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Question #40632

Difficulty: 8/10 | 🟢 Free

The Question

Let $f: \mathbb{Z} \to \mathbb{Z}$ be a function where $f(r)$ is defined as the coefficient of $x^r$ in the binomial expansion of $(1+x)^{15}$ . Let $S$ be the set of all real numbers $x$ satisfying the equation: $f([x]^2 - [x] - 4) = f([x]^2 - [x] - 5)$ where $[t]$ denotes the greatest integer less than or equal to $t$ . Which of the following represents the set $S$ ?

Options

A.
$(-\infty, -4) \cup [-3, -2) \cup [-1, 3) \cup [4, 5) \cup [6, \infty)$
B.
$(-\infty, -2) \cup [-1, 3) \cup [4, \infty)$
C.
$[-3, -2) \cup [-1, 3) \cup [4, 5)$
D.
$[-3, -2) \cup [4, 5)$

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Monster

Ultra-high difficulty questions designed to test your limit, stamina, and deep conceptual understanding.

Concepts Behind This Question

Coefficient of x^r in Expansion|Composite Injectivity/Surjectivity|Codomain vs Range Distinction

Why This Question Is Tricky

Common Mistakes

This question tests the distinction between the restricted domain of non-zero binomial coefficients and the formal codomain of the coefficient function.

Students typically equate $\binom{n}{k} = \binom{n}{k-1}$ to find $k = \frac{n+1}{2}$ , ignoring that the coefficient of $x^r$ is $0$ for $r < 0$ and $r > n$ .

The trap exploits this by embedding the binomial index in a quadratic step-function $g(x) = [x]^2 - [x] - 4$ .

Failing to account for the $f(k) = f(k-1) = 0$ cases (where $k \le -1$ or $k \ge 17$ ) leads to missing intervals.

Furthermore, a secondary trap catches students who incorrectly assume $k \ge 16$ works, forgetting that $f(16) = 0$ but $f(15) = 1$ .

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Binomial Theorem - Class 11
Concept	Coefficient of x^r in Expansion
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free