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Question #40813

Difficulty: 8/10 | 🟢 Free

The Question

Let A,B,CA, B, C be the angles of an acute-angled triangle. A point PP is defined as P(sin⁡2A+sin⁡2B+sin⁡2C,cos⁡2A+cos⁡2B+cos⁡2C)P(\sin^2 A + \sin^2 B + \sin^2 C, \cos^2 A + \cos^2 B + \cos^2 C). Let LL be the length of the chord of contact of the tangents drawn from PP to the circle x2+y2=92x^2 + y^2 = \frac{9}{2}. Find the maximum possible value of LL.

Options

  • A.

    3105\frac{3\sqrt{10}}{5}

  • B.

    3

  • C.

    355\frac{3\sqrt{5}}{5}

  • D.

    322\frac{3\sqrt{2}}{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

Chord of Contact Logic|Conditional Identity Constraints|Monotonicity and Boundedness

Why This Question Is Tricky

Common Mistakes

This question tests the integration of conditional trigonometric identities with coordinate geometry, specifically targeting the 'Monotonicity and Boundedness' trap.

Students often blindly apply derivative-based optimization to the distance function D2(x)=2x2−6x+9D^2(x) = 2x^2 - 6x + 9 without establishing the strict domain x∈(2,9/4]x \in (2, 9/4] derived from the acute triangle constraint ∑sin⁡2A=2+2cos⁡Acos⁡Bcos⁡C\sum \sin^2 A = 2 + 2\cos A\cos B\cos C.

Failing to recognize this bounded domain leads to incorrect boundary evaluations: assuming x→0x \to 0 yields a false maximum chord length of 3, while using the unconstrained vertex x=1.5x = 1.5 places the point on the circle, resulting in a chord length of 0.

The true maximum lies at the boundary x=9/4x = 9/4, requiring careful verification of the domain and monotonicity.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterConic Sections - Class 11
ConceptChord of Contact Logic
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree