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Question #41375

Difficulty: 8/10 | 🟢 Free

The Question

Let $f(x) = \begin{vmatrix} 2 & 2 & x+1 \\ 2 & x+1 & 2 \\ x+1 & 2 & 2 \end{vmatrix}$ . If $x_1$ is the point of local maximum and $x_2$ is the point of local minimum of $f(x)$ , then the value of $2x_1 - x_2$ is:

Options

A.
$5$
B.
$-7$
C.
$7$
D.
$-5$

Check Your Answer

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

Second Derivative Test Limitations|Factor Theorem in Determinants|Row/Column Exchange Parity

Why This Question Is Tricky

Common Mistakes

This question tests the intersection of determinant properties and the Second Derivative Test.

A common heuristic among JEE Advanced students is to use the Factor Theorem to quickly write the polynomial form of a determinant.

Because the standard determinant with identical off-diagonal elements often yields a positive leading coefficient, students frequently assume $f(x) = (x+5)(x-1)^2$ without checking the anti-diagonal product or tracking the parity of row/column exchanges.

This missing negative sign flips the concavity of the function, causing $f''(x)$ to have the opposite sign.

Consequently, the student misclassifies the local maximum as a local minimum (and vice versa), leading directly to the trap option $-7$ .

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Application of Derivatives - Class 12
Concept	Second Derivative Test Limitations
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free