JEE MainMathematics

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Monotonicity and Injectivity of Exponential Difference Functions

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

Yes, on any connected interval in 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.

Let f:(0,1) → ℝ be a function defined by f(x) = 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,

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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 $g^{'} (x) = \frac{1}{2} {csch}^{2} (x / 2) > 0$ 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

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 $g^{'} (x) = \frac{2 e^{x}}{(e^{x} - 1)^{2}} > 0$ 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 $x$ : $g^{'} (x) = \frac{(1 - e^{x}) (e^{x}) - (1 + e^{x}) (- e^{x})}{(1 - e^{x})^{2}} = \frac{e^{x} - e^{2 x} + e^{x} + e^{2 x}}{(1 - e^{x})^{2}} = \frac{2 e^{x}}{(1 - e^{x})^{2}}$ For all $x \in (0, 1)$ , $2 e^{x} > 0$ 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
Academic status: Answer verified against the official NTA key
Source: Previous-year question
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 $R$ , if $x_{1} < x_{2} ⟹ g (x_{1}) < g (x_{2})$ , so $g (x_{1}) = g (x_{2})$ is impossible for distinct inputs.

Answer

Simplifying $g (x)$ yields $g (x) = - \coth (x / 2)$ , whose derivative $g^{'} (x) = \frac{1}{2} {csch}^{2} (x / 2) > 0$ 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 - 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 $g^{'} (x) = \frac{2 e^{x}}{(e^{x} - 1)^{2}} > 0$ 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 $x$ : $g^{'} (x) = \frac{(1 - e^{x}) (e^{x}) - (1 + e^{x}) (- e^{x})}{(1 - e^{x})^{2}} = \frac{e^{x} - e^{2 x} + e^{x} + e^{2 x}}{(1 - e^{x})^{2}} = \frac{2 e^{x}}{(1 - e^{x})^{2}}$ For all $x \in (0, 1)$ , $2 e^{x} > 0$ 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.

Monotonicity and Injectivity of Exponential Difference Functions

What feels right?

Simplifying $g (x)$ yields $g (x) = - \coth (x / 2)$ , whose derivative $g^{'} (x) = \frac{1}{2} {csch}^{2} (x / 2) > 0$ 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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Students also ask

Monotonicity and Injectivity of Exponential Difference Functions

Simplifying g(x)g(x) yields g(x)=−coth⁡(x/2)g(x)=−coth(x/2), whose derivative g′(x)=12csch⁡2(x/2)>0g′(x)=21​csch2(x/2)>0 on (0,1)(0,1), meaning gg is strictly increasing and therefore one-one; thus both statements are true.

01Given

02Goal

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Students also ask

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

The number of points on the curve $y = 54 x^{5} - 135 x^{4} - 70 x^{3} + 180 x^{2} + 210 x$ at whic…

The area of the region enclosed by the parabola $y = 4 x - x^{2}$ and…