JEE Main 2025. Previous Year Question — exact date and shift not confidently identified.
Problem
The function $f:(-\infty,\infty)\to(-\infty,1)$, defined by
$$f(x) = \frac{2^x – 2^{-x}}{2^x + 2^{-x}}$$
is:
Key insight. This is the familiar $\tanh$-shaped function in disguise. Rewriting it as $1 – \dfrac{2}{4^x+1}$ makes both properties easy to read off at once: it’s clearly strictly increasing (so one-one), but its actual output range turns out to be a much narrower interval than the stated codomain (so not onto).
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Approach
Multiply numerator and denominator by $2^x$ to clear the negative exponents, turning the function into a clean expression in $4^x$. Splitting that expression as “1 minus something” makes the behaviour at $x\to\pm\infty$ immediate — which settles both the range (for “onto”) and the monotonicity (for “one-one”) in one move.
Solution
Step 1 — Simplify by multiplying through by 2ˣ
$$f(x) = \frac{2^x – 2^{-x}}{2^x + 2^{-x}} \times \frac{2^x}{2^x} = \frac{2^{2x} – 1}{2^{2x}+1} = \frac{4^x – 1}{4^x + 1}$$
Step 2 — Rewrite as “1 minus a fraction”
$$f(x) = \frac{(4^x+1) – 2}{4^x+1} = 1 – \frac{2}{4^x+1}$$
Step 3 — Check onto, using the limiting behaviour
As $x \to -\infty$, $4^x \to 0^+$, so $\dfrac{2}{4^x+1} \to 2$, giving $f(x) \to 1-2 = -1$ — but this limit is never actually reached, since $4^x > 0$ strictly for every real $x$.
As $x \to +\infty$, $4^x \to \infty$, so $\dfrac{2}{4^x+1} \to 0$, giving $f(x) \to 1$ — again, only approached, never reached.
So the actual range of $f$ is the open interval $(-1, 1)$.
The stated codomain is $(-\infty, 1)$, which contains plenty of values — for instance $-5$ — that $f$ never produces. Since range $\subsetneq$ codomain, $f$ is not onto.
Step 4 — Check one-one, using monotonicity
$4^x$ is strictly increasing for all real $x$, so $4^x+1$ is strictly increasing, so $\dfrac{2}{4^x+1}$ is strictly decreasing, and therefore
$$f(x) = 1 – \frac{2}{4^x+1}$$
is strictly increasing on its entire domain. A strictly monotonic function can never repeat an output value, so $f$ is one-one.
Answer
(D) One-one but not onto.
Common mistakes
- Assuming the stated codomain automatically equals the range. The problem explicitly hands you a codomain of $(-\infty,1)$ — but that’s just where the problem says outputs could land, not a guarantee that every value there is actually achieved. The actual range has to be worked out, not assumed.
- Checking monotonicity by testing a few points instead of proving it algebraically. Spotting that $4^x$ is strictly increasing (and building the rest of the argument on that single fact) is far more reliable than checking a handful of values and hoping the pattern holds everywhere.
Practise next
- Determine whether $g(x) = \dfrac{3^x – 3^{-x}}{3^x + 3^{-x}}$, with the same domain and codomain structure, is one-one and/or onto, using the identical rewriting technique.
Show answer
One-one, but not onto $\mathbb{R}$. Dividing through by $3^x$ gives $g(x)=\dfrac{9^{x}-1}{9^{x}+1}$, which is $\tanh(x\ln3)$.
Writing $u=9^{x}>0$, $g=\dfrac{u-1}{u+1}=1-\dfrac{2}{u+1}$ is strictly increasing in $u$, and $u$ is strictly increasing in $x$ — so $g$ is one-one.
As $u$ runs over $(0,\infty)$, $g$ runs over $(-1,1)$ and never reaches the endpoints. So $g$ is not onto $\mathbb{R}$; it is a bijection onto $(-1,1)$.
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