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Question #41366

Difficulty: 8/10 | 🟢 Free

The Question

Let $f(x) = \left| \det \begin{pmatrix} x^2 & x & 1 \\ 1 & x & x^2 \\ 1 & 1 & 1 \end{pmatrix} \right|$ . Let $S$ be the set of all points where $f(x)$ attains a local extremum, $D$ be the set of all points where $f(x)$ is non-differentiable, and $C$ be the set of all stationary points of $f(x)$ . If $s = \sum_{x \in S} x$ , $d = \sum_{x \in D} x$ , and $c = \sum_{x \in C} x$ , find the value of $2s - d + 4c$ .

Options

A.
2
B.
-2
C.
3
D.
-3

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

Critical Point vs Stationary Point|Differentiation of Determinants|Modulus Function Breakdown

Why This Question Is Tricky

Common Mistakes

This question tests the precise distinction between critical points, stationary points, and non-differentiable points in the context of modulus functions.

A common trap is assuming that all local extrema or roots of the modulus argument are stationary points.

Here, $x=-1$ is a local minimum but is non-differentiable, so it belongs to $S$ and $D$ but not $C$ .

Conversely, students often mistakenly classify $x=1$ as non-differentiable because it is a root of the determinant, forgetting that its multiplicity of 3 ensures $f'(1)=0$ , making it a valid stationary point.

The expression $2s - d + 4c$ is specifically designed to yield incorrect integer values if any of these sets are misidentified.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Application of Derivatives - Class 12
Concept	Critical Point vs Stationary Point
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free