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Question #40357

Difficulty: 8/10 | 🟢 Free

The Question

Consider the function f(x)=2∣x−2∣−x2+4x−3f(x) = 2|x - 2| - x^2 + 4x - 3 defined on the interval [0,4][0, 4]. Let PP be the number of points in (0,4)(0, 4) where f(x)f(x) attains a local extremum. Let QQ be the number of points in (0,4)(0, 4) where f′(x)=0f'(x) = 0. Let RR be the number of intervals from the set S={[0,2],[1,3],[2,4]}S = \{[0, 2], [1, 3], [2, 4]\} on which Rolle's Theorem is applicable to f(x)f(x). Find the value of P2+Q2+R2P^2 + Q^2 + R^2.

Options

  • A.

    14

  • B.

    17

  • C.

    18

  • D.

    21

Check Your Answer

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

Modulus Function Symmetry|Rolle's Theorem Anchors|Second Derivative Test Limitations

Why This Question Is Tricky

Common Mistakes

This question tests the limitations of the Second Derivative Test and the strict differentiability requirement of Rolle's Theorem using modulus symmetry.

The function is designed with a local minimum at a cusp (x=2x=2), where the first derivative is undefined, forcing students to use the first derivative test (sign change) rather than blindly applying f′′(x)=0f''(x)=0.

Furthermore, it sets a sophisticated Rolle's Theorem trap: while f(1)=f(3)f(1)=f(3), the presence of the non-differentiable cusp at x=2x=2 within the interval (1,3)(1, 3) disqualifies the theorem.

Students who memorize Rolle's conditions without verifying differentiability across the entire open interval will incorrectly assume it applies to [1,3][1, 3], leading to an erroneous count.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterRelations and Functions - Class 11
ConceptModulus Function Symmetry
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree