Q64 / 180 · COMEDK 2024 (Morning Shift) · Mathematics
Let $f: R \rightarrow R$ be a function defined by $f=\dfrac{e^{|x|}-e^{-x}}{e^x+e^{-x}}$ then
- A$f$ is injection and surjection
- B$f$ is a surjection but not an injection function
- C$f$ is an injection but not a surjection function
- Correct answerD$f$ is neither an injection nor a surjection
Solution
The function is defined as $f(x) = \dfrac{e^{|x|} - e^{-x}}{e^x + e^{-x}}$.
Case 1: $x \ge 0$. Then $|x| = x$, so $f(x) = \dfrac{e^x - e^{-x}}{e^x + e^{-x}} = \tanh(x)$. For $x \ge 0$, $\tanh(x)$ ranges from $0$ to $1$ (exclusive of $1$ as $x \to \infty$).
Case 2: $x < 0$. Then $|x| = -x$, so $f(x) = \dfrac{e^{-x} - e^{-x}}{e^x + e^{-x}} = \dfrac{0}{e^x + e^{-x}} = 0$.
Thus, the function is $f(x) = 0$ for all $x \le 0$ and $f(x) = \tanh(x)$ for $x > 0$.
Since $f(x) = 0$ for all $x \in (-\infty, 0]$, the function is not injective because $f(0) = f(-1) = 0$.
The range of the function is $[0, 1)$. Since the codomain is $R$, the range is not equal to the codomain, so the function is not surjective.
Answer: $f$ is neither an injection nor a surjection
Case 1: $x \ge 0$. Then $|x| = x$, so $f(x) = \dfrac{e^x - e^{-x}}{e^x + e^{-x}} = \tanh(x)$. For $x \ge 0$, $\tanh(x)$ ranges from $0$ to $1$ (exclusive of $1$ as $x \to \infty$).
Case 2: $x < 0$. Then $|x| = -x$, so $f(x) = \dfrac{e^{-x} - e^{-x}}{e^x + e^{-x}} = \dfrac{0}{e^x + e^{-x}} = 0$.
Thus, the function is $f(x) = 0$ for all $x \le 0$ and $f(x) = \tanh(x)$ for $x > 0$.
Since $f(x) = 0$ for all $x \in (-\infty, 0]$, the function is not injective because $f(0) = f(-1) = 0$.
The range of the function is $[0, 1)$. Since the codomain is $R$, the range is not equal to the codomain, so the function is not surjective.
Answer: $f$ is neither an injection nor a surjection
Hence, the correct answer is option: 'D'.