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Question #63611

Difficulty: 8/10 | 🟢 Free

The Question

Let SS be the set of all points of local extrema of the function f(x)=sin⁡−1(1−x21+x2)+2x5f(x) = \sin^{-1}\left( \frac{1 - x^2}{1 + x^2} \right) + \frac{2x}{5} on R\mathbb{R}. If n(S)n(S) denotes the number of elements in SS, then the value of 3n(S)+∑x∈Sx3^{n(S)} + \sum_{x \in S} x is equal to:

Options

  • A.

    11

  • B.

    9

  • C.

    5

  • D.

    7

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

Algebraic-Trigonometric Graphical Mapping|Tangent Half-Angle Substitution|First Derivative Test Trap

Why This Question Is Tricky

Examiner Levers

Recursive Logic

The answer to part A is required to solve part B (Step-wise failure test).

Difficulty Boost : +5

Common Mistakes

This question tests the critical evaluation of Inverse Trigonometric Functions (ITFs) and the First Derivative Test at non-differentiable boundaries.

Students commonly fall into the trap of blindly applying the identity sin⁡−1(1−x21+x2)=π2−2tan⁡−1x\sin^{-1}\left(\frac{1-x^2}{1+x^2}\right) = \frac{\pi}{2} - 2\tan^{-1}x for all x∈Rx \in \mathbb{R}, ignoring the absolute value requirement ∣2tan⁡−1x∣|2\tan^{-1}x|.

This false assumption yields a smooth rational derivative 2x2−85(1+x2)\frac{2x^2-8}{5(1+x^2)}, producing a deceptive local maximum at x=−2x = -2.

By correctly identifying the piecewise nature of the function, the false stationary point is eliminated, and the true local maximum is revealed at the non-differentiable corner point x=0x = 0, validating the necessity of checking boundary conditions in substitution domains.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterInverse Trigonometric Functions - Class 12
ConceptAlgebraic-Trigonometric Graphical Mapping
Difficulty8
Cognitive Load7
Target Solve Time170 sec
AccessFree