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Question #40353

Difficulty: 8/10 | 🟢 Free

The Question

Let SS be the set of all real solutions to the equation cos⁡−1(16x5−20x3+5x)=2cos⁡−1x\cos^{-1}(16x^5 - 20x^3 + 5x) = 2\cos^{-1}x Find the value of ∑x∈Sx\sum_{x \in S} x.

Options

  • A.

    1+cos⁡2π71 + \cos\frac{2\pi}{7}

  • B.

    00

  • C.

    12\frac{1}{2}

  • D.

    cos⁡2π7\cos\frac{2\pi}{7}

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Specialist

Targeted practice for niche, high-difficulty concepts that frequently appear in JEE Advanced exams.

Concepts Behind This Question

Polynomial Function Parity|Extraneous Solution Rejection|The Quadrant Adjustment Logic

Why This Question Is Tricky

Common Mistakes

This question tests the Quadrant Adjustment Logic and extraneous solution rejection in trigonometric polynomial equations.

Students typically map the polynomial to a multiple-angle identity (Chebyshev polynomial) and solve the algebraic equation 16x5−20x3+5x=2x2−116x^5 - 20x^3 + 5x = 2x^2 - 1, yielding five roots.

The critical trap lies in failing to apply the principal value range of the inverse cosine function.

Because the LHS cos⁡−1(...)\cos^{-1}(...) is strictly bounded in [0,π][0, \pi], the RHS 2cos⁡−1x2\cos^{-1}x must also lie in [0,π][0, \pi], which enforces the hidden domain restriction x≥0x \ge 0.

Students who ignore this will incorrectly sum all algebraic roots (yielding 0) or fail to reject the negative roots cos⁡(4π/7)\cos(4\pi/7) and cos⁡(6π/7)\cos(6\pi/7).

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterRelations and Functions - Class 11
ConceptPolynomial Function Parity
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree