Sets, Relations & Functions
Composition — One-One / Onto
nta_pyq_2024_apr
Grade 11
Question:
Let $f,g:\mathbb{R}\to\mathbb{R}$ be defined as $f(x)=|x-1|$ and $g(x)=\begin{cases}e^x, & x\geq0\\ x+1, & x\leq0\end{cases}$. Then the function $f(g(x))$ is:
neither one-one nor onto.
one-one but not onto.
onto but not one-one.
both one-one and onto.
Step-by-Step Solution
Key Concept: $f(g(x))=\begin{cases}e^x-1, & x\geq0\\ -x, & x\leq0\end{cases}$. Both pieces are injective and the combined function covers $[0,\infty)$.
One-one but not onto (range $=[0,\infty)$).
Correct Answer: 1