Question:
If a real valued function \( f:(1, 2] \rightarrow B \) defined by \( f(x) = \log_{10}(x-1) \) is a bijection, then \( B = \)
If a real valued function \( f:(1, 2] \rightarrow B \) defined by \( f(x) = \log_{10}(x-1) \) is a bijection, then \( B = \)
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To find the range of a composite function, determine the output interval of the inner function first, then apply the outer function over that specific interval.
Updated On: Jun 9, 2026
- \([0, \infty) \)
- \( \mathbb{R} \)
- \( (100, 0] \)
- \( (-\infty, 0] \)
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The Correct Option is D
Solution and Explanation
Concept:
A function is a bijection if it is both injective (one-to-one) and surjective (onto). For a function \( f: A \rightarrow B \) to be a bijection, the set \( B \) must be equal to the range of the function \( f(x) \).
\[
\text{Range}(f) = \{ f(x) : x \in \text{Domain}(f) \}
\]
Step 1: Identify the domain of the function.
The given domain is \( x \in (1, 2] \).
Step 2: Determine the range of \( x-1 \).
Since \( 1 < x \le 2 \), subtracting 1 from all parts gives: \( 0 < x - 1 \le 1 \).
Step 3: Apply the \( \log_{10} \) function to the inequality.
Since \( \log_{10} \) is a strictly increasing function: \( \log_{10}(0+) < \log_{10}(x-1) \le \log_{10}(1) \) \( -\infty < f(x) \le 0 \)
Step 4: Define the set \( B \).
For the function to be surjective, \( B \) must be the range of \( f(x) \), which is \((-\infty, 0]\). B = (-, 0]
Step 1: Identify the domain of the function.
The given domain is \( x \in (1, 2] \).
Step 2: Determine the range of \( x-1 \).
Since \( 1 < x \le 2 \), subtracting 1 from all parts gives: \( 0 < x - 1 \le 1 \).
Step 3: Apply the \( \log_{10} \) function to the inequality.
Since \( \log_{10} \) is a strictly increasing function: \( \log_{10}(0+) < \log_{10}(x-1) \le \log_{10}(1) \) \( -\infty < f(x) \le 0 \)
Step 4: Define the set \( B \).
For the function to be surjective, \( B \) must be the range of \( f(x) \), which is \((-\infty, 0]\). B = (-, 0]
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