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Question #41258

Difficulty: 8/10 | 🟢 Free

The Question

Let A=(113526−2−1−3)A = \begin{pmatrix} 1 & 1 & 3 \\ 5 & 2 & 6 \\ -2 & -1 & -3 \end{pmatrix} and let f(x)=cos⁡(sin⁡−1x)+ln⁡(1+tan⁡−1x)f(x) = \cos(\sin^{-1} x) + \ln(1 + \tan^{-1} x). Evaluate Tr(ATf(A))\text{Tr}(A^T f(A)), where Tr(M)\text{Tr}(M) denotes the trace of matrix MM and ATA^T is the transpose of AA.

Options

  • A.

    0

  • B.

    3

  • C.

    87

  • D.

    90

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

Composition Inverse Identity Traps|Matrix Trace Properties|Nilpotent Matrix Logic

Why This Question Is Tricky

Common Mistakes

This question triggers the 'Identity Trap' by presenting a matrix with elements >1>1 inside inverse trigonometric and logarithmic functions.

Students often incorrectly assume the matrix Maclaurin series diverges, forgetting that convergence depends strictly on the spectral radius (eigenvalues).

Since AA is nilpotent, its eigenvalues are 0, guaranteeing convergence and terminating the infinite series exactly at A2A^2.

Furthermore, the problem tests trace properties via the Frobenius inner product Tr(ATB)=∑AijBij\text{Tr}(A^T B) = \sum A_{ij}B_{ij}.

Students frequently misapply trace linearity or erroneously assume Tr(f(A))=f(Tr(A))\text{Tr}(f(A)) = f(\text{Tr}(A)), leading to distractors like 0 or 90.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterInverse Trigonometric Functions - Class 12
ConceptComposition Inverse Identity Traps
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree