Question

Let \( a>1 \) and \( 0<b<1 \). -∞\( f : \mathbb{R} \to [0, 1] \) is defined by \( f(x) = \begin{cases} a^x & \text{if } x<0 b^x & \text{if } 0 \leq x \leq 1 \end{cases} \), then \( f(x) \) is:

• A. A bijection
• B. One-one but not onto
• C. Onto but not one-one
• D. Neither one-one nor onto

Hint:

To determine whether a piecewise function is one-one or onto, analyze each piece of the function separately, and check if the function covers the entire range and if distinct inputs lead to distinct outputs.

Answer 1

The Correct Option is D

We are given the function \( f(x) \) defined as: \[ f(x) = \begin{cases} a^x & \text{if } x<0 b^x & \text{if } 0 \leq x \leq 1 \end{cases} \] with \( a>1 \) and \( 0<b<1 \). 
Step 1: 
We need to analyze whether \( f(x) \) is one-one (injective) and onto (surjective).
- Checking if \( f(x) \) is one-one (injective):
A function is one-one if distinct inputs lead to distinct outputs, i.e., \( f(x_1) = f(x_2) \Rightarrow x_1 = x_2 \).
For \( x<0 \), \( f(x) = a^x \), and since \( a>1 \), the function \( f(x) = a^x \) is strictly decreasing. 
Therefore, for \( x_1<x_2 \), \( f(x_1)>f(x_2) \), so the function is injective for \( x<0 \).
For \( 0 \leq x \leq 1 \), \( f(x) = b^x \), and since \( 0<b<1 \), the function \( f(x) = b^x \) is strictly decreasing. 
Therefore, for \( x_1<x_2 \), \( f(x_1)>f(x_2) \), so the function is also injective for \( 0 \leq x \leq 1 \).
- Checking if \( f(x) \) is onto (surjective):
For \( f(x) \) to be onto, for every \( y \in [0, 1] \), there must be an \( x \in \mathbb{R} \) such that \( f(x) = y \).
- For \( x<0 \), \( f(x) = a^x \) takes values in \( (0, 1) \), but does not cover the entire range \( [0, 1] \) because the function does not include 0.
- For \( 0 \leq x \leq 1 \), \( f(x) = b^x \) takes values in \( (0, 1) \), but also does not cover the entire range \( [0, 1] \) because the function does not reach 1.
Thus, the function is not onto because it does not cover the entire range \( [0, 1] \). 
Step 2: 
Since \( f(x) \) is injective but not surjective, we conclude that the function is neither one-one nor onto. 
Thus, the correct answer is: \( \boxed{\text{Neither one-one nor onto}} \).

Answer 2

Given \( a > 1 \) and \( 0 < b < 1 \), the function \( f : \mathbb{R} \to [0, 1] \) is defined as:
\[ f(x) = \begin{cases} a^x & \text{if } x < 0 \\[6pt] b^x & \text{if } 0 \leq x \leq 1 \end{cases} \]

Step 1: Consider the behavior for \( x < 0 \):
Since \( a > 1 \), \( a^x \) is a decreasing function for negative \( x \), with range \( (0, 1) \) as \( x \to 0^- \) gives \( a^0 = 1 \), and as \( x \to -\infty \), \( a^x \to 0 \).

Step 2: Consider the behavior for \( 0 \leq x \leq 1 \):
Since \( 0 < b < 1 \), \( b^x \) is decreasing on \( [0, 1] \), with:
\[ f(0) = b^0 = 1, \quad f(1) = b^1 = b \in (0, 1) \]

Step 3: Note that \( f \) is continuous at \( x=0 \) because:
\[ \lim_{x \to 0^-} f(x) = a^0 = 1 = f(0) = b^0 \]

Step 4: Check if \( f \) is one-one (injective):
- On \( (-\infty, 0) \), \( f \) is strictly decreasing from 0 to 1.
- On \( [0,1] \), \( f \) is strictly decreasing from 1 to \( b \).
But the value 1 is attained at both \( x=0^- \) and \( x=0 \). Thus, \( f \) is not one-one over \( \mathbb{R} \).

Step 5: Check if \( f \) is onto \( [0,1] \):
- On \( (-\infty,0) \), \( f \) takes all values in \( (0,1) \).
- On \( [0,1] \), \( f \) takes values in \( [b,1] \subset (0,1] \).
- The value 0 is not attained by \( f \) (only approached as \( x \to -\infty \)), so \( f \) is not onto \( [0,1] \).

Therefore,
\[ \boxed{\text{Neither one-one nor onto}} \]