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Question #40303

Difficulty: 8/10 | 🟢 Free

The Question

Let SS be the set of all odd polynomial functions f(x)f(x) of degree 5 such that f(x)f(x) is strictly increasing on R\mathbb{R}. If f(1)=20f(1) = 20 and f′(1)=262f'(1) = 262, find the number of elements in the set SS.

Options

  • A.

    0

  • B.

    1

  • C.

    2

  • D.

    ∞\infty

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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

Universal Set Relativity|Polynomial Function Parity|Increasing/Decreasing Strictness|Constant Function Logic

Why This Question Is Tricky

Common Mistakes

This problem tests the 'Universal Set Relativity' and 'Constant Function Logic' by evaluating the boundary conditions of strict monotonicity.

Students often assume that a system of linear constraints on a polynomial's coefficients will yield a valid function family, focusing only on the algebraic substitution.

The trap lies in the strict monotonicity condition f′(x)≥0f'(x) \ge 0, which translates to a quadratic inequality in x2x^2.

By setting the boundary conditions using specific cognitive load anchors (262 and 20), the problem forces the vertex of the derivative parabola into the positive domain (tv>0t_v > 0).

Students who fail to check the global minimum on the restricted domain t≥0t \ge 0, or who incorrectly assume f′(x)>0f'(x) > 0 implies a simple global discriminant condition (Δ≤0\Delta \le 0 for the biquadratic), will falsely conclude a function exists.

The negative discriminant of the final constraint proves the function set effectively 'leaves' the universal set of strictly increasing functions, resulting in an empty set.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterSets - Class 11
ConceptUniversal Set Relativity
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree