \frac{2 x + 3}{5 x + 2}$ and $g (x) = \frac{2 - 3 x}{1 - x}$ . If the range of the function $f \circ g : [2, 4] \to R$ is $[α, β]$ , then $\frac{1}{β - α}$ is equal to

Select Answer:

Visualized Solution

Understanding the Goal

Formulating $f (g (x))$

Simplifying the Numerator

Simplifying the Denominator

The Final Form of $h (x)$

Derivative of $h (x)$

Calculating $α = h (2)$

Calculating $β = h (4)$

Finding $β - α$

Computing $\frac{1}{β - α}$

The Sigma Insight: Composite Functions

Analyzing the Setup

The Algebraic Transformation

The Power of Monotonicity

The Final Calculation

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Select Answer:

Visualized Solution

Understanding the Goal

Formulating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(g(x))

Simplifying the Numerator

Simplifying the Denominator

The Final Form of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } h(x)

Derivative of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } h(x)

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α=h(2)

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } β=h(4)

Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } β−α

Computing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } β−α1​