Sets, Relations & Functions
Injectivity and Surjectivity
nta_pyq_2025_apr
Grade 11
Question:
The function $f : (-\infty, \infty) \to (-\infty, 1)$, defined by $f(x) = \dfrac{2^x - 2^{-x}}{2^x + 2^{-x}}$ is:
Neither one-one nor onto
Onto but not one-one
Both one-one and onto
One-one but not onto
Step-by-Step Solution
Key Concept: $f'(x) > 0$ (always positive) $\to$ strictly increasing $\to$ one-one. The actual range is $(-1,1)$ (since $f\to-1$ as $x\to-\infty$ and $f\to1$ as $x\to+\infty$), which does not equal the codomain $(-\infty,1)$ $\to$ not onto.
$f'(x) = \frac{2\ln2 \cdot 2^{2x}}{(2^{2x}+1)^2} > 0$ $\to$ one-one. As $x\to\pm\infty$, $f\to\pm1$. Range $=(-1,1) \neq (-\infty,1)$ $\to$ not onto. One-one but not onto.
Correct Answer: One-one but not onto