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Limits, Continuity and Differentiability: Mathematics | JEE Main

Q: Does strictly increasing always guarantee a function is one-one?

Yes, on any connected interval in , if , so is impossible for distinct inputs.

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,

Your answer stays private

What feels right?

No score. Commit to your first instinct. We’ll show what your mind noticed and what it missed.

ABoth (I) and (II) are trueBNeither (I) nor (II) is trueCOnly (I) is trueDOnly (II) is true

Step-by-step solutionView

Correct answer

Simplifying $g(x)$ yields $g(x) = -\coth(x/2)$ , whose derivative $g'(x) = \frac{1}{2}\operatorname{csch}^2(x/2) > 0$ on $(0,1)$ , meaning $g$ is strictly increasing and therefore one-one; thus both statements are true.

Option analysis

Why each option works or fails

A · Both (I) and (II) are true

None. The student correctly analyzed the monotonicity and injectivity of $g(x)$ . Simplifying $g(x) = f(-x) - f(x)$ gives $g(x) = \frac{1}{1-e^x} - \frac{1}{1-e^{-x}} = -\frac{e^x+1}{e^x-1}$ . Its derivative is strictly positive on $(0,1)$ , proving $g$ is strictly increasing (I) and hence injective (II).

B · Neither (I) nor (II) is true

The student made a sign error when computing $f(-x)$ or differentiated without accounting for the negative sign in the exponent, concluding $g'(x) < 0$ or that $g$ is non-monotonic. Check the algebra: $f(-x) = \frac{1}{1-e^x} = -\frac{e^{-x}}{1-e^{-x}}$ . Then $g(x) = -\frac{1+e^{-x}}{1-e^{-x}}$ , which increases strictly as $x$ increases from $0$ to $1$ .

C · Only (I) is true

The student believed that a strictly increasing function on an open interval can fail to be one-one (perhaps confusing injectivity with surjectivity). Any strictly monotonic function defined on an interval is necessarily one-to-one (injective). If (I) holds, (II) must follow immediately.

D · Only (II) is true

The student verified injectivity algebraically (e.g., setting $g(x_1) = g(x_2)$ ) but thought $g(x)$ was decreasing rather than increasing due to misinterpreting the sign of $g'(x)$ . Differentiate $g(x) = -1 - \frac{2}{e^x-1}$ to get $g'(x) = \frac{2e^x}{(e^x-1)^2} > 0$ for all $x \in (0,1)$ , confirming it is strictly increasing.

Reviewed route

Solution

StepWorking

01given

$f(x) = \frac{1}{1-e^{-x}}$ on $(0,1)$ , and $g(x) = f(-x) - f(x)$ .

02goal

Determine whether $g(x)$ is strictly increasing and whether $g(x)$ is one-one on the interval $(0,1)$ .

03approach

Simplify the expression for $g(x)$ , compute its first derivative $g'(x)$ , and check the sign of $g'(x)$ on $(0,1)$ . A strictly monotonic function on an interval is necessarily one-one.

04execute

Substitute $f(-x) = \frac{1}{1-e^x}$ and $f(x) = \frac{e^x}{e^x-1}$ into $g(x)$ : $g(x) = \frac{1}{1-e^x} - \frac{e^x}{e^x-1} = \frac{1}{1-e^x} + \frac{e^x}{1-e^x} = \frac{1+e^x}{1-e^x}$ Differentiate with respect to $x$ : $g'(x) = \frac{(1-e^x)(e^x) - (1+e^x)(-e^x)}{(1-e^x)^2} = \frac{e^x - e^{2x} + e^x + e^{2x}}{(1-e^x)^2} = \frac{2e^x}{(1-e^x)^2}$ For all $x \in (0,1)$ , $2e^x > 0$ and $(1-e^x)^2 > 0$ , so $g'(x) > 0$ .

✓verify

Since $g'(x) > 0$ for all $x \in (0,1)$ , $g(x)$ is strictly increasing in $(0,1)$ , making Statement (I) true. Any strictly monotonic function on an interval is strictly injective (one-one), making Statement (II) true as well. Thus, both statements are true.

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Concepts assessed: Mathematics
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 8 September 2026

Quick checks

Students also ask

Does strictly increasing always guarantee a function is one-one?

Yes, on any connected interval in $\mathbb{R}$ , if $x_1 < x_2 \implies g(x_1) < g(x_2)$ , so $g(x_1) = g(x_2)$ is impossible for distinct inputs.

Answer

Simplifying $g(x)$ yields $g(x) = -\coth(x/2)$ , whose derivative $g'(x) = \frac{1}{2}\operatorname{csch}^2(x/2) > 0$ on $(0,1)$ , meaning $g$ is strictly increasing and therefore one-one; thus both statements are true.

