Before You Begin

Question #40975

Difficulty: 8/10 | 🟢 Free

The Question

Let $A(\theta) = \begin{pmatrix} \sin\theta & 1 & \cos\theta \\ \cos\theta & 2 & \sin\theta \\ -\sin\theta & 1 & \cos\theta \end{pmatrix}$ . For all $\theta \in [0, 2\pi]$ where $A(\theta)$ is invertible, find the range of the function $f(\theta) = \det(A^{-1}(\theta))$ .

Options

A.
$\left(-\infty, \frac{1-\sqrt{5}}{4}\right] \cup \left[\frac{1+\sqrt{5}}{4}, \infty\right)$
B.
$\left[\frac{1-\sqrt{5}}{4}, \frac{1+\sqrt{5}}{4}\right]$
C.
$\left(-\infty, \frac{-1-\sqrt{5}}{4}\right] \cup \left[\frac{\sqrt{5}-1}{4}, \infty\right)$
D.
$\left[\frac{-1-\sqrt{5}}{4}, \frac{\sqrt{5}-1}{4}\right]$

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

Range and Coefficient of Range|Inversion and Determinant Product|Minors and Cofactors Sign Convention

Why This Question Is Tricky

Common Mistakes

This question tests two lethal JEE traps.

First, the Minors and Cofactors Sign Convention: forgetting the negative sign for the $a_{12}$ cofactor leads to an incorrect determinant $2\sin 2\theta + \cos 2\theta + 1$ , yielding Option C.

Second, the Reciprocal Range Trap: students often incorrectly assume the range of $1/f(x)$ is $[1/\max, 1/\min]$ .

When $f(x)$ crosses zero, the reciprocal function becomes unbounded, splitting into two intervals $(-\infty, 1/\min] \cup [1/\max, \infty)$ .

Failing to account for the zero-crossing yields the bounded interval in Option B.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Statistics - Class 11
Concept	Range and Coefficient of Range
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free