Start Learning Free

Before You Begin

Question #41167

Difficulty: 8/10 | 🟢 Free

The Question

Let f(x)=sin⁡−1(sin⁡x)+cos⁡−1(cos⁡x)+tan⁡−1(tan⁡x)f(x) = \sin^{-1}(\sin x) + \cos^{-1}(\cos x) + \tan^{-1}(\tan x) for x∈(0,π)∖{π2}x \in (0, \pi) \setminus \left\{\frac{\pi}{2}\right\}. Let L(x)L(x) be the linear approximation of f(x)f(x) at x=29π60x = \frac{29\pi}{60}. If EE is the absolute error when L(x)L(x) is used to approximate the value of f(31π60)f\left(\frac{31\pi}{60}\right), then EE equals:

Options

  • A.

    31π30\frac{31\pi}{30}

  • B.

    29π30\frac{29\pi}{30}

  • C.

    π30\frac{\pi}{30}

  • D.

    π\pi

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

Composition Inverse Identity Traps|The Quadrant Adjustment Logic|Approximation via Differentials

Why This Question Is Tricky

Common Mistakes

This question tests the 'Identity Trap' in inverse trigonometric functions combined with linear approximations.

Students often incorrectly assume f−1(f(x))=xf^{-1}(f(x)) = x universally, which would lead to f(x)=3xf(x) = 3x everywhere and an error of 0.

Furthermore, applying a differential approximation across a threshold (x=π/2x = \pi/2) where the function undergoes a jump discontinuity (due to quadrant adjustments like π−x\pi - x and x−πx - \pi) compounds the error.

The linear model L(x)L(x) assumes a continuous smooth slope and fails to capture the sign flips and domain restrictions, resulting in a massive deviation from the true value.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterInverse Trigonometric Functions - Class 12
ConceptComposition Inverse Identity Traps
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree