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: \( \boxed{\text{Neither one-one nor onto}} \).