Why each option works or fails

A · correct: Both (I) and (II) are true - None. The student correctly analyzed the monotonicity and injectivity of $g(x)$ . Simplifying $g(x) = f(-x) - f(x)$ gives $g(x) = \frac{1}{1-e^x} - \frac{1}{1-e^{-x}} = -\frac{e^x+1}{e^x-1}$ . Its derivative is strictly positive on $(0,1)$ , proving $g$ is strictly increasing (I) and hence injective (II).
B: Neither (I) nor (II) is true - The student made a sign error when computing $f(-x)$ or differentiated without accounting for the negative sign in the exponent, concluding $g'(x) < 0$ or that $g$ is non-monotonic. Check the algebra: $f(-x) = \frac{1}{1-e^x} = -\frac{e^{-x}}{1-e^{-x}}$ . Then $g(x) = -\frac{1+e^{-x}}{1-e^{-x}}$ , which increases strictly as $x$ increases from $0$ to $1$ .
C: Only (I) is true - The student believed that a strictly increasing function on an open interval can fail to be one-one (perhaps confusing injectivity with surjectivity). Any strictly monotonic function defined on an interval is necessarily one-to-one (injective). If (I) holds, (II) must follow immediately.
D: Only (II) is true - The student verified injectivity algebraically (e.g., setting $g(x_1) = g(x_2)$ ) but thought $g(x)$ was decreasing rather than increasing due to misinterpreting the sign of $g'(x)$ . Differentiate $g(x) = -1 - \frac{2}{e^x-1}$ to get $g'(x) = \frac{2e^x}{(e^x-1)^2} > 0$ for all $x \in (0,1)$ , confirming it is strictly increasing.

Step-by-step solution

given: $f(x) = \frac{1}{1-e^{-x}}$ on $(0,1)$ , and $g(x) = f(-x) - f(x)$ .
goal: Determine whether $g(x)$ is strictly increasing and whether $g(x)$ is one-one on the interval $(0,1)$ .
approach: Simplify the expression for $g(x)$ , compute its first derivative $g'(x)$ , and check the sign of $g'(x)$ on $(0,1)$ . A strictly monotonic function on an interval is necessarily one-one.
execute: Substitute $f(-x) = \frac{1}{1-e^x}$ and $f(x) = \frac{e^x}{e^x-1}$ into $g(x)$ : $g(x) = \frac{1}{1-e^x} - \frac{e^x}{e^x-1} = \frac{1}{1-e^x} + \frac{e^x}{1-e^x} = \frac{1+e^x}{1-e^x}$ Differentiate with respect to $x$ : $g'(x) = \frac{(1-e^x)(e^x) - (1+e^x)(-e^x)}{(1-e^x)^2} = \frac{e^x - e^{2x} + e^x + e^{2x}}{(1-e^x)^2} = \frac{2e^x}{(1-e^x)^2}$ For all $x \in (0,1)$ , $2e^x > 0$ and $(1-e^x)^2 > 0$ , so $g'(x) > 0$ .
verify: Since $g'(x) > 0$ for all $x \in (0,1)$ , $g(x)$ is strictly increasing in $(0,1)$ , making Statement (I) true. Any strictly monotonic function on an interval is strictly injective (one-one), making Statement (II) true as well. Thus, both statements are true.

Shortcut:

When to use it: When recognising standard combinations of exponentials as hyperbolic functions.

given: $g(x) = \frac{1+e^x}{1-e^x}$ on $(0,1)$ .

approach: Divide numerator and denominator by $e^{x/2}$ to recognize hyperbolic cotangent: $g(x) = -\coth(x/2)$ .

execute: $g(x) = -\frac{e^{x/2}+e^{-x/2}}{e^{x/2}-e^{-x/2}} = -\coth(x/2)$ . Since $\coth(u)$ is strictly decreasing for $u > 0$ , $-\coth(u)$ is strictly increasing for $u > 0$ . Thus $g(x)$ is strictly increasing on $(0,1)$ , which directly implies it is one-one.

verify: At $x \to 0^+$ , $-\coth(x/2) \to -\infty$ . At $x = 1$ , $-\coth(1/2) \approx -2.16$ . The function smoothly increases from $-\infty$ to $-2.16$ , confirming it is strictly increasing and one-one.

Limits, Continuity and Differentiability: Mathematics | JEE Main

What feels right?

Simplifying $g(x)$ yields $g(x) = -\coth(x/2)$ , whose derivative $g'(x) = \frac{1}{2}\operatorname{csch}^2(x/2) > 0$ on $(0,1)$ , meaning $g$ is strictly increasing and therefore one-one; thus both statements are true.

Why each option works or fails

Solution

01given

02goal

03approach

04execute

✓verify

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