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Question #41468

Difficulty: 8/10 | 🟢 Free

The Question

Let $f: [-1, 1] \setminus \{0\} \to (0, \infty)$ be a differentiable function such that $x^2 f(x) + \frac{1}{f(x)} \left( \arctan \frac{1}{x} \right)^2 \le 2x \arctan \frac{1}{x}$ for all $x \in [-1, 1] \setminus \{0\}$ . Evaluate the integral $I = \int_{-1}^1 \left( f(x) + xf'(x) \right) dx$

Options

A.
$-\frac{\pi}{2}$
B.
$\frac{\pi}{2}$
C.
0
D.
$-\pi$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Bridge

Practice problems that bridge the gap between foundational theory and advanced practical application.

Concepts Behind This Question

Integrating f(x) + xf'(x) forms|Arithmetic Mean Inequality|Closed vs Open Intervals

Why This Question Is Tricky

Common Mistakes

This question tests the synthesis of the AM-GM inequality to determine a function's explicit form, followed by the integration of an exact derivative $f(x) + xf'(x) = (xf(x))'$ .

The primary trap lies in the blind application of the Fundamental Theorem of Calculus (FTC).

Students will correctly identify $xf(x) = \arctan(1/x)$ but will incorrectly evaluate the integral as $[xf(x)]_{-1}^1 = \pi/2$ .

FTC requires the antiderivative to be continuous on the entire interval.

Here, $\arctan(1/x)$ has a jump discontinuity at $x=0$ .

The correct approach is to evaluate the derivative explicitly as $-1/(1+x^2)$ , which has a removable discontinuity at $x=0$ , yielding a proper Riemann integral of $-\pi/2$ .

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Integrals - Class 12
Concept	Integrating f(x) + xf'(x) forms
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free