Let f : (0, 1) → ℝ be a function defined by f(x) = 1/(1 - e⁻ˣ), and g(x) = (f(-x) - f(x)). Consider

Q: Wait, why do we need to simplify $g(x)$? Can't we just differentiate $f(-x) - f(x)$ directly?

You absolutely can! However, differentiating the raw expression $f(-x) - f(x)$ involves the chain rule on two separate, slightly messy terms. By simplifying $g(x)$ first, you transform a complex difference into a single, clean rational function involving $e^x$. This makes the subsequent differentiation much faster and significantly reduces the chance of algebraic errors.

Q: How does showing $g'(x) > 0$ prove that the function is one-one?

This is a fundamental theorem in calculus! If a function is continuous and its derivative is strictly positive (or strictly negative) over an interval, the function is strictly monotonic. A strictly monotonic function is always injective, meaning it passes the horizontal line test. Since $g'(x) = \frac{2e^x}{(1 - e^x)^2}$ is always positive for $x \in (0, 1)$, the function is strictly increasing, which guarantees it is one-one.

Q: Is there any risk of the denominator becoming zero in the domain $(0, 1)$?

That is a great observation! We must always check the domain. The function $f(x)$ is defined for $1 - e^{-x} \neq 0$, which means $e^{-x} \neq 1$, or $x \neq 0$. In our domain $(0, 1)$, $x$ is never zero, so $e^x$ is never $1$. Thus, the denominator $1 - e^x$ is never zero, and our function is well-defined and continuous throughout the entire interval.

Question

Let $f : (0, 1) \rightarrow \mathbb{R}$ be a function defined by $f(x) = \frac{1}{1 - e^{-x}}$, and $g(x) = (f(-x) - f(x))$. Consider two statements (I) $g$ is an increasing function in $(0, 1)$ (II) $g$ is one-one in $(0, 1)$ Then,

Options:

(A) Only (I) is true

(B) Only (II) is true

(C) Neither (I) nor (II) is true

(D) Both (I) and (II) are true

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler on the path to JEE mastery! Today, we are going to dissect a problem that, at first glance, looks like a messy algebraic trap.

We are given a function $f(x) = \frac{1}{1 - e^{-x}}$ and asked to analyze $g(x) = f(-x) - f(x)$. Many students see this and immediately reach for the quotient rule, ready to differentiate the raw expression.

But wait! Before we rush into the heavy machinery of calculus, let us pause and appreciate the beauty of algebraic manipulation.

### The Art of Simplification

The expression $f(x) = \frac{1}{1 - e^{-x}}$ is slightly clunky because of the negative exponent. Let us make it elegant by multiplying the numerator and denominator by $e^x$:

$$f(x) = \frac{e^x}{e^x - 1}$$

Now, let us find $f(-x)$. Substituting $-x$ for $x$, we get:

$$f(-x) = \frac{1}{1 - e^x}$$

Now, look at our target function $g(x) = f(-x) - f(x)$. Substituting our new forms, we have:

$$g(x) = \frac{1}{1 - e^x} - \frac{e^x}{e^x - 1}$$

Notice the denominators: $1 - e^x$ and $e^x - 1$. They are perfect opposites! If we pull a negative sign out of the second denominator, the expression becomes:

$$g(x) = \frac{1}{1 - e^x} + \frac{e^x}{1 - e^x}$$

Adding these together, we get the beautifully compact form:

$$g(x) = \frac{1 + e^x}{1 - e^x}$$

This is the power of simplification; we have turned a daunting difference into a single, manageable fraction.

### The Engine of the Function

Now that we have $g(x) = \frac{1 + e^x}{1 - e^x}$, checking if it is increasing is a breeze. We need the derivative $g'(x)$.

Using the quotient rule, where $u = 1 + e^x$ and $v = 1 - e^x$, we have $g'(x) = \frac{v u' - u v'}{v^2}$. Calculating the derivatives, $u' = e^x$ and $v' = -e^x$.

Substituting these into our formula:

$$g'(x) = \frac{(1 - e^x)(e^x) - (1 + e^x)(-e^x)}{(1 - e^x)^2}$$

Expanding the numerator, we get $e^x - e^{2x} + e^x + e^{2x}$. The $e^{2x}$ terms vanish into thin air, leaving us with:

$$g'(x) = \frac{2e^x}{(1 - e^x)^2}$$

This is the moment of truth. Since $e^x$ is always positive and the denominator is a squared term (and thus always positive for $x 
eq 0$), $g'(x)$ is strictly positive for all $x \in (0, 1)$.

### The Logical Conclusion

Because $g'(x) > 0$, we know $g(x)$ is a **strictly increasing function**.

And here is the final piece of the puzzle: any strictly monotonic function is inherently **one-one (injective)**. If a function is always climbing, it can never return to a previous value, meaning no two distinct inputs can yield the same output.

It passes the horizontal line test with flying colors. Thus, both statements (I) and (II) are **true**.

We didn't just solve a problem; we navigated the elegant structure of the function. Keep this mindset—simplify first, differentiate second—and you will conquer any function analysis problem the JEE throws at you!

Animated Solution for Mathematics - Differentiation: Let $f : (0, 1) \to R$ be a function defined by $f (x) = \frac{1}{1 - e ^{- x}}$ , and $g (x) = (f (- x) - f (x))$ . Consider two statements (I) $g$ is an increasing function in $(0, 1)$ (II) $g$ is one-one in $(0, 1)$ Then,

Select Answer:

Visualized Solution

Define $f (x)$ and $f (- x)$

Construct $g (x)$

Simplify the Second Term

Combine Fractions

Differentiate $g (x)$

Simplify $g^{'} (x)$

Analyze the Sign of $g^{'} (x)$

Check One-One Property

Final Conclusion

The Sigma Insight: Monotonicity

Analyzing the Setup

The Art of Simplification

The Engine of the Function

The Logical Conclusion

Similar Questions

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Comprehension Passage

Consider the statements: $P$ : There exists some $x \in R$ such that $f (x) + 2 x = 2 (1 + x^{2})$ , $Q$ : There exists some $x \in R$ such that $2 f (x) + 1 = 2 x (1 + x)$

Which of the following is true?

If $f (x) = x e^{x (1 - x)}$ , then $f (x)$ is

The function defined by $f (x) = (x + 2) e^{- x}$ is

Select Answer:

Visualized Solution

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

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