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Question #41157

Difficulty: 8/10 | 🟢 Free

The Question

Let f(x)=sin⁡−1(sin⁡x)f(x) = \sin^{-1}(\sin x) and g(x)=cos⁡−1(cos⁡x)g(x) = \cos^{-1}(\cos x). Define the function Δ(x)\Delta(x) as the determinant Δ(x)=∣f(x)g(x)x−11020−11∣\Delta(x) = \begin{vmatrix} f(x) & g(x) & x \\ -1 & 10 & 2 \\ 0 & -1 & 1 \end{vmatrix} Let SS be the set of all x∈(0,2π)x \in (0, 2\pi) such that Δ(x)=0\Delta(x) = 0. Then the value of ∑x∈SxΔ′(x)\sum_{x \in S} x \Delta'(x) is equal to:

Options

  • A.

    8π8\pi

  • B.

    4π4\pi

  • C.

    −8π-8\pi

  • D.

    0

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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

Principal Value Branch Boundaries|Differentiation of Determinants|Singular vs. Non-Singular Threshold

Why This Question Is Tricky

Common Mistakes

This question tests the student's mastery of the principal value branches of inverse trigonometric functions and their piecewise derivatives.

A common trap is mechanically assuming sin⁡−1(sin⁡x)=x\sin^{-1}(\sin x) = x and cos⁡−1(cos⁡x)=x\cos^{-1}(\cos x) = x globally, which leads to the false conclusion that the determinant has no roots in the given interval.

By forcing the student to find the singular points of the matrix, they must correctly apply the piecewise definitions f(x)=π−xf(x) = \pi - x or x−2πx - 2\pi and g(x)=2π−xg(x) = 2\pi - x.

Furthermore, evaluating the derivative of the determinant requires recognizing that the derivatives of these composite functions are step functions (±1\pm 1) depending on the quadrant, penalizing students who blindly differentiate without checking the domain validity at the specific roots.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterInverse Trigonometric Functions - Class 12
ConceptPrincipal Value Branch Boundaries
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree