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Question #40435

Difficulty: 8/10 | 🟢 Free

The Question

Let $\alpha = \left( \sum_{r=0}^{11} \binom{11}{r} \right) \left( \prod_{k=1}^{10} \cos \frac{2^{k-1} \pi}{1025} \right)$ and $\beta = \sum_{k=0}^{5} \binom{5}{k} (-1)^k \left(\frac{3}{2}\right)^k \left(\frac{1}{2}\right)^{5-k}$ . Find the sum of all integers $x$ satisfying the inequality $(x - \alpha)^2 (x - \beta)^2 (x - 4) (x - 6) \le 0$

Options

A.
14
B.
15
C.
16
D.
17

Check Your Answer

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

Factor-Product Conversion|Binomial Coefficient Summation Properties|Even Multiplicity Roots

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to synthesize combinatorial identities and trigonometric telescoping products to construct a polynomial inequality.

The primary trap lies in the application of the Wavy Curve Method (Sign Scheme).

Students frequently alternate signs across all roots, failing to recognize that roots with even multiplicity (here, $x=2$ and $x=-1$ ) do not cause a sign change.

Consequently, they either incorrectly include the intervals between these roots (yielding a sum of 17) or divide out the squared terms entirely, missing the isolated valid points $x=-1$ and $x=2$ (yielding a sum of 15).

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Trigonometric Functions - Class 11
Concept	Factor-Product Conversion
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free