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Question #41353

Difficulty: 8/10 | 🟢 Free

The Question

Let f(x)=∣x2−4x∣x−2∣1−311021∣f(x) = \begin{vmatrix} x^2-4x & |x-2| & 1 \\ -3 & 1 & 1 \\ 0 & 2 & 1 \end{vmatrix}. Consider the application of Rolle's Theorem for the function f(x)f(x) on the interval [0,4][0, 4]. If Rolle's Theorem is applicable, let SS be the sum of all valid values of c∈(0,4)c \in (0, 4) such that f′(c)=0f'(c) = 0. If Rolle's Theorem is not applicable, let S=−1S = -1. Find the value of SS.

Options

  • A.

    4

  • B.

    -1

  • C.

    2

  • D.

    0

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Monster

Ultra-high difficulty questions designed to test your limit, stamina, and deep conceptual understanding.

Concepts Behind This Question

Rolle's Theorem Anchors|Differentiation of Determinants|Modulus Function Breakdown|Domain Restrictions

Why This Question Is Tricky

Common Mistakes

This question tests the strict prerequisites of Rolle's Theorem, specifically the differentiability constraint on the open interval (a, b).

The determinant is designed to simplify to a function containing a modulus term, 3|x-2|, which introduces a point of non-differentiability at x=2.

A common trap is for students to blindly differentiate the function for x ≠ 2, find the roots c = 1/2 and c = 7/2, and incorrectly assume Rolle's Theorem is applicable since these points lie in (0, 4) and sum to 4 (Option A).

By forcing the evaluation of differentiability at x=2, the problem penalizes superficial application of the theorem and rewards rigorous verification of its conditions.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterApplication of Derivatives - Class 12
ConceptRolle's Theorem Anchors
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree