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Question #41133

Difficulty: 8/10 | 🟢 Free

The Question

Let $f: \mathbb{R} \to \mathbb{R}$ and $g: \mathbb{R} \to \mathbb{R}$ be defined as: $f(x) = \frac{x^2 - x + 1}{x^2 + x + 1}$ $g(x) = x^3 - 3x^2 + 3x + 2$ Let $A = [\alpha, \beta]$ be the range of $h(x) = g(f(x))$ and $B = [\gamma, \delta]$ be the range of $k(x) = f(g(x))$ . Let $n_1$ be the number of real roots of the equation $h(x) = 4$ , and $n_2$ be the number of real roots of the equation $k(x) = 1$ . Find the value of $27\alpha + \beta + 3\gamma + \delta + n_1 + n_2$ .

Options

A.
91
B.
88
C.
90
D.
93

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

Composition Non-Commutativity|Surjective Mapping (Onto)|Function Definition (The 'One-to-One' Constraint)

Why This Question Is Tricky

Common Mistakes

This question tests the non-commutativity of function composition and the retention of surjectivity/injectivity.

A common trap is assuming that if the outer function $g(x)$ is a bijective polynomial (onto $\mathbb{R}$ ), the composition $g(f(x))$ will also be onto $\mathbb{R}$ or share the range of $f(g(x))$ .

However, the non-injectivity of $f(x)$ creates a collapse in the domain of $g$ , restricting the range of $h(x)$ to a strict subset of $\mathbb{R}$ .

Furthermore, students often make erroneous inverse assumptions, believing $h(x)$ might be injective because $g(x)$ is.

Solving $h(x) = 4$ yields multiple roots ( $n_1 = 2$ ), explicitly demonstrating the failure of the horizontal line test and punishing the assumption of injectivity.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Relations and Functions - Class 12
Concept	Composition Non-Commutativity
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free