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Question #41914

Difficulty: 8/10 | 🟢 Free

The Question

Let f(x)=∣x2∣x−1∣3xx−11101∣f(x) = \begin{vmatrix} x^2 & |x-1| & 3 \\ x & x-1 & 1 \\ 1 & 0 & 1 \end{vmatrix}. Consider the application of Rolle's Theorem for the function f(x)f(x) on the interval [−1,3][-1, 3]. The value of c∈(−1,3)c \in (-1, 3) such that f′(c)=0f'(c) = 0 is:

Options

  • A.

    2+73\frac{2 + \sqrt{7}}{3}

  • B.

    2−73\frac{2 - \sqrt{7}}{3}

  • C.

    53\sqrt{\frac{5}{3}}

  • D.

    1

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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

Mean Value Theorem Applicability|Differentiation of Determinants|Modulus Sign Dependency|Modulus Function Breakdown

Why This Question Is Tricky

Common Mistakes

The problem tests the applicability of Rolle's Theorem on a determinant containing a modulus function.

Students often blindly differentiate the determinant or ignore the modulus sign dependency, leading to incorrect critical points from the wrong piecewise branch (e.g., (2−7)/3(2-\sqrt{7})/3 or 5/3\sqrt{5/3}).

The trap forces them to perform a 'Modulus Function Breakdown' to ensure global differentiability and select the valid root strictly within the correct domain interval.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterContinuity and Differentiability - Class 12
ConceptMean Value Theorem Applicability
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree