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Question #41376

Difficulty: 8/10 | 🟢 Free

The Question

Let $f: \mathbb{R} \to \mathbb{R}$ be a function defined by $f(x) = x^3 - 3x^2 + 6x + k \tan^{-1}(x)$ where $k \in \mathbb{Z}$ . Find the minimum integral value of $k$ such that $f(x)$ is an injective function.

Options

A.
-6
B.
-5
C.
-4
D.
-3

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

Second Derivative Test Limitations|Injective Mapping (One-One)|Algebraic-Trigonometric Graphical Mapping

Why This Question Is Tricky

Common Mistakes

This question tests the limitations of relying solely on the second derivative test for global monotonicity.

The trap lies in the student's attempt to find the minimum of $f'(x)$ by solving $f''(x) = 0$ .

Because $f''(x) = 6(x-1) - \frac{2kx}{(1+x^2)^2} = 0$ is a complex equation, it cannot be solved directly.

A common heuristic mistake is to independently check the minima of the algebraic part ( $x=1$ ) and the trigonometric derivative part ( $x=0$ ), both of which yield $k \ge -6$ , leading students to incorrectly choose option A.

By isolating the parameter $k$ and finding the global extremum of the bounding polynomial, the student bypasses local concavity traps to find the true critical threshold at $x=1/2$ .

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Application of Derivatives - Class 12
Concept	Second Derivative Test Limitations
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free