Before You Begin

Question #41218

Difficulty: 8/10 | 🟢 Free

The Question

Let the function $f: (-\infty, -3] \to \left(0, \frac{\pi}{2}\right]$ be defined as $f(x) = \sin^{-1}\left( \frac{2|x+2|}{x^2 + 4x + 5} \right)$ . If $f^{-1}(x)$ is the inverse function of $f(x)$ , then $f^{-1}(x)$ is equal to:

Options

A.
$-2 - \cot\left(\frac{x}{2}\right)$
B.
$-2 - \tan\left(\frac{x}{2}\right)$
C.
$-2 + \cot\left(\frac{x}{2}\right)$
D.
$-2 + \tan\left(\frac{x}{2}\right)$

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

Domain Constraint Intersections|Bijectivity and Inverse|Modulus Function Symmetry|Principal Value Branch Boundaries

Why This Question Is Tricky

Common Mistakes

This question tests the intersection of modulus properties and inverse trigonometric Principal Value Branches (PVBs).

The primary trap lies in the piecewise definition of $\sin^{-1}\left(\frac{2t}{1+t^2}\right)$ .

Students often blindly apply the standard identity $2\tan^{-1}t$ , ignoring that it is only valid for $|t| \le 1$ .

Because the domain is restricted to $(-\infty, -3]$ , the argument $t = |x+2|$ is strictly $\ge 1$ , forcing the use of the branch $\pi - 2\tan^{-1}t$ .

Furthermore, the modulus function $|x+2|$ evaluates to $-(x+2)$ in this domain.

Failing to apply either the correct PVB or the correct modulus sign leads directly to the carefully constructed distractor options.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Inverse Trigonometric Functions - Class 12
Concept	Domain Constraint Intersections
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free