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Question #41552

Difficulty: 8/10 | 🟢 Free

The Question

Let M(x)=(x3−λxe∣x∣sin⁡xcos⁡x−e∣x∣sin⁡xx3−λx1+x2−cos⁡x−1+x2x3−λx)M(x) = \begin{pmatrix} x^3 - \lambda x & e^{|x|} \sin x & \cos x \\ -e^{|x|} \sin x & x^3 - \lambda x & \sqrt{1+x^2} \\ -\cos x & -\sqrt{1+x^2} & x^3 - \lambda x \end{pmatrix}, where λ\lambda is a real parameter. Let f(x)=det⁡(M(x))f(x) = \det(M(x)). If the rows of M(x)M(x) are linearly dependent at x=2x = 2, then the value of the integral ∫−22(f(x)2+cos⁡x+3λ2x2)dx\int_{-2}^{2} \left( \frac{f(x)}{2 + \cos x} + \frac{3\lambda}{2} x^2 \right) dx is equal to:

Options

  • A.

    16

  • B.

    32

  • C.

    64

  • D.

    0

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

Symmetry Utilization|Skew-Symmetric Determinant|Linear Dependency of Rows

Why This Question Is Tricky

Common Mistakes

This question tests the ability to identify the characteristic polynomial of a skew-symmetric matrix disguised as a complex determinant.

The 'Linear Dependency' trap forces students to evaluate the determinant at x=2x=2.

Brute-force expansion leads to an unmanageable equation, but recognizing the M(x)=yI+SM(x) = yI + S structure immediately reveals that y(2)=0y(2)=0.

Furthermore, students must utilize the skew-symmetric property to deduce that the determinant is an odd function, turning an analytically impossible integration into a trivial polynomial evaluation.

Failing to check for rank-deficiency or symmetry traps the student in a calculus nightmare.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterApplication of Integrals - Class 12
ConceptSymmetry Utilization
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree