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Question #41273

Difficulty: 8/10 | 🟢 Free

The Question

Let n≥2n \ge 2 be an integer. Consider the determinant Δ(y)=∣(n0)y2(n1)(n2)(n+10)y2(n+11)(n+12)(n+20)y2(n+21)(n+22)+1y3∣\Delta(y) = \begin{vmatrix} \binom{n}{0}y^2 & \binom{n}{1} & \binom{n}{2} \\ \binom{n+1}{0}y^2 & \binom{n+1}{1} & \binom{n+1}{2} \\ \binom{n+2}{0}y^2 & \binom{n+2}{1} & \binom{n+2}{2} + \frac{1}{y^3} \end{vmatrix} Let TT be the term independent of yy in the binomial expansion of (Δ(y))9(\Delta(y))^9. For an integer x∈[1,T]x \in [1, T], a matrix M(x)M(x) is defined as: M(x)=(sin⁡(πx2)1−cos⁡(πx)xsin⁡(πx2)sin⁡(πx3)cos⁡(2πx3)−1sin⁡(2πx3)sin⁡(πx7)sin⁡(2πx7)1−cos⁡(2πx7))M(x) = \begin{pmatrix} \sin\left(\frac{\pi x}{2}\right) & 1 - \cos(\pi x) & x \sin\left(\frac{\pi x}{2}\right) \\ \sin\left(\frac{\pi x}{3}\right) & \cos\left(\frac{2\pi x}{3}\right) - 1 & \sin\left(\frac{2\pi x}{3}\right) \\ \sin\left(\frac{\pi x}{7}\right) & \sin\left(\frac{2\pi x}{7}\right) & 1 - \cos\left(\frac{2\pi x}{7}\right) \end{pmatrix} Find the number of valid values of xx such that no row of M(x)M(x) is a zero vector.

Options

  • A.

    24

  • B.

    2

  • C.

    60

  • D.

    82

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

Elementary Row Operation Invariance|Term Independent of x|Complementary Counting|Mutually Exclusive vs Independent

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to combine determinant simplification using elementary row operations with set theory and probability concepts.

The first phase requires recognizing Pascal's identity to reduce the determinant to a simple binomial, leading to the term independent of y (T=84T=84).

The core trap lies in the second phase: counting the valid integers xx.

Students must find the complement of the union of three events (rows being zero vectors).

A common fatal mistake is assuming these events are 'Mutually Exclusive' rather than 'Independent'.

If they simply add the individual counts (42+28+12=8242 + 28 + 12 = 82) and subtract from 84, they arrive at the trap answer 2.

Recognizing that divisibility by coprime numbers forms independent events allows the correct use of the inclusion-exclusion principle or direct multiplication of independent probabilities to find the true count of 24.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterDeterminants - Class 12
ConceptElementary Row Operation Invariance
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree