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Question #41162

Difficulty: 8/10 | 🟢 Free

The Question

Let $x = \tan\left(\frac{17\pi}{20}\right)$ . If $f(x) = \sin^{-1}\left(\frac{2x}{1+x^2}\right) + \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right) + \tan^{-1}\left(\frac{2x}{1-x^2}\right) + \tan^{-1}\left(\frac{3x-x^3}{1-3x^2}\right)$ then find the value of $\left\{ \frac{7}{\pi} f(x) \right\}$ , where $\{ \cdot \}$ denotes the fractional part function.

Options

A.
$\frac{3}{4}$
B.
$\frac{1}{4}$
C.
$\frac{11}{20}$
D.
$\frac{17}{20}$

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 Quadrant Adjustment Logic|Double and Triple Angle Ratios|Integral Part and Fractional Part

Why This Question Is Tricky

Common Mistakes

This question tests multiple layers of domain-restricted trigonometric identities and principal value branches.

The primary trap is the assumption that $\cos^{-1}((1-x^2)/(1+x^2)) = 2\tan^{-1}x$ universally; for $x < 0$ , it is actually $-2\tan^{-1}x$ .

Failing to adjust this sign, or failing to map $\tan^{-1}(\tan(17\pi/20))$ to its principal branch $-\frac{3\pi}{20}$ , leads to an incorrect sum.

A secondary trap lies in the fractional part function applied to a negative number: students often incorrectly compute $\{-5.25\}$ as $0.25$ instead of $0.75$ , catching those who survive the trigonometric gauntlet but stumble on basic function definitions.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Inverse Trigonometric Functions - Class 12
Concept	The Quadrant Adjustment Logic
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free