Question

Let $f: R \rightarrow R$ be defined by $f(x) = 5^{|x|} + \text{sgn}(5^{-x})$, where sgn x denotes signum function of x. Then f is

• A. one-one but not onto
• B. onto but not one-one
• C. both one-one and onto
• D. neither one-one nor onto

Hint:

To test if a function is one-one, check if $f(x_1) = f(x_2)$ implies $x_1 = x_2$. A quick way is to find a counterexample, like $f(a) = f(-a)$ for functions involving $|x|$ or $x^2$. To test for onto, find the range of the function and compare it with the given co-domain.

Answer 1

The Correct Option is D

The given function is $f(x) = 5^{|x|} + \text{sgn}(5^{-x})$.
The signum function, $\text{sgn}(y)$, is defined as 1 for $y>0$, 0 for $y = 0$, and -1 for $y<0$.
For any real number $x$, the term $5^{-x}$ is always a positive value, i.e., $5^{-x}>0$.
Therefore, $\text{sgn}(5^{-x}) = 1$ for all $x \in R$.
The function simplifies to $f(x) = 5^{|x|} + 1$.
To check for one-one (injective):
Let's consider $f(1)$ and $f(-1)$.
$f(1) = 5^{|1|} + 1 = 5^{1} + 1 = 6$.
$f(-1) = 5^{|-1|} + 1 = 5^{1} + 1 = 6$.
Since $f(1) = f(-1)$ but $1 \neq -1$, the function is not one-one.
To check for onto (surjective):
The co-domain of the function is $R$ (all real numbers). Let's find the range.
For any real number $x$, we have $|x| \geq 0$.
This implies $5^{|x|} \geq 5^{0}$, which means $5^{|x|} \geq 1$.
Therefore, $f(x) = 5^{|x|} + 1 \geq 1 + 1 = 2$.
The range of the function is $[2, \infty)$.
Since the range $[2, \infty)$ is a proper subset of the co-domain $R$, the function is not onto.
Thus, the function is neither one-one nor onto.