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Question #41038

Difficulty: 8/10 | 🟢 Free

The Question

Let SS be the set of all 3×33 \times 3 matrices of the form B=(ab0cd0001)B = \begin{pmatrix} a & b & 0 \\ c & d & 0 \\ 0 & 0 & 1 \end{pmatrix}, where a,b,c,d∈{1,2,3,4,5,6}a, b, c, d \in \{1, 2, 3, 4, 5, 6\}. A matrix BB is chosen at random from SS. Let PP be the probability that BB is non-singular.

Consider the matrix A(x)=(0x2−xsin⁡(πx)x−x20ex−1−1−sin⁡(πx)1−ex−10)A(x) = \begin{pmatrix} 0 & x^2 - x & \sin(\pi x) \\ x - x^2 & 0 & e^{x-1} - 1 \\ -\sin(\pi x) & 1 - e^{x-1} & 0 \end{pmatrix} and let f(x)=det⁡(BA(x)BT)f(x) = \det(B A(x) B^T).

Find the value of 648P+f′(2024)648 P + f'(2024).

Options

  • A.

    605

  • B.

    648

  • C.

    562

  • D.

    691

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

Multiplication Theorem|Complementary Counting|Skew-Symmetric Determinant|Differentiation of Determinants

Why This Question Is Tricky

Common Mistakes

This question tests the Multiplication Theorem of Determinants and Complementary Counting, while embedding a classic JEE trap: differentiating a determinant of a skew-symmetric matrix.

Students often attempt to apply Jacobi's formula or differentiate the determinant column-by-column, failing to recognize that a skew-symmetric matrix of odd order has a determinant identically equal to zero.

Identifying this 'kill-switch' immediately reduces f′(2024)f'(2024) to 0.

The second phase requires complementary counting to find the probability of a matrix being non-singular by calculating the number of solutions to ad=bcad = bc for dice rolls, rewarding systematic frequency grouping over brute-force counting.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterProbability - Class 11
ConceptMultiplication Theorem
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree