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Question #40295

Difficulty: 8/10 | 🟢 Free

The Question

Let $f: \mathbb{R} \to \mathbb{R}$ be defined by $f(x) = |x^2 - 4x| - 2x$ . Let $S$ be the set of all closed intervals $A \subset \mathbb{R}$ such that the restricted function $f: A \to [-5, 0]$ is bijective. Which of the following intervals is an element of $S$ ?

Options

A.
$[0, 1]$
B.
$[2, 1+\sqrt{6}]$
C.
$[1, 1+\sqrt{6}]$
D.
$[4, 6]$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Bridge

Practice problems that bridge the gap between foundational theory and advanced practical application.

Concepts Behind This Question

Subset vs. Proper Subset|Surjective Mapping (Onto)|Modulus Function Breakdown

Why This Question Is Tricky

Common Mistakes

This question tests the 'Modulus Breakdown' trap in the context of domain restriction for bijectivity.

Students who ignore the absolute value will incorrectly assume $f(x) = x^2 - 6x$ everywhere.

Solving $x^2 - 6x \in [-5, 0]$ yields the intervals $[0, 1]$ and $[5, 6]$ .

This leads them to incorrectly choose $[0, 1]$ as a valid domain.

However, on $[0, 1]$ , the actual function is $f(x) = -x^2 + 2x$ , which maps to $[0, 1]$ , failing the surjectivity condition.

The correct approach requires piecewise analysis to find the true intervals where the function is monotonic and spans the exact codomain.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Sets - Class 11
Concept	Subset vs. Proper Subset
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free