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Question #40452

Difficulty: 8/10 | 🟢 Free

The Question

Let $f(x) = \begin{vmatrix} \sin x & \cos x & \sec x \\ \sin\left(x+\frac{\pi}{3}\right) & \cos\left(x+\frac{\pi}{3}\right) & \sec\left(x+\frac{\pi}{3}\right) \\ \sin\left(x+\frac{2\pi}{3}\right) & \cos\left(x+\frac{2\pi}{3}\right) & \sec\left(x+\frac{2\pi}{3}\right) \end{vmatrix}$ .

Find the value of $\sum_{n=1}^{2025} f\left(\frac{n\pi}{3}\right)$ .

Options

A.
$-\frac{3\sqrt{3}}{2}$
B.
$\frac{6075\sqrt{3}}{2}$
C.
$-\frac{6075\sqrt{3}}{2}$
D.
$\frac{3\sqrt{3}}{2}$

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Bridge

Practice problems that bridge the gap between foundational theory and advanced practical application.

Concepts Behind This Question

Symmetry and Periodicity|Row/Column Exchange Parity|Elementary Row Operation Invariance

Why This Question Is Tricky

Common Mistakes

This question tests the invariance of determinants under row operations and permutations, specifically targeting the parity of cyclic row shifts.

A common cognitive trap is assuming that a function defined by a determinant with shifted periodic entries is strictly periodic with the shift as its period.

Students often incorrectly assume a cyclic permutation of 3 rows flips the determinant's sign, which, combined with the anti-periodicity $\sec(x+\pi) = -\sec(x)$ , would falsely imply $f(x+\pi/3) = f(x)$ .

In reality, a 3-row cyclic shift is an even permutation (2 swaps), so the determinant alternates sign $f(x+\pi/3) = -f(x)$ , collapsing the massive sum to a single negative term.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Trigonometric Functions - Class 11
Concept	Symmetry and Periodicity
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free