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Question #41380

Difficulty: 8/10 | 🟢 Free

The Question

Let $f(x) = \begin{vmatrix} 0 & x^2-4x+1 & x^2-4x+1 & x^2-4x+1 \\ -x^2+4x-1 & 0 & 2x^2-8x+2 & 4x^2-16x+4 \\ -x^2+4x-1 & -2x^2+8x-2 & 0 & x^2-4x+1 \\ -x^2+4x-1 & -4x^2+16x-4 & -x^2+4x-1 & 0 \end{vmatrix}$ . Find the global maximum value of $f(x)$ on the interval $[0, 5]$ .

Options

A.
6
B.
36
C.
216
D.
1296

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

Global Extrema on Intervals|Skew-Symmetric Determinant|Determinant Scaling Property

Why This Question Is Tricky

Common Mistakes

This question tests the determinant scaling property $\det(kA) = k^n \det(A)$ combined with the properties of skew-symmetric matrices and global extrema.

A common trap is factoring out the polynomial $g(x)$ only once (treating the determinant as a linear operator rather than multilinear), leading to the incorrect function $f(x) = g(x)$ or $g(x)^2$ , which yields the trap answers 6 or 36.

Furthermore, recognizing the matrix as skew-symmetric allows the use of the Pfaffian to quickly evaluate the constant determinant, bypassing a brutal polynomial expansion.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Application of Derivatives - Class 12
Concept	Global Extrema on Intervals
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free