Let $f, g: \mathbb{R} \rightarrow \mathbb{R}$ be defined as: $f(x) = |x - 1|$ and $g(x) = \begin{cases} e^x, & x \geq 0 \\ x + 1, & x \leq 0 \end{cases}$ Then the function $f(g(x))$ is
- neither one-one nor onto.
- one-one but not onto.
- both one-one and onto.
- onto but not one-one.
The Correct Option is A
Solution and Explanation
To find \( f(g(x)) \), we first evaluate \( g(x) \) based on the value of \( x \).
\[ g(x) = \begin{cases} e^x, & x \geq 0 \\ x + 1, & x \leq 0 \end{cases} \]
The function \( f \) is defined as \( f(x) = |x - 1| \). Therefore, we have: \[ f(g(x)) = |g(x) - 1|. \]
Case 1: When \( x \geq 0 \)
\[ f(g(x)) = |e^x - 1|. \]
Case 2: When \( x \leq 0 \)
\[ f(g(x)) = |x + 1 - 1| = |x| = -x \quad \text{(since \( x \leq 0 \))}. \]
Analysis of \( f(g(x)) \) - For \( x \geq 0 \), \( f(g(x)) = |e^x - 1| \) is neither one-one nor onto because it cannot cover all values in the codomain (as it is non-negative). For \( x \leq 0 \), \( f(g(x)) = -x \) is also neither one-one nor onto due to its behavior as a non-injective transformation on the interval.
Therefore, the function \( f(g(x)) \) is neither one-one nor onto.
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