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Question #40506

Difficulty: 8/10 | 🟢 Free

The Question

Let f(x)=∣x2+2i∣x∣−4+i∣f(x) = |x^2 + 2i|x| - 4 + i| for x∈Rx \in \mathbb{R}, where i=−1i = \sqrt{-1}. Find the minimum value of f(x)f(x).

Options

  • A.

    17\sqrt{17}

  • B.

    323\sqrt{2}

  • C.

    25−552\sqrt{\frac{25 - 5\sqrt{5}}{2}}

  • D.

    25+552\sqrt{\frac{25 + 5\sqrt{5}}{2}}

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Specialist

Targeted practice for niche, high-difficulty concepts that frequently appear in JEE Advanced exams.

Concepts Behind This Question

Maximum/Minimum of Quadratic Expressions|Quadratic Equations with Complex Coefficients|Modulus Function Breakdown

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to handle complex quadratics transformed into real-valued modulus expressions.

The trap lies in the non-differentiable point (kink) introduced by ∣x∣|x|.

Students who blindly replace ∣x∣|x| with xx and differentiate over all real numbers will find a false global minimum at x=−1−52x = \frac{-1-\sqrt{5}}{2}, leading to the incorrect option C.

By enforcing the domain constraint u=∣x∣≥0u = |x| \ge 0, the true global minimum is forced onto the boundary u=0u=0, highlighting the danger of relying solely on f′(x)=0f'(x)=0 without checking domain boundaries and points of non-differentiability.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterComplex Numbers and Quadratic Equations - Class 11
ConceptMaximum/Minimum of Quadratic Expressions
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree