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Question #41221

Difficulty: 8/10 | 🟢 Free

The Question

Let a curve be defined implicitly by the equation x=sin⁡−1(2y1+y2)+cos⁡−1(1−y21+y2)+ln⁡yx = \sin^{-1}\left(\frac{2y}{1+y^2}\right) + \cos^{-1}\left(\frac{1-y^2}{1+y^2}\right) + \ln y The value of dydx\frac{dy}{dx} at x=π+12ln⁡3x = \pi + \frac{1}{2}\ln 3 is

Options

  • A.

    3\sqrt{3}

  • B.

    3−32\frac{3-\sqrt{3}}{2}

  • C.

    13\frac{1}{\sqrt{3}}

  • D.

    3+32\frac{3+\sqrt{3}}{2}

Check Your Answer

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

The Sum-Difference Formula Constraints|Derivative of Implicit Inverse Forms|Increasing/Decreasing Strictness

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to correctly apply piecewise definitions of inverse trigonometric functions before implicit differentiation.

The 'Lethal Collision' occurs because students often blindly differentiate using the standard chain rule (forgetting that (1−y2)2=∣1−y2∣\sqrt{(1-y^2)^2} = |1-y^2|) or by incorrectly assuming both inverse terms equal 2tan⁡−1y2\tan^{-1}y for all y>0y > 0.

This ignores the strictness of the interval; for y>1y > 1, the derivative of the sine inverse term flips sign, causing the inverse trigonometric terms to completely cancel out.

Failing to recognize this piecewise strictness leads to an incorrect algebraic derivative expression dxdy=41+y2+1y\frac{dx}{dy} = \frac{4}{1+y^2} + \frac{1}{y}, resulting in the trap answer 3−32\frac{3-\sqrt{3}}{2}.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterInverse Trigonometric Functions - Class 12
ConceptThe Sum-Difference Formula Constraints
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree