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Question #40377

Difficulty: 8/10 | 🟢 Free

The Question

Let $f(x) = \frac{x^2 - 3x + 2}{x^2 - 4}$ and $g(x) = \frac{2x^2 - 5x - 3}{-x^2 + 4x - 3}$ . Let $S \subset \mathbb{R}$ be the maximal domain on which the composite function $h(x) = f(g(x))$ is defined. Which of the following statements is correct regarding the injectivity of $h(x)$ on $S$ and the sum of the elements not in $S$ ?

Options

A.
$h$ is injective on $S$ , and $\sum_{x \notin S} x = \frac{17}{4}$
B.
$h$ is injective on $S$ , and $\sum_{x \notin S} x = 1$
C.
$h$ is not injective on $S$ , and $\sum_{x \notin S} x = \frac{17}{4}$
D.
$h$ is injective on $S$ , and $\sum_{x \notin S} x = 4$

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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

Identity Function Behavior|Injective Mapping (One-One)|The Zero-Factor Cancellation Trap|Domain Constraint Intersections

Why This Question Is Tricky

Common Mistakes

This question tests the 'Zero-Factor Cancellation Trap' within the context of composite functions and injectivity.

Students frequently simplify rational functions by canceling common factors (e.g., $(x-2)$ in $f(x)$ and $(x-3)$ in $g(x)$ ) without explicitly recording the domain restrictions $x \neq 2$ and $x \neq 3$ .

When forming the composite function $h(x) = f(g(x))$ , failing to account for these hidden discontinuities leads to missing the points where $g(x)$ is undefined ( $x=3$ ) and where $g(x)$ hits the excluded domain of $f$ ( $g(x)=2 \implies x=1/4$ ).

This causes students to incorrectly evaluate the domain $S$ as merely $\mathbb{R} \setminus \{1\}$ , leading to the trap option B, or missing the $g(x)=2$ constraint leading to option D.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Relations and Functions - Class 11
Concept	Identity Function Behavior
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free